#set document(title: "6.2 Factor Trinomials", 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")) == 6.2#h(0.6em)Factor Trinomials #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Find all the factors of 72. #linebreak() If you missed this problem, review . #solutionbox[ 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the product: #math.equation(block: false, alt: "open parenthesis 3 y plus 4 close parenthesis open parenthesis 2 y plus 5 close parenthesis .")[$( 3 y + 4 ) ( 2 y + 5 ) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "6 y squared plus 23 y plus 20")[$6 y^(2) + 23 y + 20$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−9 open parenthesis 6 close parenthesis ;")[$−9 ( 6 ) ