#set document(title: "6.5 Polynomial Equations", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 6.5#h(0.6em)Polynomial Equations #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Solve: #math.equation(block: false, alt: "5 y minus 3 equals 0 .")[$5 y − 3 = 0 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "y equals the fraction 3 over 5")[$y = frac(3, 5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "n cubed minus 9 n squared minus 22 n .")[$n^(3) − 9 n^(2) − 22 n .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "n open parenthesis n minus 11 close parenthesis open parenthesis n plus 2 close parenthesis")[$n ( n − 11 ) ( n + 2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 8 x minus 16 ,")[$f ( x ) = 8 x − 16 ,$] find #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis")[$f ( 3 )$] and solve #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0 .")[$f ( x ) = 0 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "8 ; x equals 2")[$8 ; #h(0.5em) x = 2$] ] ] We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible. A #strong[polynomial equation] is an equation that contains a polynomial expression. The #strong[degree of the polynomial equation] is the degree of the polynomial. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Polynomial Equation] A #strong[polynomial equation] is an equation that contains a polynomial expression. The #strong[degree of the polynomial equation] is the degree of the polynomial. ] We have already solved polynomial equations of #strong[degree one]. Polynomial equations of degree one are linear equations are of the form #math.equation(block: false, alt: "a x plus b equals c .")[$a x + b = c .$] We are now going to solve polynomial equations of #strong[degree two]. A polynomial equation of degree two is called a #strong[quadratic equation]. Listed below are some examples of quadratic equations: #math.equation(block: true, alt: "x squared plus 5 x plus 6 equals 0 3 y squared plus 4 y equals 10 64 u squared minus 81 equals 0 n open parenthesis n plus 1 close parenthesis equals 42")[$x^(2) + 5 x + 6 = 0 #h(3em) 3 y^(2) + 4 y = 10 #h(3em) 64 u^(2) − 81 = 0 #h(3em) n ( n + 1 ) = 42$]The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get #math.equation(block: false, alt: "n squared plus n .")[$n^(2) + n .$] The general form of a quadratic equation is #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 ,")[$a x^(2) + b x + c = 0 ,$] with #math.equation(block: false, alt: "a not equal to 0 .")[$a ≠ 0 .$] (If #math.equation(block: false, alt: "a equals 0 ,")[$a = 0 ,$] then #math.equation(block: false, alt: "0 times x squared equals 0")[$0 · x^(2) = 0$] and we are left with no quadratic term.) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quadratic Equation] An equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] is called a quadratic equation. #math.equation(block: true, alt: "a , b , and c are real numbers and a not equal to 0")[$a , b , #h(0.2em) "and" #h(0.2em) c #h(0.2em) "are real numbers and" #h(0.2em) a ≠ 0$] ] To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter. === Use the Zero Product Property We will first solve some quadratic equations by using the #strong[Zero Product Property]. The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero Product Property] If #math.equation(block: false, alt: "a times b equals 0 ,")[$a · b = 0 ,$] then either #math.equation(block: false, alt: "a equals 0")[$a = 0$] or #math.equation(block: false, alt: "b equals 0")[$b = 0$] or both. ] We will now use the Zero Product Property, to solve a #strong[quadratic equation]. #examplebox("Example 1")[How to Solve a Quadratic Equation Using the Zero Product Property][ Solve: #math.equation(block: false, alt: "open parenthesis 5 n minus 2 close parenthesis open parenthesis 6 n minus 1 close parenthesis equals 0 .")[$( 5 n − 2 ) ( 6 n − 1 ) = 0 .$] #solutionbox[ #figure(figph[The equation is open parentheses 5n minus 2 close parentheses open parentheses 6n minus 1 close parentheses equals 0. The product equals zero, so at least one factor must equal zero. Step 1 is set each factor equal to zero. So, 5n minus 2 equals 0 and 6n minus 1 equals 0.], alt: "The equation is open parentheses 5n minus 2 close parentheses open parentheses 6n minus 1 close parentheses equals 0. The product equals zero, so at least one factor must equal zero. Step 1 is set each factor equal to zero. So, 5n minus 2 equals 0 and 6n minus 1 equals 0.", caption: none) #figure(figph[Step 2 is to solve the linear equations. So, we get n equal to 2 by 5 and n equal to 1 by 6.], alt: "Step 2 is to solve the linear equations. So, we get n equal to 2 by 5 and n equal to 1 by 6.", caption: none) #figure(figph[Step 3 is to check by substituting each solution separately into the original equation.], alt: "Step 3 is to check by substituting each solution separately into the original equation.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "open parenthesis 3 m minus 2 close parenthesis open parenthesis 2 m plus 1 close parenthesis equals 0 .")[$( 3 m − 2 ) ( 2 m + 1 ) = 0 .$] #solutionbox[ #math.equation(block: true, alt: "m equals the fraction 2 over 3 , m equals minus the fraction 1 over 2")[$m = frac(2, 3) , m = − frac(1, 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "open parenthesis 4 p plus 3 close parenthesis open parenthesis 4 p minus 3 close parenthesis equals 0 .")[$( 4 p + 3 ) ( 4 p − 3 ) = 0 .$] #solutionbox[ #math.equation(block: true, alt: "p equals minus the fraction 3 over 4 , p equals the fraction 3 over 4")[$p = − frac(3, 4) , p = frac(3, 4)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Use the Zero Product Property.] + Set each factor equal to zero. + Solve the linear equations. + Check. ] === Solve Quadratic Equations by Factoring The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in #strong[standard form], #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 .")[$a x^(2) + b x + c = 0 .$] Then we must factor the expression on the left. #examplebox("Example 2")[How to Solve a Quadratic Equation by Factoring][ Solve: #math.equation(block: false, alt: "2 y squared equals 13 y plus 45 .")[$2 y^(2) = 13 y + 45 .$] #solutionbox[ #figure(figph[The equation is 2 y squared equals 13y plus 45. Step 1 is to write it in standard form a x squared plus bx plus c. So we have 2 y squared minus 13y minus 45 equals 0.], alt: "The equation is 2 y squared equals 13y plus 45. Step 1 is to write it in standard form a x squared plus bx plus c. So we have 2 y squared minus 13y minus 45 equals 0.", caption: none) #figure(figph[Step 2 is to factor the quadratic expression. So we have 2y plus 5, y minus 9 equals 0.], alt: "Step 2 is to factor the quadratic expression. So we have 2y plus 5, y minus 9 equals 0.", caption: none) #figure(figph[Step 3 is to use the zero product property. Setting each factor equal to zero, we have two linear equations: 2y plus 5 equals 0 and y minus 9 equals 0.], alt: "Step 3 is to use the zero product property. Setting each factor equal to zero, we have two linear equations: 2y plus 5 equals 0 and y minus 9 equals 0.", caption: none) #figure(figph[Step 4 is to solve the linear equations. We get, y equals minus 5 by 2 and y equals 9.], alt: "Step 4 is to solve the linear equations. We get, y equals minus 5 by 2 and y equals 9.", caption: none) #figure(figph[Step 5 is to check by substituting each solution separately into the original equation], alt: "Step 5 is to check by substituting each solution separately into the original equation", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "3 c squared equals 10 c minus 8 .")[$3 c^(2) = 10 c − 8 .$] #solutionbox[ #math.equation(block: true, alt: "c equals 2 , c equals the fraction 4 over 3")[$c = 2 , c = frac(4, 3)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "2 d squared minus 5 d equals 3 .")[$2 d^(2) − 5 d = 3 .$] #solutionbox[ #math.equation(block: true, alt: "d equals 3 , d equals minus the fraction 1 over 2")[$d = 3 , d = − frac(1, 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a quadratic equation by factoring.] + Write the quadratic equation in standard form, #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 .")[$a x^(2) + b x + c = 0 .$] + Factor the quadratic expression. + Use the Zero Product Property. + Solve the linear equations. + Check. Substitute each solution separately into the original equation. ] Before we factor, we must make sure the #strong[quadratic equation] is in #strong[standard form]. Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example? #examplebox("Example 3")[][ Solve: #math.equation(block: false, alt: "169 x squared equals 49 .")[$169 x^(2) = 49 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "169 x squared equals 49")[$#h(5.5em) 169 x^(2) = 49$]]), [Write the quadratic equation in standard form.], [#math.equation(block: false, alt: "169 x squared minus 49 equals 0")[$#h(3.3em) 169 x^(2) − 49 = 0$]], [Factor. It is a difference of squares.], [#math.equation(block: false, alt: "open parenthesis 13 x minus 7 close parenthesis open parenthesis 13 x plus 7 close parenthesis equals 0")[$#h(0.1em) ( 13 x − 7 ) ( 13 x + 7 ) = 0$]], [Use the Zero Product Property to set each factor to #math.equation(block: false, alt: "0")[$0$]. #linebreak() Solve each equation.], [#math.equation(block: false, alt: "13 x minus 7, equals, 0, 13 x plus 7, equals, 0; 13 x, equals, 7, 13 x, equals, −7; x, equals, the fraction 7 over 13, x, equals, minus the fraction 7 over 13")[$13 x − 7 & = & 0 & & & & 13 x + 7 & = & 0 \ 13 x & = & 7 & & & & 13 x & = & −7 \ x & = & frac(7, 13) & & & & x & = & − frac(7, 13)$]], )) Check: We leave the check up to you. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "25 p squared equals 49 .")[$25 p^(2) = 49 .$] #solutionbox[ #math.equation(block: true, alt: "p equals the fraction 7 over 5 , p equals minus the fraction 7 over 5")[$p = frac(7, 5) , "p" = − frac(7, 5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "36 x squared equals 121 .")[$36 x^(2) = 121 .$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction 11 over 6 , x equals minus the fraction 11 over 6")[$x = frac(11, 6) , x = − frac(11, 6)$] ] ] In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the #strong[Zero Product Property], one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form. #examplebox("Example 4")[][ Solve: #math.equation(block: false, alt: "open parenthesis 3 x minus 8 close parenthesis open parenthesis x minus 1 close parenthesis equals 3 x .")[$( 3 x − 8 ) ( x − 1 ) = 3 x .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis 3 x minus 8 close parenthesis open parenthesis x minus 1 close parenthesis equals 3 x")[$#h(0.3em) ( 3 x − 8 ) ( x − 1 ) = 3 x$]]), [Multiply the binomials.], [#math.equation(block: false, alt: "3 x squared minus 11 x plus 8 equals 3 x")[$#h(0.7em) 3 x^(2) − 11 x + 8 = 3 x$]], [Write the quadratic equation in standard form.], [#math.equation(block: false, alt: "3 x squared minus 14 x plus 8 equals 0")[$#h(0.7em) 3 x^(2) − 14 x + 8 = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "open parenthesis 3 x minus 2 close parenthesis open parenthesis x minus 4 close parenthesis equals 0")[$#h(0.25em) ( 3 x − 2 ) ( x − 4 ) = 0$]], [Use the Zero Product Property to set each factor to 0. #linebreak() Solve each equation.], [#math.equation(block: false, alt: "3 x minus 2, equals, 0, x minus 4, equals, 0; 3 x, equals, 2, x, equals, 4")[$3 x − 2 & = & 0 & & & & x − 4 & = & 0 \ 3 x & = & 2 & & & & x & = & 4$]], [], [#math.equation(block: false, alt: "x equals the fraction 2 over 3")[$#h(2.9em) x = frac(2, 3)$]], [Check your answers.], [The check is left to you.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "open parenthesis 2 m plus 1 close parenthesis open parenthesis m plus 3 close parenthesis equals 1 2 m .")[$( 2 m + 1 ) ( m + 3 ) = 1 2 m .$] #solutionbox[ #math.equation(block: true, alt: "m equals 1 , m equals the fraction 3 over 2")[$m = 1 , m = frac(3, 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "open parenthesis k plus 1 close parenthesis open parenthesis k minus 1 close parenthesis equals 8 .")[$( k + 1 ) ( k − 1 ) = 8 .$] #solutionbox[ #math.equation(block: true, alt: "k equals 3 , k equals −3")[$k = 3 , k = −3$] ] ] In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0. #examplebox("Example 5")[][ Solve: #math.equation(block: false, alt: "3 x squared equals 12 x plus 63 .")[$3 x^(2) = 12 x + 63 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "3 x squared equals 12 x plus 63")[$#h(7.1em) 3 x^(2) = 12 x + 63$]]), [Write the quadratic equation in standard form.], [#math.equation(block: false, alt: "3 x squared minus 12 x minus 63 equals 0")[$#h(2.1em) 3 x^(2) − 12 x − 63 = 0$]], [Factor the greatest common factor first.], [#math.equation(block: false, alt: "3 open parenthesis x squared minus 4 x minus 21 close parenthesis equals 0")[$#h(1.7em) 3 ( x^(2) − 4 x − 21 ) = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "3 open parenthesis x minus 7 close parenthesis open parenthesis x plus 3 close parenthesis equals 0")[$#h(2.1em) 3 ( x − 7 ) ( x + 3 ) = 0$]], [Use the Zero Product Property to set each factor to 0. #linebreak() Solve each equation.], [#math.equation(block: false, alt: "3 not equal to 0, x minus 7, equals, 0, x plus 3, equals, 0; 3 not equal to 0, x, equals, 7, x, equals, −3")[$#h(0.3em) 3 ≠ 0 & & & & x − 7 & = & 0 & & & & x + 3 & = & 0 \ 3 ≠ 0 & & & & x & = & 7 & & & & x & = & −3$]], [Check your answers.], [The check is left to you.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "18 a squared minus 30 equals −33 a .")[$18 a^(2) − 30 = −33 a .$] #solutionbox[ #math.equation(block: true, alt: "a equals minus the fraction 5 over 2 , a equals the fraction 2 over 3")[$a = − frac(5, 2) , a = frac(2, 3)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "123 b equals −6 minus 60 b squared .")[$123 b = −6 − 60 b^(2) .$] #solutionbox[ #math.equation(block: true, alt: "b equals −2 , b equals minus the fraction 1 over 20")[$b = −2 , b = − frac(1, 20)$] ] ] The #strong[Zero Product Property] also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations. #examplebox("Example 6")[][ Solve: #math.equation(block: false, alt: "9 m cubed plus 100 m equals 60 m squared .")[$9 m^(3) + 100 m = 60 m^(2) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "9 m cubed plus 100 m equals 60 m squared")[$#h(5.1em) 9 m^(3) + 100 m = 60 m^(2)$]]), [Bring all the terms to one side so that the other side is zero.], [#math.equation(block: false, alt: "9 m cubed minus 60 m squared plus 100 m equals 0")[$#h(1.6em) 9 m^(3) − 60 m^(2) + 100 m = 0$]], [Factor the greatest common factor first.], [#math.equation(block: false, alt: "m open parenthesis 9 m squared minus 60 m plus 100 close parenthesis equals 0")[$#h(1.2em) m ( 9 m^(2) − 60 m + 100 ) = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "m open parenthesis 3 m minus 10 close parenthesis open parenthesis 3 m minus 10 close parenthesis equals 0")[$#h(1.1em) m ( 3 m − 10 ) ( 3 m − 10 ) = 0$]], [Use the Zero Product Property to set each factor to 0. #linebreak() Solve each equation.], [#math.equation(block: false, alt: "m equals 0, 3 m minus 10, equals, 0, 3 m minus 10, equals, 0; m equals 0, m, equals, the fraction 10 over 3, m, equals, the fraction 10 over 3")[$#h(0.5em) m = 0 & & & & 3 m − 10 & = & 0 & & & & 3 m − 10 & = & 0 \ m = 0 & & & & m & = & frac(10, 3) & & & & m & = & frac(10, 3)$]], [Check your answers.], [The check is left to you.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "8 x cubed equals 24 x squared minus 18 x .")[$8 x^(3) = 24 x^(2) − 18 x .$] #solutionbox[ #math.equation(block: true, alt: "x equals 0 , x equals the fraction 3 over 2")[$x = 0 , x = frac(3, 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "16 y squared equals 32 y cubed plus 2 y .")[$16 y^(2) = 32 y^(3) + 2 y .$] #solutionbox[ #math.equation(block: true, alt: "y equals 0 , y equals the fraction 1 over 4")[$y = 0 , y = frac(1, 4)$] ] ] === Solve Equations with Polynomial Functions As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the #strong[Zero Product Property] will be help us find these answers. #examplebox("Example 7")[][ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 2 x minus 2 ,")[$f ( x ) = x^(2) + 2 x − 2 ,$] ⓐ find #emph[x] when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 6")[$f ( x ) = 6$] ⓑ find two points that lie on the graph of the function. #solutionbox[ ⓐ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 2 x minus 2")[$f ( x ) = x^(2) + 2 x − 2$]]), [Substitute 6 for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].], [#math.equation(block: false, alt: "6 equals x squared plus 2 x minus 2")[$#h(1.2em) 6 = x^(2) + 2 x − 2$]], [Put the quadratic in standard form.], [#math.equation(block: false, alt: "x squared plus 2 x minus 8 equals 0")[$#h(0.8em) x^(2) + 2 x − 8 = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x minus 2 close parenthesis equals 0")[$( x + 4 ) ( x − 2 ) = 0$]], [Use the zero product property. #linebreak() Solve.], [#math.equation(block: false, alt: "x plus 4, equals, 0, or, x minus 2, equals, 0; x, equals, −4, or, x, equals, 2")[$x + 4 & = & 0 & & & "or" & & & x − 2 & = & 0 \ x & = & −4 & & & "or" & & & x & = & 2$]], [Check:], [], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 2 x minus 2, f open parenthesis x close parenthesis equals x squared plus 2 x minus 2; f open parenthesis − 4 close parenthesis equals open parenthesis − 4 close parenthesis squared plus 2 open parenthesis − 4 close parenthesis minus 2, f open parenthesis 2 close parenthesis equals 2 squared plus 2 times 2 minus 2; f open parenthesis − 4 close parenthesis equals 16 minus 8 minus 2, f open parenthesis 2 close parenthesis equals 4 plus 4 minus 2; f open parenthesis − 4 close parenthesis equals 6 ✓, f open parenthesis 2 close parenthesis equals 6 ✓")[$#h(4em) #h(0.7em) f ( x ) = x^(2) + 2 x − 2 & & & & f ( x ) = x^(2) + 2 x − 2 \ f ( "−" 4 ) = attach(( "−" 4 ), t: 2) + 2 ( "−" 4 ) − 2 & & & & f ( 2 ) = 2^(2) + 2 · 2 − 2 \ f ( "−" 4 ) = 16 − 8 − 2 & & & & f ( 2 ) = 4 + 4 − 2 \ f ( "−" 4 ) = 6 ✓ & & & & f ( 2 ) = 6 ✓$]], [], )) #linebreak() ⓑ Since #math.equation(block: false, alt: "f open parenthesis −4 close parenthesis equals 6")[$f ( −4 ) = 6$] and #math.equation(block: false, alt: "f open parenthesis 2 close parenthesis equals 6 ,")[$f ( 2 ) = 6 ,$] the points #math.equation(block: false, alt: "open parenthesis −4 , 6 close parenthesis")[$( −4 , 6 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 6 close parenthesis")[$( 2 , 6 )$] lie on the graph of the function. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 2 x minus 8 ,")[$f ( x ) = x^(2) − 2 x − 8 ,$] ⓐ find #emph[x] when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 7")[$f ( x ) = 7$] ⓑ Find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals −3")[$x = −3$] or #math.equation(block: false, alt: "x equals 5")[$x = 5$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis −3 , 7 close parenthesis")[$( −3 , 7 )$] #math.equation(block: false, alt: "open parenthesis 5 , 7 close parenthesis")[$( 5 , 7 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 8 x plus 3 ,")[$f ( x ) = x^(2) − 8 x + 3 ,$] ⓐ find #emph[x] when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −4")[$f ( x ) = −4$] ⓑ Find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 1")[$x = 1$] or #math.equation(block: false, alt: "x equals 7")[$x = 7$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 1 , −4 close parenthesis")[$( 1 , −4 )$] #math.equation(block: false, alt: "open parenthesis 7 , −4 close parenthesis")[$( 7 , −4 )$] ] ] The #strong[Zero Product Property] also helps us determine where the function is zero. A value of #emph[x] where the function is 0, is called a #strong[zero of the function]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero of a Function] For any function #emph[f], if #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0 ,")[$f ( x ) = 0 ,$] then #emph[x] is a #strong[zero of the function]. ] When #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0 ,")[$f ( x ) = 0 ,$] the point #math.equation(block: false, alt: "open parenthesis x , 0 close parenthesis")[$( x , 0 )$] is a point on the graph. This point is an #strong[#emph[x]-intercept] of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later. #examplebox("Example 8")[][ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared plus 10 x minus 8 ,")[$f ( x ) = 3 x^(2) + 10 x − 8 ,$] find ⓐ the zeros of the function, ⓑ any #emph[x]-intercepts of the graph of the function, ⓒ any #emph[y]-intercepts of the graph of the function #solutionbox[ ⓐ To find the zeros of the function, we need to find when the function value is 0. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared plus 10 x minus 8")[$f ( x ) = 3 x^(2) + 10 x − 8$]]), [Substitute 0 for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].], [#math.equation(block: false, alt: "0 equals 3 x squared plus 10 x minus 8")[$#h(1.3em) 0 = 3 x^(2) + 10 x − 8$]], [Factor the trinomial.], [#math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis 3 x minus 2 close parenthesis equals 0")[$( x + 4 ) ( 3 x − 2 ) = 0$]], [Use the zero product property. #linebreak() Solve.], [#math.equation(block: false, alt: "x plus 4, equals, 0, or, 3 x minus 2, equals, 0; x, equals, −4, or, x, equals, the fraction 2 over 3")[$x + 4 & = & 0 & & & "or" & & & 3 x − 2 & = & 0 \ x & = & −4 & & & "or" & & & x & = & frac(2, 3)$]], )) #linebreak() ⓑ An #emph[x]-intercept occurs when #math.equation(block: false, alt: "y equals 0 .")[$y = 0 .$] Since #math.equation(block: false, alt: "f open parenthesis −4 close parenthesis equals 0")[$f ( −4 ) = 0$] and #math.equation(block: false, alt: "f open parenthesis the fraction 2 over 3 close parenthesis equals 0 ,")[$f ( frac(2, 3) ) = 0 ,$] the points #math.equation(block: false, alt: "open parenthesis −4 , 0 close parenthesis")[$( −4 , 0 )$] and #math.equation(block: false, alt: "open parenthesis the fraction 2 over 3 , 0 close parenthesis")[$( frac(2, 3) , 0 )$] lie on the graph. These points are #emph[x]-intercepts of the function. #linebreak() #linebreak() #linebreak() ⓒ A #emph[y]-intercept occurs when #math.equation(block: false, alt: "x equals 0 .")[$x = 0 .$] To find the #emph[y]-intercepts we need to find #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis .")[$f ( 0 ) .$] #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared plus 10 x minus 8")[$f ( x ) = 3 x^(2) + 10 x − 8$]]), [Find #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis")[$f ( 0 )$] by substituting 0 for #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals 3 times 0 squared plus 10 times 0 minus 8")[$f ( 0 ) = 3 · 0^(2) + 10 · 0 − 8$]], [Simplify.], [#math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals −8")[$f ( 0 ) = −8$]], )) Since #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals −8 ,")[$f ( 0 ) = −8 ,$] the point #math.equation(block: false, alt: "open parenthesis 0 , −8 close parenthesis")[$( 0 , −8 )$] lies on the graph. This point is the #emph[y]-intercept of the function. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x squared minus 7 x plus 5 ,")[$f ( x ) = 2 x^(2) − 7 x + 5 ,$] find ⓐ the zeros of the function, ⓑ any #emph[x]-intercepts of the graph of the function, ⓒ any #emph[y]-intercepts of the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 1")[$x = 1$] or #math.equation(block: false, alt: "x equals the fraction 5 over 2")[$x = frac(5, 2)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis ,")[$( 1 , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis the fraction 5 over 2 , 0 close parenthesis")[$( frac(5, 2) , 0 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 0 , 5 close parenthesis")[$( 0 , 5 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 6 x squared plus 13 x minus 15 ,")[$f ( x ) = 6 x^(2) + 13 x − 15 ,$] find ⓐ the zeros of the function, ⓑ any #emph[x]-intercepts of the graph of the function, ⓒ any #emph[y]-intercepts of the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals −3")[$x = −3$] or #math.equation(block: false, alt: "x equals the fraction 5 over 6")[$x = frac(5, 6)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis −3 , 0 close parenthesis ,")[$( −3 , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis the fraction 5 over 6 , 0 close parenthesis")[$( frac(5, 6) , 0 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 0 , −15 close parenthesis")[$( 0 , −15 )$] ] ] === Solve Applications Modeled by Polynomial Equations The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Use a problem solving strategy to solve word problems.] + #strong[Read] the problem. Make sure all the words and ideas are understood. + #strong[Identify] what we are looking for. + #strong[Name] what we are looking for. Choose a variable to represent that quantity. + #strong[Translate] into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation. + #strong[Solve] the equation using appropriate algebra techniques. + #strong[Check] the answer in the problem and make sure it makes sense. + #strong[Answer] the question with a complete sentence. ] We will start with a number problem to get practice translating words into a polynomial equation. #examplebox("Example 9")[][ The product of two consecutive odd integers is 323. Find the integers. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, [#strong[Step 1. Read] the problem.], [], [#strong[Step 2. Identify] what we are looking for.], [We are looking for two consecutive integers.], [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "n equals the first integer.")[$n = #h(0.2em) "the first integer."$]], [], [#math.equation(block: false, alt: "n plus 2 equals next consecutive odd integer")[$n + 2 = #h(0.2em) "next consecutive odd integer"$]], [#strong[Step 4. Translate] into an equation. Restate the problem in a sentence.], [The product of the two consecutive odd integers is 323.], [], [#math.equation(block: false, alt: "n open parenthesis n plus 2 close parenthesis equals 323")[$#h(3.47em) n ( n + 2 ) = 323$]], [#strong[Step 5. Solve] the equation.], [#math.equation(block: false, alt: "n squared plus 2 n equals 323")[$#h(3.7em) n^(2) + 2 n = 323$]], [Bring all the terms to one side.], [#math.equation(block: false, alt: "n squared plus 2 n minus 323 equals 0")[$#h(1em) n^(2) + 2 n − 323 = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "open parenthesis n minus 17 close parenthesis open parenthesis n plus 19 close parenthesis equals 0")[$( n − 17 ) ( n + 19 ) = 0$]], [Use the Zero Product Property. #linebreak() Solve the equations.], [#math.equation(block: false, alt: "n minus 17, equals, 0, n plus 19, equals, 0; n, equals, 17, n, equals, −19")[$n − 17 & = & 0 & & & & n + 19 & = & 0 \ n & = & 17 & & & & n & = & −19$]], [There are two values for #emph[n] that are solutions to this problem. So there are two sets of consecutive odd integers that will work.], [], [If the first integer is #math.equation(block: false, alt: "n equals 17")[$n = 17$]], [If the first integer is #math.equation(block: false, alt: "n equals −19")[$n = −19$]], [then the next odd integer is], [then the next odd integer is], [#math.equation(block: false, alt: "n plus 2")[$n + 2$]], [#math.equation(block: false, alt: "n plus 2")[$n + 2$]], [#math.equation(block: false, alt: "17 plus 2")[$17 + 2$]], [#math.equation(block: false, alt: "minus 19 plus 2")[$− 19 + 2$]], [#math.equation(block: false, alt: "19")[$19$]], [#math.equation(block: false, alt: "minus 17")[$− 17$]], [#math.equation(block: false, alt: "17 , 19")[$17 , 19$]], [#math.equation(block: false, alt: "minus 17 , −19")[$− 17 , −19$]], [#strong[Step 6. Check] the answer.], [], [The results are consecutive odd integers], [], [#math.equation(block: false, alt: "17 , 19 and − 19 , −17 .")[$17 , 19 #h(0.2em) "and" #h(0.2em) "−" 19 , −17 .$]], [], [#math.equation(block: false, alt: "17 times 19 equals 323 ✓ − 19 open parenthesis − 17 close parenthesis equals 323 ✓")[$17 · 19 = 323 ✓ #h(4em) "−" 19 ( "−" 17 ) = 323 ✓$]], [], [Both pairs of consecutive integers are solutions.], [], [#strong[Step 7. Answer] the question], [The consecutive integers are 17, 19 and #math.equation(block: false, alt: "minus 19 , −17 .")[$− 19 , −17 .$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The product of two consecutive odd integers is 255. Find the integers. #solutionbox[ #math.equation(block: false, alt: "−15 , −17")[$−15 , −17$] and 15, 17 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The product of two consecutive odd integers is 483 Find the integers. #solutionbox[ #math.equation(block: false, alt: "−23 , −21")[$−23 , −21$] and 21, 23 ] ] Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results. In some applications, negative solutions will result from the algebra, but will not be realistic for the situation. #examplebox("Example 10")[][ A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem. In problems involving #linebreak() geometric figures, a sketch can help you visualize #linebreak() the situation.], [#figure(figph[A rectangle is shown with its width labeled 'w' and its length labeled 'w + 4', indicating the dimensions are expressed in terms of variable 'w'.], alt: "A rectangle is shown with its width labeled 'w' and its length labeled 'w + 4', indicating the dimensions are expressed in terms of variable 'w'.", caption: none)]), [#strong[Step 2. Identify] what you are looking for.], [We are looking for the length and width.], [#strong[Step 3. Name] what you are looking for.], [Let #math.equation(block: false, alt: "w equals")[$w =$] the width of the bedroom.], [The length is four feet more than the width.], [#math.equation(block: false, alt: "w plus 4 equals")[$w + 4 =$] the length of the garden], [#strong[Step 4. Translate] into an equation.], [], [Restate the important information in a sentence.], [The area of the bedroom is 117 square feet.], [Use the formula for the area of a rectangle.], [#math.equation(block: false, alt: "A equals l times w")[$A = l · w #h(1.2em)$]], [Substitute in the variables.], [#math.equation(block: false, alt: "117 equals open parenthesis w plus 4 close parenthesis w")[$#h(0.2em) 117 = ( w + 4 ) w$]], [#strong[Step 5. Solve] the equation Distribute first.], [#math.equation(block: false, alt: "117 equals w squared plus 4 w")[$117 = w^(2) + 4 w$]], [Get zero on one side.], [#math.equation(block: false, alt: "117 equals w squared plus 4 w")[$117 = w^(2) + 4 w$]], [Factor the trinomial.], [#math.equation(block: false, alt: "0 equals w squared plus 4 w minus 117")[$#h(3.5em) 0 = w^(2) + 4 w − 117$]], [Use the Zero Product Property.], [#math.equation(block: false, alt: "0 equals open parenthesis w plus 13 close parenthesis open parenthesis w minus 9 close parenthesis")[$#h(4.4em) 0 = ( w + 13 ) ( w − 9 )$]], [Solve each equation.], [#math.equation(block: false, alt: "0 equals w plus 13 0 equals w minus 9")[$#h(5.6em) 0 = w + 13 #h(1.5em) 0 = w − 9$]], [Since #emph[w] is the width of the bedroom, it does not #linebreak() make sense for it to be negative. We eliminate that value for #emph[w].], [#math.equation(block: false, alt: "−13 equals w 9 equals w")[$#h(3em) cancel(−13 = w) #h(3.5em) 9 = w$]], [], [#math.equation(block: false, alt: "w equals 9")[$w = 9$] Width is 9 feet.], [Find the value of the length.], [#math.equation(block: false, alt: "w plus 4")[$w + 4 #h(1.5em)$] #linebreak() #math.equation(block: false, alt: "9 plus 4")[$9 + 4 #h(1.5em)$] #linebreak() $#h(7em)$13  Length is 13 feet.], [#strong[Step 6. Check] the answer. #linebreak() Does the answer make sense? #linebreak() #linebreak() #figure(figph[An image illustrating the calculation of a rectangle's area. The width is 9 and the length is 13 (derived from w+4, where w=9). The area is calculated as 13 \* 9 = 117.], alt: "An image illustrating the calculation of a rectangle's area. The width is 9 and the length is 13 (derived from w+4, where w=9). The area is calculated as 13 * 9 = 117.", caption: none) #linebreak() Yes, this makes sense.], [], [#strong[Step 7. Answer] the question.], [The width of the bedroom is 9 feet and #linebreak() the length is 13 feet.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign. #solutionbox[ The width is 5 feet and length is 6 feet. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio. #solutionbox[ The width of the patio is 12 feet and the length is 15 feet. ] ] In the next example, we will use the Pythagorean Theorem #math.equation(block: false, alt: "open parenthesis a squared plus b squared equals c squared close parenthesis .")[$( a^(2) + b^(2) = c^(2) ) .$] This formula gives the relation between the legs and the hypotenuse of a right triangle. #figure(figph[Figure shows a right triangle with the shortest side being a, the second side being b and the hypotenuse being c.], alt: "Figure shows a right triangle with the shortest side being a, the second side being b and the hypotenuse being c.", caption: none) We will use this formula in the next example. #examplebox("Example 11")[][ A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail. #figure(figph[Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.], alt: "Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.", caption: none) #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem], []), [#strong[Step 2. Identify] what you are looking for.], [We are looking for the lengths of the #linebreak() sides of the sail.], [#strong[Step 3. Name] what you are looking for. #linebreak() One side is 7 less than the other.], [Let #math.equation(block: false, alt: "x equals")[$x =$] length of a side of the sail. #linebreak() #math.equation(block: false, alt: "x minus 7 equals")[$x − 7 =$] length of other side], [#strong[Step 4. Translate] into an equation. Since this is a #linebreak() right triangle we can use the Pythagorean Theorem.], [#math.equation(block: false, alt: "a squared plus b squared equals c squared")[$#h(4.8em) a^(2) + b^(2) = c^(2)$]], [Substitute in the variables.], [#math.equation(block: false, alt: "x squared plus open parenthesis x minus 7 close parenthesis squared equals 17 squared")[$#h(2.5em) x^(2) + attach(( x − 7 ), t: 2) = 17^(2)$]], [#strong[Step 5. Solve] the equation #linebreak() Simplify.], [#math.equation(block: false, alt: "x squared plus x squared minus 14 x plus 49 equals 289")[$x^(2) + x^(2) − 14 x + 49 = 289$]], [], [#math.equation(block: false, alt: "2 x squared minus 14 x plus 49 equals 289")[$#h(1.5em) 2 x^(2) − 14 x + 49 = 289$]], [It is a quadratic equation, so get zero on one side.], [#math.equation(block: false, alt: "2 x squared minus 14 x minus 240 equals 0")[$#h(1em) 2 x^(2) − 14 x − 240 = 0$]], [Factor the greatest common factor.], [#math.equation(block: false, alt: "2 open parenthesis x squared minus 7 x minus 120 close parenthesis equals 0")[$#h(0.8em) 2 ( x^(2) − 7 x − 120 ) = 0$]], [Factor the trinomial.], [#math.equation(block: false, alt: "2 open parenthesis x minus 15 close parenthesis open parenthesis x plus 8 close parenthesis equals 0")[$#h(1.1em) 2 ( x − 15 ) ( x + 8 ) = 0$]], [Use the Zero Product Property.], [#math.equation(block: false, alt: "2 not equal to 0 x minus 15 equals 0 x plus 8 equals 0")[$2 ≠ 0 #h(1.5em) x − 15 = 0 #h(0.6em) x + 8 = 0$]], [Solve.], [#math.equation(block: false, alt: "2 not equal to 0 x equals 15 x equals −8")[$2 ≠ 0 #h(1.5em) x = 15 #h(2em) x = −8$]], [Since #emph[x] is a side of the triangle, #math.equation(block: false, alt: "x equals −8")[$x = −8$] does not #linebreak() make sense.], [#math.equation(block: false, alt: "2 not equal to 0 x equals 15 x equals −8")[$2 ≠ 0 #h(1.5em) x = 15 #h(2em) cancel(x = −8)$]], [Find the length of the other side.], [], [If the length of one side is #linebreak()    then the length of the other side is], [#figure(figph[The image shows the mathematical equation 'x = 15' centered on a plain white background.], alt: "The image shows the mathematical equation 'x = 15' centered on a plain white background.", caption: none) #linebreak() #figure(figph[The mathematical expression 'X - 7' is displayed in grayscale against a white background.], alt: "The mathematical expression 'X - 7' is displayed in grayscale against a white background.", caption: none) #linebreak() #figure(figph[The numbers '15' in red and '7' in black, separated by a black dash, on a white background. This appears to be a mathematical expression or part of a scoring display.], alt: "The numbers '15' in red and '7' in black, separated by a black dash, on a white background. This appears to be a mathematical expression or part of a scoring display.", caption: none) #linebreak() 8 is the length of the other side.], [#strong[Step 6. Check] the answer in the problem #linebreak() Do these numbers make sense? #linebreak() #linebreak() #figure(figph[This image illustrates the Pythagorean theorem in action with a right triangle. The sides 8, 15, and 17 form a Pythagorean triple, as 8^2 + 15^2 = 64 + 225 = 289, which equals 17^2.], alt: "This image illustrates the Pythagorean theorem in action with a right triangle. The sides 8, 15, and 17 form a Pythagorean triple, as 8^2 + 15^2 = 64 + 225 = 289, which equals 17^2.", caption: none)], [], [#strong[Step 7. Answer] the question], [The sides of the sail are 8, 15 and 17 feet.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck. #solutionbox[ 5 feet and 12 feet ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg. #solutionbox[ The other leg is 24 feet and the hypotenuse is 25 feet. ] ] The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground. #examplebox("Example 12")[][ Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 64 t plus 80")[$h ( t ) = −16 t^(2) + 64 t + 80$] models the height, #emph[h], of the ball above the ground as a function of time, #emph[t.] Find: ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at #math.equation(block: false, alt: "t equals 2")[$t = 2$] seconds. #solutionbox[ ⓐ The zeros of this function are found by solving #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals 0 .")[$h ( t ) = 0 .$] This will tell us when the ball will hit the ground. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "h open parenthesis t close parenthesis equals 0")[$h ( t ) = 0 #h(4em)$]]), [Substitute in the polynomial for #math.equation(block: false, alt: "h open parenthesis t close parenthesis .")[$h ( t ) .$]], [#math.equation(block: false, alt: "−16 t squared plus 64 t plus 80 equals 0")[$−16 t^(2) + 64 t + 80 = 0 #h(4em)$]], [Factor the GCF, #math.equation(block: false, alt: "−16 .")[$−16 .$]], [#math.equation(block: false, alt: "−16 open parenthesis t squared minus 4 t minus 5 close parenthesis equals 0")[$−16 ( t^(2) − 4 t − 5 ) = 0 #h(4em)$]], [Factor the trinomial.], [#math.equation(block: false, alt: "−16 open parenthesis t minus 5 close parenthesis open parenthesis t plus 1 close parenthesis equals 0")[$−16 ( t − 5 ) ( t + 1 ) = 0 #h(4em)$]], [Use the Zero Product Property. #linebreak() Solve.], [#math.equation(block: false, alt: "t minus 5, equals, 0, t plus 1, equals, 0; t, equals, 5, t, equals, −1")[$t − 5 & = & 0 & & & & & & #h(2em) t + 1 & = & 0 \ t & = & 5 & & & & & & #h(2em) t & = & −1$]], )) The result #math.equation(block: false, alt: "t equals 5")[$t = 5$] tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result #math.equation(block: false, alt: "t equals −1")[$t = −1$] is discarded. ⓑ The ball will be 80 feet above the ground when #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals 80 .")[$h ( t ) = 80 .$] #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "h open parenthesis t close parenthesis equals 80")[$h ( t ) = 80 #h(3.8em)$]]), [Substitute in the polynomial for #math.equation(block: false, alt: "h open parenthesis t close parenthesis .")[$h ( t ) .$]], [#math.equation(block: false, alt: "−16 t squared plus 64 t plus 80 equals 80")[$−16 t^(2) + 64 t + 80 = 80 #h(3.8em)$]], [Subtract 80 from both sides.], [#math.equation(block: false, alt: "−16 t squared plus 64 t equals 0")[$−16 t^(2) + 64 t = 0 #h(4.2em)$]], [Factor the GCF, #math.equation(block: false, alt: "−16 t .")[$−16 t .$]], [#math.equation(block: false, alt: "−16 t open parenthesis t minus 4 close parenthesis equals 0")[$−16 t ( t − 4 ) = 0 #h(4.2em)$]], [Use the Zero Product Property. #linebreak() Solve.], [#math.equation(block: false, alt: "−16 t, equals, 0, t minus 4, equals, 0; t, equals, 0, t, equals, 4")[$−16 t & = & 0 & & & & t − 4 & = & 0 \ t & = & 0 & & & & t & = & 4 #h(2em)$]], [], [The ball will be at 80 feet the moment Dennis tosses the ball and then 4 seconds later, when the ball is falling.], )) ⓒ To find the height ball at #math.equation(block: false, alt: "t equals 2")[$t = 2$] seconds we find #math.equation(block: false, alt: "h open parenthesis 2 close parenthesis .")[$h ( 2 ) .$] #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 64 t plus 80")[$h ( t ) = −16 t^(2) + 64 t + 80$]]), [To find #math.equation(block: false, alt: "h open parenthesis 2 close parenthesis")[$h ( 2 )$] substitute 2 for #math.equation(block: false, alt: "t .")[$t .$]], [#math.equation(block: false, alt: "h open parenthesis 2 close parenthesis equals −16 open parenthesis 2 close parenthesis squared plus 64 times 2 plus 80")[$h ( 2 ) = −16 attach(( 2 ), t: 2) + 64 · 2 + 80$]], [Simplify.], [#math.equation(block: false, alt: "h open parenthesis 2 close parenthesis equals 144")[$h ( 2 ) = 144$]], [], [After 2 seconds, the ball will be at 144 feet.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 48 t plus 160")[$h ( t ) = −16 t^(2) + 48 t + 160$] models the height, #emph[h], of the rock above the ocean as a function of time, #emph[t]. Find: ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at #math.equation(block: false, alt: "t equals 1.5")[$t = 1.5$] seconds. #solutionbox[ ⓐ 5 seconds; ⓑ 0 and 3 seconds; ⓒ 196 feet ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 32 t plus 128")[$h ( t ) = −16 t^(2) + 32 t + 128$] models the height, #emph[h], of the penny above the ocean as a function of time, #emph[t]. Find: ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at #math.equation(block: false, alt: "t equals 1")[$t = 1$] seconds which is when the penny will be at its highest point. #solutionbox[ ⓐ 4 seconds; ⓑ 0 and 2 seconds; ⓒ 144 feet ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access this online resource for additional instruction and practice with quadratic equations. - #link("https://openstax.org/l/37ZeroProperty")[Beginning Algebra & Solving Quadratics with the Zero Property] ] === Key Concepts - #strong[Polynomial Equation:] A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial. - #strong[Quadratic Equation:] An equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] is called a quadratic equation. #linebreak() #math.equation(block: true, alt: "a , b , c are real numbers and a not equal to 0")[$a , b , c #h(0.2em) "are real numbers and" #h(0.2em) a ≠ 0$] - #strong[Zero Product Property:] If #math.equation(block: false, alt: "a times b equals 0 ,")[$a · b = 0 ,$] then either #math.equation(block: false, alt: "a equals 0")[$a = 0$] or #math.equation(block: false, alt: "b equals 0")[$b = 0$] or both. - #strong[How to use the Zero Product Property] + Set each factor equal to zero. + Solve the linear equations. + Check. - #strong[How to solve a quadratic equation by factoring.] + Write the quadratic equation in standard form, #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 .")[$a x^(2) + b x + c = 0 .$] + Factor the quadratic expression. + Use the Zero Product Property. #linebreak() + Solve the linear equations. + Check. Substitute each solution separately into the original equation. - #strong[Zero of a Function:] For any function #emph[f], if #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0 ,")[$f ( x ) = 0 ,$] then #emph[x] is a zero of the function. - #strong[How to use a problem solving strategy to solve word problems.] + #strong[Read] the problem. Make sure all the words and ideas are understood. + #strong[Identify] what we are looking for. + #strong[Name] what we are looking for. Choose a variable to represent that quantity. + #strong[Translate] into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation. + #strong[Solve] the equation using appropriate algebra techniques. + #strong[Check] the answer in the problem and make sure it makes sense. + #strong[Answer] the question with a complete sentence. === Section Exercises ==== Practice Makes Perfect #strong[Use the Zero Product Property] In the following exercises, solve. #math.equation(block: true, alt: "open parenthesis 3 a minus 10 close parenthesis open parenthesis 2 a minus 7 close parenthesis equals 0")[$( 3 a − 10 ) ( 2 a − 7 ) = 0$] #solutionbox[ #math.equation(block: true, alt: "a equals 10 / 3 , a equals 7 / 2")[$a = 10 "/" 3 , a = 7 "/" 2$] ] #math.equation(block: true, alt: "open parenthesis 5 b plus 1 close parenthesis open parenthesis 6 b plus 1 close parenthesis equals 0")[$( 5 b + 1 ) ( 6 b + 1 ) = 0$] #math.equation(block: true, alt: "6 m open parenthesis 12 m minus 5 close parenthesis equals 0")[$6 m ( 12 m − 5 ) = 0$] #solutionbox[ #math.equation(block: true, alt: "m equals 0 , m equals 5 / 12")[$m = 0 , m = 5 "/" 12$] ] #math.equation(block: true, alt: "2 x open parenthesis 6 x minus 3 close parenthesis equals 0")[$2 x ( 6 x − 3 ) = 0$] #math.equation(block: true, alt: "open parenthesis 2 x minus 1 close parenthesis squared equals 0")[$attach(( 2 x − 1 ), t: 2) = 0$] #solutionbox[ #math.equation(block: true, alt: "x equals 1 / 2")[$x = 1 "/" 2$] ] #math.equation(block: true, alt: "open parenthesis 3 y plus 5 close parenthesis squared equals 0")[$attach(( 3 y + 5 ), t: 2) = 0$] #strong[Solve Quadratic Equations by Factoring] In the following exercises, solve. #math.equation(block: true, alt: "5 a squared minus 26 a equals 24")[$5 a^(2) − 26 a = 24$] #solutionbox[ #math.equation(block: true, alt: "a equals the fraction −4 over 5 , a equals 6")[$a = frac(−4, 5) , a = 6$] ] #math.equation(block: true, alt: "4 b squared plus 7 b equals −3")[$4 b^(2) + 7 b = −3$] #math.equation(block: true, alt: "4 m squared equals 17 m minus 15")[$4 m^(2) = 17 m − 15$] #solutionbox[ #math.equation(block: true, alt: "m equals 5 / 4 , m equals 3")[$m = 5 "/" 4 , m = 3$] ] #math.equation(block: true, alt: "n squared equals 5 n minus 6")[$n^(2) = 5 n − 6$] #math.equation(block: true, alt: "7 a squared plus 14 a equals 7 a")[$7 a^(2) + 14 a = 7 a$] #solutionbox[ #math.equation(block: true, alt: "a equals −1 , a equals 0")[$a = −1 , a = 0$] ] #math.equation(block: true, alt: "12 b squared minus 15 b equals −9 b")[$12 b^(2) − 15 b = −9 b$] #math.equation(block: true, alt: "49 m squared equals 144")[$49 m^(2) = 144$] #solutionbox[ #math.equation(block: true, alt: "m equals 12 / 7 , m equals −12 / 7")[$m = 12 "/" 7 , m = −12 "/" 7$] ] #math.equation(block: true, alt: "625 equals x squared")[$625 = x^(2)$] #math.equation(block: true, alt: "16 y squared equals 81")[$16 y^(2) = 81$] #solutionbox[ #math.equation(block: true, alt: "y equals −9 / 4 , y equals 9 / 4")[$y = −9 "/" 4 , y = 9 "/" 4$] ] #math.equation(block: true, alt: "64 p squared equals 225")[$64 p^(2) = 225$] #math.equation(block: true, alt: "121 n squared equals 36")[$121 n^(2) = 36$] #solutionbox[ #math.equation(block: true, alt: "n equals −6 / 11 , n equals 6 / 11")[$n = −6 "/" 11 , "n" = 6 "/" 11$] ] #math.equation(block: true, alt: "100 y squared equals 9")[$100 y^(2) = 9$] #math.equation(block: true, alt: "open parenthesis x plus 6 close parenthesis open parenthesis x minus 3 close parenthesis equals −8")[$( x + 6 ) ( x − 3 ) = −8$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals −5")[$x = 2 , x = −5$] ] #math.equation(block: true, alt: "open parenthesis p minus 5 close parenthesis open parenthesis p plus 3 close parenthesis equals −7")[$( p − 5 ) ( p + 3 ) = −7$] #math.equation(block: true, alt: "open parenthesis 2 x plus 1 close parenthesis open parenthesis x minus 3 close parenthesis equals −4 x")[$( 2 x + 1 ) ( x − 3 ) = −4 x$] #solutionbox[ #math.equation(block: true, alt: "x equals 3 / 2 , x equals −1")[$x = 3 "/" 2 , x = −1$] ] #math.equation(block: true, alt: "open parenthesis y minus 3 close parenthesis open parenthesis y plus 2 close parenthesis equals 4 y")[$( y − 3 ) ( y + 2 ) = 4 y$] #math.equation(block: true, alt: "open parenthesis 3 x minus 2 close parenthesis open parenthesis x plus 4 close parenthesis equals 12 x")[$( 3 x − 2 ) ( x + 4 ) = 12 x$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals −4 / 3")[$x = 2 , x = −4 "/" 3$] ] #math.equation(block: true, alt: "open parenthesis 2 y minus 3 close parenthesis open parenthesis 3 y minus 1 close parenthesis equals 8 y")[$( 2 y − 3 ) ( 3 y − 1 ) = 8 y$] #math.equation(block: true, alt: "20 x squared minus 60 x equals −45")[$20 x^(2) − 60 x = −45$] #solutionbox[ #math.equation(block: true, alt: "x equals 3 / 2")[$x = 3 "/" 2$] ] #math.equation(block: true, alt: "3 y squared minus 18 y equals −27")[$3 y^(2) − 18 y = −27$] #math.equation(block: true, alt: "15 x squared minus 10 x equals 40")[$15 x^(2) − 10 x = 40$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals −4 / 3")[$x = 2 , x = −4 "/" 3$] ] #math.equation(block: true, alt: "14 y squared minus 77 y equals −35")[$14 y^(2) − 77 y = −35$] #math.equation(block: true, alt: "18 x squared minus 9 equals −21 x")[$18 x^(2) − 9 = −21 x$] #solutionbox[ #math.equation(block: true, alt: "x equals −3 / 2 , x equals 1 / 3")[$x = −3 "/" 2 , x = 1 "/" 3$] ] #math.equation(block: true, alt: "16 y squared plus 12 equals −32 y")[$16 y^(2) + 12 = −32 y$] #math.equation(block: true, alt: "16 p cubed equals 24 p squared minus 9 p")[$16 p^(3) = 24 p^(2) − 9 p$] #solutionbox[ #math.equation(block: true, alt: "p equals 0 , p equals ¾")[$p = 0 , p = ¾$] ] #math.equation(block: true, alt: "m cubed minus 2 m squared equals − m")[$m^(3) − 2 m^(2) = "−" m$] #math.equation(block: true, alt: "2 x cubed plus 72 x equals 24 x squared")[$2 x^(3) + 72 x = 24 x^(2)$] #solutionbox[ #math.equation(block: true, alt: "x equals 0 , x equals 6")[$x = 0 , x = 6$] ] #math.equation(block: true, alt: "3 y cubed plus 48 y equals 24 y squared")[$3 y^(3) + 48 y = 24 y^(2)$] #math.equation(block: true, alt: "36 x cubed plus 24 x squared equals −4 x")[$36 x^(3) + 24 x^(2) = −4 x$] #solutionbox[ #math.equation(block: true, alt: "x equals 0 , x equals –1 / 3")[$x = 0 , x = –1 "/" 3$] ] #math.equation(block: true, alt: "2 y cubed plus 2 y squared equals 12 y")[$2 y^(3) + 2 y^(2) = 12 y$] #strong[Solve Equations with Polynomial Functions] In the following exercises, solve. For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 8 x plus 8 ,")[$f ( x ) = x^(2) − 8 x + 8 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −4")[$f ( x ) = −4$] ⓑ Use this information to find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 2")[$x = 2$] or #math.equation(block: false, alt: "x equals 6")[$x = 6$] ⓑ #math.equation(block: false, alt: "open parenthesis 2 , −4 close parenthesis")[$( 2 , −4 )$] #math.equation(block: false, alt: "open parenthesis 6 , −4 close parenthesis")[$( 6 , −4 )$] ] For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 11 x plus 20 ,")[$f ( x ) = x^(2) + 11 x + 20 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −8")[$f ( x ) = −8$] ⓑ Use this information to find two points that lie on the graph of the function. For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 8 x squared minus 18 x plus 5 ,")[$f ( x ) = 8 x^(2) − 18 x + 5 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −4")[$f ( x ) = −4$] ⓑ Use this information to find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals the fraction 3 over 2")[$x = frac(3, 2)$] or #math.equation(block: false, alt: "x equals the fraction 3 over 4")[$x = frac(3, 4)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 3 over 2 , −4 close parenthesis")[$( frac(3, 2) , −4 )$] #math.equation(block: false, alt: "open parenthesis the fraction 3 over 4 , −4 close parenthesis")[$( frac(3, 4) , −4 )$] ] For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 18 x squared plus 15 x minus 10 ,")[$f ( x ) = 18 x^(2) + 15 x − 10 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 15")[$f ( x ) = 15$] ⓑ Use this information to find two points that lie on the graph of the function. In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the #emph[x]-intercepts of the graph of the function ⓒ the #emph[y]-intercept of the graph of the function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 9 x squared minus 4")[$f ( x ) = 9 x^(2) − 4$] #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals the fraction 2 over 3")[$x = frac(2, 3)$] or #math.equation(block: false, alt: "x equals minus the fraction 2 over 3")[$x = − frac(2, 3)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 2 over 3 , 0 close parenthesis")[$( frac(2, 3) , 0 )$], #math.equation(block: false, alt: "open parenthesis minus the fraction 2 over 3 , 0 close parenthesis")[$( − frac(2, 3) , 0 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 0 , −4 close parenthesis")[$( 0 , −4 )$] ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 25 x squared minus 49")[$f ( x ) = 25 x^(2) − 49$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 x squared minus 7 x minus 5")[$f ( x ) = 6 x^(2) − 7 x − 5$] #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals the fraction 5 over 3")[$x = frac(5, 3)$] or #math.equation(block: false, alt: "x equals minus the fraction 1 over 2")[$x = − frac(1, 2)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 5 over 3 , 0 close parenthesis")[$( frac(5, 3) , 0 )$], #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 , 0 close parenthesis")[$( − frac(1, 2) , 0 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 0 , −5 close parenthesis")[$( 0 , −5 )$] ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 12 x squared minus 11 x plus 2")[$f ( x ) = 12 x^(2) − 11 x + 2$] #strong[Solve Applications Modeled by Quadratic Equations] In the following exercises, solve. The product of two consecutive odd integers is 143. Find the integers. #solutionbox[ #math.equation(block: false, alt: "−13 , −11")[$−13 , −11$] and 11, 13 ] The product of two consecutive odd integers is 195. Find the integers. The product of two consecutive even integers is 168. Find the integers. #solutionbox[ #math.equation(block: false, alt: "−14 , −12")[$−14 , −12$] and 12, 14 ] The product of two consecutive even integers is 288. Find the integers. The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet. #solutionbox[ Width: 4 feet; Length: 7 feet. ] A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall. The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board. #solutionbox[ Width: 5 feet; Length: 11 feet. ] A rectangular carport has area 150 square feet. The width of the carport is five feet less than twice its length. Find the width and the length of the carport. A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant. #solutionbox[ The sides are 6 feet and 8 feet. ] A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs. A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool. #solutionbox[ The building side is 8 feet, the hypotenuse is 17 feet, and the third side is 15 feet. ] A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure. Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 32 t")[$h ( t ) = −16 t^(2) + 32 t$] models the height, #emph[h], of the rocket above the ground as a function of time, #emph[t]. Find: ⓐ the zeros of this function, which tell us when the rocket will be on the ground. ⓑ the time the rocket will be 16 feet above the ground. #solutionbox[ ⓐ 0 seconds and 2 seconds ⓑ 1 second ] Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 32 t plus 48")[$h ( t ) = −16 t^(2) + 32 t + 48$] models the height, #emph[h], of the ball above the ground as a function of time, #emph[t]. Find: ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at #math.equation(block: false, alt: "t equals 1")[$t = 1$] seconds which is when the ball will be at its highest point. ==== Writing Exercises Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation? #solutionbox[ Answers will vary. ] Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero. ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve quadratic equations by using the zero product property, solve quadratic equations by factoring and solve applications modeled by quadratic equations.], alt: "This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve quadratic equations by using the zero product property, solve quadratic equations by factoring and solve applications modeled by quadratic equations.", caption: none) ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not? === Chapter Review Exercises ==== Greatest Common Factor and Factor by Grouping #strong[Find the Greatest Common Factor of Two or More Expressions] In the following exercises, find the greatest common factor. #math.equation(block: true, alt: "12 a squared b cubed , 15 a b squared")[$12 a^(2) b^(3) , 15 a b^(2)$] #solutionbox[ #math.equation(block: true, alt: "3 a b squared")[$3 a b^(2)$] ] #math.equation(block: true, alt: "12 m squared n cubed , 42 m to the power 5 n cubed")[$12 m^(2) n^(3) , 42 m^(5) n^(3)$] #math.equation(block: true, alt: "15 y cubed , 21 y squared , 30 y")[$15 y^(3) , 21 y^(2) , 30 y$] #solutionbox[ #math.equation(block: true, alt: "3 y")[$3 y$] ] #math.equation(block: true, alt: "45 x cubed y squared , 15 x to the power 4 y , 10 x to the power 5 y cubed")[$45 x^(3) y^(2) , 15 x^(4) y , 10 x^(5) y^(3)$] #strong[Factor the Greatest Common Factor from a Polynomial] In the following exercises, factor the greatest common factor from each polynomial. #math.equation(block: true, alt: "35 y plus 84")[$35 y + 84$] #solutionbox[ #math.equation(block: true, alt: "7 open parenthesis 5 y plus 12 close parenthesis")[$7 ( 5 y + 12 )$] ] #math.equation(block: true, alt: "6 y squared plus 12 y minus 6")[$6 y^(2) + 12 y − 6$] #math.equation(block: true, alt: "18 x cubed minus 15 x")[$18 x^(3) − 15 x$] #solutionbox[ #math.equation(block: true, alt: "3 x open parenthesis 6 x squared minus 5 close parenthesis")[$3 x ( 6 x^(2) − 5 )$] ] #math.equation(block: true, alt: "15 m to the power 4 plus 6 m squared n")[$15 m^(4) + 6 m^(2) n$] #math.equation(block: true, alt: "4 x cubed minus 12 x squared plus 16 x")[$4 x^(3) − 12 x^(2) + 16 x$] #solutionbox[ #math.equation(block: true, alt: "4 x open parenthesis x squared minus 3 x plus 4 close parenthesis")[$4 x ( x^(2) − 3 x + 4 )$] ] #math.equation(block: true, alt: "−3 x plus 24")[$−3 x + 24$] #math.equation(block: true, alt: "−3 x cubed plus 27 x squared minus 12 x")[$−3 x^(3) + 27 x^(2) − 12 x$] #solutionbox[ #math.equation(block: true, alt: "−3 x open parenthesis x squared minus 9 x plus 4 close parenthesis")[$−3 x ( x^(2) − 9 x + 4 )$] ] #math.equation(block: true, alt: "3 x open parenthesis x minus 1 close parenthesis plus 5 open parenthesis x minus 1 close parenthesis")[$3 x ( x − 1 ) + #h(0.2em) 5 ( x − 1 )$] #strong[Factor by Grouping] In the following exercises, factor by grouping. #math.equation(block: true, alt: "a x minus a y plus b x minus b y")[$a x − a y + b x − b y$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis open parenthesis x minus y close parenthesis")[$( a + b ) ( x − y )$] ] #math.equation(block: true, alt: "x squared y minus x y squared plus 2 x minus 2 y")[$x^(2) y − x y^(2) + 2 x − 2 y$] #math.equation(block: true, alt: "x squared plus 7 x minus 3 x minus 21")[$x^(2) + 7 x − 3 x − 21$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 7 close parenthesis")[$( x − 3 ) ( x + 7 )$] ] #math.equation(block: true, alt: "4 x squared minus 16 x plus 3 x minus 12")[$4 x^(2) − 16 x + 3 x − 12$] #math.equation(block: true, alt: "m cubed plus m squared plus m plus 1")[$m^(3) + m^(2) + m + 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m squared plus 1 close parenthesis open parenthesis m plus 1 close parenthesis")[$( m^(2) + 1 ) ( m + 1 )$] ] #math.equation(block: true, alt: "5 x minus 5 y minus y plus x")[$5 x − 5 y − y + x$] ==== Factor Trinomials #strong[Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]] In the following exercises, factor each trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c .")[$x^(2) + b x + c .$] #math.equation(block: true, alt: "a squared plus 14 a plus 33")[$a^(2) + 14 a + 33$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 3 close parenthesis open parenthesis a plus 11 close parenthesis")[$( a + 3 ) ( a + 11 )$] ] #math.equation(block: true, alt: "k squared minus 16 k plus 60")[$k^(2) − 16 k + 60$] #math.equation(block: true, alt: "m squared plus 3 m minus 54")[$m^(2) + 3 m − 54$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m plus 9 close parenthesis open parenthesis m minus 6 close parenthesis")[$( m + 9 ) ( m − 6 )$] ] #math.equation(block: true, alt: "x squared minus 3 x minus 10")[$x^(2) − 3 x − 10$] In the following examples, factor each trinomial of the form #math.equation(block: false, alt: "x squared plus b x y plus c y squared .")[$x^(2) + b x y + c y^(2) .$] #math.equation(block: true, alt: "x squared plus 12 x y plus 35 y squared")[$x^(2) + 12 x y + 35 y^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 5 y close parenthesis open parenthesis x plus 7 y close parenthesis")[$( x + 5 y ) ( x + 7 y )$] ] #math.equation(block: true, alt: "r squared plus 3 r s minus 28 s squared")[$r^(2) + 3 r s − 28 s^(2)$] #math.equation(block: true, alt: "a squared plus 4 a b minus 21 b squared")[$a^(2) + 4 a b − 21 b^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 7 b close parenthesis open parenthesis a minus 3 b close parenthesis")[$( a + 7 b ) ( a − 3 b )$] ] #math.equation(block: true, alt: "p squared minus 5 p q minus 36 q squared")[$p^(2) − 5 p q − 36 q^(2)$] #math.equation(block: true, alt: "m squared minus 5 m n plus 30 n squared")[$m^(2) − 5 m n + 30 n^(2)$] #solutionbox[ Prime ] #strong[Factor Trinomials of the Form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] Using Trial and Error] In the following exercises, factor completely using trial and error. #math.equation(block: true, alt: "x cubed plus 5 x squared minus 24 x")[$x^(3) + 5 x^(2) − 24 x$] #math.equation(block: true, alt: "3 y cubed minus 21 y squared plus 30 y")[$3 y^(3) − 21 y^(2) + 30 y$] #solutionbox[ #math.equation(block: true, alt: "3 y open parenthesis y minus 5 close parenthesis open parenthesis y minus 2 close parenthesis")[$3 y ( y − 5 ) ( y − 2 )$] ] #math.equation(block: true, alt: "5 x to the power 4 plus 10 x cubed minus 75 x squared")[$5 x^(4) + 10 x^(3) − 75 x^(2)$] #math.equation(block: true, alt: "5 y squared plus 14 y plus 9")[$5 y^(2) + 14 y + 9$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 y plus 9 close parenthesis open parenthesis y plus 1 close parenthesis")[$( 5 y + 9 ) ( y + 1 )$] ] #math.equation(block: true, alt: "8 x squared plus 25 x plus 3")[$8 x^(2) + 25 x + 3$] #math.equation(block: true, alt: "10 y squared minus 53 y minus 11")[$10 y^(2) − 53 y − 11$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 y plus 1 close parenthesis open parenthesis 2 y minus 11 close parenthesis")[$( 5 y + 1 ) ( 2 y − 11 )$] ] #math.equation(block: true, alt: "6 p squared minus 19 p q plus 10 q squared")[$6 p^(2) − 19 p q + 10 q^(2)$] #math.equation(block: true, alt: "−81 a squared plus 153 a plus 18")[$−81 a^(2) + 153 a + 18$] #solutionbox[ #math.equation(block: true, alt: "−9 open parenthesis 9 a plus 1 close parenthesis open parenthesis a minus 2 close parenthesis")[$−9 ( 9 a + 1 ) ( a − 2 )$] ] #strong[Factor Trinomials of the Form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] using the ‘ac’ Method] In the following exercises, factor. #math.equation(block: true, alt: "2 x squared plus 9 x plus 4")[$2 x^(2) + 9 x + 4$] #math.equation(block: true, alt: "18 a squared minus 9 a plus 1")[$18 a^(2) − 9 a + 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 a minus 1 close parenthesis open parenthesis 6 a minus 1 close parenthesis")[$( 3 a − 1 ) ( 6 a − 1 )$] ] #math.equation(block: true, alt: "15 p squared plus 2 p minus 8")[$15 p^(2) + 2 p − 8$] #math.equation(block: true, alt: "15 x squared plus 6 x minus 2")[$15 x^(2) + 6 x − 2$] #solutionbox[ Prime ] #math.equation(block: true, alt: "8 a squared plus 32 a plus 24")[$8 a^(2) + 32 a + 24$] #math.equation(block: true, alt: "3 x squared plus 3 x minus 36")[$3 x^(2) + 3 x − 36$] #solutionbox[ #math.equation(block: true, alt: "3 open parenthesis x plus 4 close parenthesis open parenthesis x minus 3 close parenthesis")[$3 ( x + 4 ) ( x − 3 )$] ] #math.equation(block: true, alt: "48 y squared plus 12 y minus 36")[$48 y^(2) + 12 y − 36$] #math.equation(block: true, alt: "18 a squared minus 57 a minus 21")[$18 a^(2) − 57 a − 21$] #solutionbox[ #math.equation(block: true, alt: "3 open parenthesis 2 a minus 7 close parenthesis open parenthesis 3 a plus 1 close parenthesis")[$3 ( 2 a − 7 ) ( 3 a + 1 )$] ] #math.equation(block: true, alt: "3 n to the power 4 minus 12 n cubed minus 96 n squared")[$3 n^(4) − 12 n^(3) − 96 n^(2)$] #strong[Factor using substitution] In the following exercises, factor using substitution. #math.equation(block: true, alt: "x to the power 4 minus 13 x squared minus 30")[$x^(4) − 13 x^(2) − 30$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x squared minus 15 close parenthesis open parenthesis x squared plus 2 close parenthesis")[$( x^(2) − 15 ) ( x^(2) + 2 )$] ] #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared minus 5 open parenthesis x minus 3 close parenthesis minus 36")[$attach(( x − 3 ), t: 2) − 5 ( x − 3 ) − 36$] ==== Factor Special Products #strong[Factor Perfect Square Trinomials] In the following exercises, factor completely using the perfect square trinomials pattern. #math.equation(block: true, alt: "25 x squared plus 30 x plus 9")[$25 x^(2) + 30 x + 9$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 x plus 3 close parenthesis squared")[$attach(( 5 x + 3 ), t: 2)$] ] #math.equation(block: true, alt: "36 a squared minus 84 a b plus 49 b squared")[$36 a^(2) − 84 a b + 49 b^(2)$] #math.equation(block: true, alt: "40 x squared plus 360 x plus 810")[$40 x^(2) + 360 x + 810$] #solutionbox[ #math.equation(block: true, alt: "10 open parenthesis 2 x plus 9 close parenthesis squared")[$10 attach(( 2 x + 9 ), t: 2)$] ] #math.equation(block: true, alt: "5 k cubed minus 70 k squared plus 245 k")[$5 k^(3) − 70 k^(2) + 245 k$] #math.equation(block: true, alt: "75 u to the power 4 minus 30 u cubed v plus 3 u squared v squared")[$75 u^(4) − 30 u^(3) v + 3 u^(2) v^(2)$] #solutionbox[ #math.equation(block: true, alt: "3 u squared open parenthesis 5 u minus v close parenthesis squared")[$3 u^(2) attach(( 5 u − v ), t: 2)$] ] #strong[Factor Differences of Squares] In the following exercises, factor completely using the difference of squares pattern, if possible. #math.equation(block: true, alt: "81 r squared minus 25")[$81 r^(2) − 25$] #math.equation(block: true, alt: "169 m squared minus n squared")[$169 m^(2) − n^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 13 m plus n close parenthesis open parenthesis 13 m minus n close parenthesis")[$( 13 m + n ) ( 13 m − n )$] ] #math.equation(block: true, alt: "25 p squared minus 1")[$25 p^(2) − 1$] #math.equation(block: true, alt: "9 minus 121 y squared")[$9 − 121 y^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 plus 11 y close parenthesis open parenthesis 3 minus 11 y close parenthesis")[$( 3 + 11 y ) ( 3 − 11 y )$] ] #math.equation(block: true, alt: "20 x squared minus 125")[$20 x^(2) − 125$] #math.equation(block: true, alt: "169 n cubed minus n")[$169 n^(3) − n$] #solutionbox[ #math.equation(block: true, alt: "n open parenthesis 13 n plus 1 close parenthesis open parenthesis 13 n minus 1 close parenthesis")[$n ( 13 n + 1 ) ( 13 n − 1 )$] ] #math.equation(block: true, alt: "6 p squared q squared minus 54 p squared")[$6 p^(2) q^(2) − 54 p^(2)$] #math.equation(block: true, alt: "24 p squared plus 54")[$24 p^(2) + 54$] #solutionbox[ #math.equation(block: true, alt: "6 open parenthesis 4 p squared plus 9 close parenthesis")[$6 ( 4 p^(2) + 9 )$] ] #math.equation(block: true, alt: "49 x squared minus 81 y squared")[$49 x^(2) − 81 y^(2)$] #math.equation(block: true, alt: "16 z to the power 4 minus 1")[$16 z^(4) − 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 z minus 1 close parenthesis open parenthesis 2 z plus 1 close parenthesis open parenthesis 4 z squared plus 1 close parenthesis")[$( 2 z − 1 ) ( 2 z + 1 ) ( 4 z^(2) + 1 )$] ] #math.equation(block: true, alt: "48 m to the power 4 n squared minus 243 n squared")[$48 m^(4) n^(2) − 243 n^(2)$] #math.equation(block: true, alt: "a squared plus 6 a plus 9 minus 9 b squared")[$a^(2) + 6 a + 9 − 9 b^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 3 minus 3 b close parenthesis open parenthesis a plus 3 plus 3 b close parenthesis")[$( a + 3 − 3 b ) ( a + 3 + 3 b )$] ] #math.equation(block: true, alt: "x squared minus 16 x plus 64 minus y squared")[$x^(2) − 16 x + 64 − y^(2)$] #strong[Factor Sums and Differences of Cubes] In the following exercises, factor completely using the sums and differences of cubes pattern, if possible. #math.equation(block: true, alt: "a cubed minus 125")[$a^(3) − 125$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a minus 5 close parenthesis open parenthesis a squared plus 5 a plus 25 close parenthesis")[$( a − 5 ) ( a^(2) + 5 a + 25 )$] ] #math.equation(block: true, alt: "b cubed minus 216")[$b^(3) − 216$] #math.equation(block: true, alt: "2 m cubed plus 54")[$2 m^(3) + 54$] #solutionbox[ #math.equation(block: true, alt: "2 open parenthesis m plus 3 close parenthesis open parenthesis m squared minus 3 m plus 9 close parenthesis")[$2 ( m + 3 ) ( m^(2) − 3 m + 9 )$] ] #math.equation(block: true, alt: "81 m cubed plus 3")[$81 m^(3) + 3$] ==== General Strategy for Factoring Polynomials #strong[Recognize and Use the Appropriate Method to Factor a Polynomial Completely] In the following exercises, factor completely. #math.equation(block: true, alt: "24 x cubed plus 44 x squared")[$24 x^(3) + 44 x^(2)$] #solutionbox[ #math.equation(block: true, alt: "4 x squared open parenthesis 6 x plus 11 close parenthesis")[$4 x^(2) ( 6 x + 11 )$] ] #math.equation(block: true, alt: "24 a to the power 4 minus 9 a cubed")[$24 a^(4) − 9 a^(3)$] #math.equation(block: true, alt: "16 n squared minus 56 m n plus 49 m squared")[$16 n^(2) − 56 m n + 49 m^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 n minus 7 m close parenthesis squared")[$attach(( 4 n − 7 m ), t: 2)$] ] #math.equation(block: true, alt: "6 a squared minus 25 a minus 9")[$6 a^(2) − 25 a − 9$] #math.equation(block: true, alt: "5 u to the power 4 minus 45 u squared")[$5 u^(4) − 45 u^(2)$] #solutionbox[ #math.equation(block: true, alt: "5 u squared open parenthesis u plus 3 close parenthesis open parenthesis u minus 3 close parenthesis")[$5 u^(2) ( u + 3 ) ( u − 3 )$] ] #math.equation(block: true, alt: "n to the power 4 minus 81")[$n^(4) − 81$] #math.equation(block: true, alt: "64 j squared plus 225")[$64 j^(2) + 225$] #solutionbox[ prime ] #math.equation(block: true, alt: "5 x squared plus 5 x minus 60")[$5 x^(2) + 5 x − 60$] #math.equation(block: true, alt: "b cubed minus 64")[$b^(3) − 64$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis b minus 4 close parenthesis open parenthesis b squared plus 4 b plus 16 close parenthesis")[$( b − 4 ) ( b^(2) + 4 b + 16 )$] ] #math.equation(block: true, alt: "m cubed plus 125")[$m^(3) + 125$] #math.equation(block: true, alt: "2 b squared minus 2 b c plus 5 c b minus 5 c squared")[$2 b^(2) − 2 b c + 5 c b − 5 c^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 b plus 5 c close parenthesis open parenthesis b minus c close parenthesis")[$( 2 b + 5 c ) ( b − c )$] ] #math.equation(block: true, alt: "48 x to the power 5 y squared minus 243 x y squared")[$48 x^(5) y^(2) − 243 x y^(2)$] #math.equation(block: true, alt: "5 q squared minus 15 q minus 90")[$5 q^(2) − 15 q − 90$] #solutionbox[ #math.equation(block: true, alt: "5 open parenthesis q plus 3 close parenthesis open parenthesis q minus 6 close parenthesis")[$5 ( q + 3 ) ( q − 6 )$] ] #math.equation(block: true, alt: "4 u to the power 5 v plus 4 u squared v cubed")[$4 u^(5) v + 4 u^(2) v^(3)$] #math.equation(block: true, alt: "10 m to the power 4 minus 6250")[$10 m^(4) − 6250$] #solutionbox[ #math.equation(block: true, alt: "10 open parenthesis m minus 5 close parenthesis open parenthesis m plus 5 close parenthesis open parenthesis m squared plus 25 close parenthesis")[$10 ( m − 5 ) ( m + 5 ) ( m^(2) + 25 )$] ] #math.equation(block: true, alt: "60 x squared y minus 75 x y plus 30 y")[$60 x^(2) y − 75 x y + 30 y$] #math.equation(block: true, alt: "16 x squared minus 24 x y plus 9 y squared minus 64")[$16 x^(2) − 24 x y + 9 y^(2) − 64$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 x minus 3 y plus 8 close parenthesis open parenthesis 4 x minus 3 y minus 8 close parenthesis")[$( 4 x − 3 y + 8 ) ( 4 x − 3 y − 8 )$] ] ==== Polynomial Equations #strong[Use the Zero Product Property] In the following exercises, solve. #math.equation(block: true, alt: "open parenthesis a minus 3 close parenthesis open parenthesis a plus 7 close parenthesis equals 0")[$( a − 3 ) ( a + 7 ) = 0$] #math.equation(block: true, alt: "open parenthesis 5 b plus 1 close parenthesis open parenthesis 6 b plus 1 close parenthesis equals 0")[$( 5 b + 1 ) ( 6 b + 1 ) = 0$] #solutionbox[ #math.equation(block: true, alt: "b equals −1 / 5 , b equals −1 / 6")[$b = −1 "/" 5 , b = −1 "/" 6$] ] #math.equation(block: true, alt: "6 m open parenthesis 12 m minus 5 close parenthesis equals 0")[$6 m ( 12 m − 5 ) = 0$] #math.equation(block: true, alt: "open parenthesis 2 x minus 1 close parenthesis squared equals 0")[$attach(( 2 x − 1 ), t: 2) = 0$] #solutionbox[ #math.equation(block: true, alt: "x equals 1 / 2")[$x = 1 "/" 2$] ] #math.equation(block: true, alt: "3 m open parenthesis 2 m minus 5 close parenthesis open parenthesis m plus 6 close parenthesis equals 0")[$3 m ( 2 m − 5 ) ( m + 6 ) = 0$] #strong[Solve Quadratic Equations by Factoring] In the following exercises, solve. #math.equation(block: true, alt: "x squared plus 9 x plus 20 equals 0")[$x^(2) + 9 x + 20 = 0$] #solutionbox[ #math.equation(block: true, alt: "x equals −4 , x equals −5")[$x = −4 , x = −5$] ] #math.equation(block: true, alt: "y squared minus y minus 72 equals 0")[$y^(2) − y − 72 = 0$] #math.equation(block: true, alt: "2 p squared minus 11 p equals 40")[$2 p^(2) − 11 p = 40$] #solutionbox[ #math.equation(block: true, alt: "p equals minus the fraction 5 over 2 , p equals 8")[$p = − frac(5, 2) , p = 8$] ] #math.equation(block: true, alt: "q cubed plus 3 q squared plus 2 q equals 0")[$q^(3) + 3 q^(2) + 2 q = 0$] #math.equation(block: true, alt: "144 m squared minus 25 equals 0")[$144 m^(2) − 25 = 0$] #solutionbox[ #math.equation(block: true, alt: "m equals the fraction 5 over 12 , m equals minus the fraction 5 over 12")[$m = frac(5, 12) , m = − frac(5, 12)$] ] #math.equation(block: true, alt: "4 n squared equals 36")[$4 n^(2) = 36$] #math.equation(block: true, alt: "open parenthesis x plus 6 close parenthesis open parenthesis x minus 3 close parenthesis equals −8")[$( x + 6 ) ( x − 3 ) = −8$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals −5")[$x = 2 , x = −5$] ] #math.equation(block: true, alt: "open parenthesis 3 x minus 2 close parenthesis open parenthesis x plus 4 close parenthesis equals 12 x")[$( 3 x − 2 ) ( x + 4 ) = 12 x$] #math.equation(block: true, alt: "16 p cubed equals 24 p squared minus 9 p")[$16 p^(3) = 24 p^(2) − 9 p$] #solutionbox[ #math.equation(block: true, alt: "p equals 0 , p equals ¾")[$p = 0 , p = ¾$] ] #math.equation(block: true, alt: "2 y cubed plus 2 y squared equals 12 y")[$2 y^(3) + 2 y^(2) = 12 y$] #strong[Solve Equations with Polynomial Functions] In the following exercises, solve. For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 11 x plus 20 ,")[$f ( x ) = x^(2) + 11 x + 20 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −8")[$f ( x ) = −8$] ⓑ Use this information to find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals −7")[$x = −7$] or #math.equation(block: false, alt: "x equals −4")[$x = −4$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis −7 , −8 close parenthesis")[$( −7 , −8 )$] #math.equation(block: false, alt: "open parenthesis −4 , −8 close parenthesis")[$( −4 , −8 )$] ] For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 9 x squared minus 18 x plus 5 ,")[$f ( x ) = 9 x^(2) − 18 x + 5 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −3")[$f ( x ) = −3$] ⓑ Use this information to find two points that lie on the graph of the function. In each function, find: ⓐ the zeros of the function ⓑ the #emph[x]-intercepts of the graph of the function ⓒ the #emph[y]-intercept of the graph of the function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 64 x squared minus 49")[$f ( x ) = 64 x^(2) − 49$] #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals the fraction 7 over 8")[$x = frac(7, 8)$] or #math.equation(block: false, alt: "x equals minus the fraction 7 over 8")[$x = − frac(7, 8)$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis the fraction 7 over 8 , 0 close parenthesis ,")[$( frac(7, 8) , 0 ) ,$] #math.equation(block: false, alt: "open parenthesis minus the fraction 7 over 8 , 0 close parenthesis")[$( − frac(7, 8) , 0 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 0 , −49 close parenthesis")[$( 0 , −49 )$] ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 x squared minus 13 x minus 5")[$f ( x ) = 6 x^(2) − 13 x − 5$] #strong[Solve Applications Modeled by Quadratic Equations] In the following exercises, solve. The product of two consecutive odd numbers is 399. Find the numbers. #solutionbox[ The numbers are #math.equation(block: false, alt: "−21")[$−21$] and #math.equation(block: false, alt: "−19")[$−19$] or 19 and 21. ] The area of a rectangular shaped patio 432 square feet. The length of the patio is 6 feet more than its width. Find the length and width. A ladder leans against the wall of a building. The length of the ladder is 9 feet longer than the distance of the bottom of the ladder from the building. The distance of the top of the ladder reaches up the side of the building is 7 feet longer than the distance of the bottom of the ladder from the building. Find the lengths of all three sides of the triangle formed by the ladder leaning against the building. #solutionbox[ The lengths are 8, 15, and 17 ft. ] Shruti is going to throw a ball from the top of a cliff. When she throws the ball from 80 feet above the ground, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 64 t plus 80")[$h ( t ) = −16 t^(2) + 64 t + 80$] models the height, #emph[h], of the ball above the ground as a function of time, #emph[t]. Find: ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 80 feet above the ground. ⓒ the height the ball will be at #math.equation(block: false, alt: "t equals 2")[$t = 2$] seconds which is when the ball will be at its highest point. === Chapter Practice Test In the following exercises, factor completely. #math.equation(block: true, alt: "80 a squared plus 120 a cubed")[$80 a^(2) + 120 a^(3)$] #solutionbox[ #math.equation(block: true, alt: "40 a squared open parenthesis 2 plus 3 a close parenthesis")[$40 a^(2) ( 2 + 3 a )$] ] #math.equation(block: true, alt: "5 m open parenthesis m minus 1 close parenthesis plus 3 open parenthesis m minus 1 close parenthesis")[$5 m ( m − 1 ) + 3 ( m − 1 )$] #math.equation(block: true, alt: "x squared plus 13 x plus 36")[$x^(2) + 13 x + 36$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x plus 9 close parenthesis")[$( x + 4 ) ( x + 9 )$] ] #math.equation(block: true, alt: "p squared plus p q minus 12 q squared")[$p^(2) + p q − 12 q^(2)$] #math.equation(block: true, alt: "x y minus 8 y plus 7 x minus 56")[$x y − 8 y + 7 x − 56$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 8 close parenthesis open parenthesis y plus 7 close parenthesis")[$( x − 8 ) ( y + 7 )$] ] #math.equation(block: true, alt: "40 r squared plus 810")[$40 r^(2) + 810$] #math.equation(block: true, alt: "9 s squared minus 12 s plus 4")[$9 s^(2) − 12 s + 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 s minus 2 close parenthesis squared")[$attach(( 3 s − 2 ), t: 2)$] ] #math.equation(block: true, alt: "6 x squared minus 11 x minus 10")[$6 x^(2) − 11 x − 10$] #math.equation(block: true, alt: "3 x squared minus 75 y squared")[$3 x^(2) − 75 y^(2)$] #solutionbox[ #math.equation(block: true, alt: "3 open parenthesis x plus 5 y close parenthesis open parenthesis x minus 5 y close parenthesis")[$3 ( x + 5 y ) ( x − 5 y )$] ] #math.equation(block: true, alt: "6 u squared plus 3 u minus 18")[$6 u^(2) + 3 u − 18$] #math.equation(block: true, alt: "x cubed plus 125")[$x^(3) + 125$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 5 close parenthesis open parenthesis x squared minus 5 x plus 25 close parenthesis")[$( x + 5 ) ( x^(2) − 5 x + 25 )$] ] #math.equation(block: true, alt: "32 x to the power 5 y squared minus 162 x y squared")[$32 x^(5) y^(2) − 162 x y^(2)$] #math.equation(block: true, alt: "6 x to the power 4 minus 19 x squared plus 15")[$6 x^(4) − 19 x^(2) + 15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 x squared minus 5 close parenthesis open parenthesis 2 x squared minus 3 close parenthesis")[$( 3 x^(2) − 5 ) ( 2 x^(2) − 3 )$] ] #math.equation(block: true, alt: "3 x cubed minus 36 x squared plus 108 x")[$3 x^(3) − 36 x^(2) + 108 x$] In the following exercises, solve #math.equation(block: true, alt: "5 a squared plus 26 a equals 24")[$5 a^(2) + 26 a = 24$] #solutionbox[ #math.equation(block: true, alt: "a equals 4 / 5 , a equals −6")[$a = 4 "/" 5 , a = −6$] ] The product of two consecutive integers is 156. Find the integers. The area of a rectangular place mat is 168 square inches. Its length is two inches longer than the width. Find the length and width of the placemat. #solutionbox[ The width is 12 inches and the length is 14 inches. ] Jing is going to throw a ball from the balcony of her condo. When she throws the ball from 80 feet above the ground, the function #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals −16 t squared plus 64 t plus 80")[$h ( t ) = −16 t^(2) + 64 t + 80$] models the height, #emph[h], of the ball above the ground as a function of time, #emph[t]. Find: ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 128 feet above the ground. ⓒ the height the ball will be at #math.equation(block: false, alt: "t equals 4")[$t = 4$] seconds. For the function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 7 x plus 5 ,")[$f ( x ) = x^(2) − 7 x + 5 ,$] ⓐ find when #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −7")[$f ( x ) = −7$] ⓑ Use this information to find two points that lie on the graph of the function. #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 3")[$x = 3$] or #math.equation(block: false, alt: "x equals 4")[$x = 4$] ⓑ #math.equation(block: false, alt: "open parenthesis 3 , −7 close parenthesis")[$( 3 , −7 )$] #math.equation(block: false, alt: "open parenthesis 4 , −7 close parenthesis")[$( 4 , −7 )$] ] For the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 25 x squared minus 81 ,")[$f ( x ) = 25 x^(2) − 81 ,$] find: ⓐ the zeros of the function ⓑ the #emph[x]-intercepts of the graph of the function ⓒ the #emph[y]-intercept of the graph of the function.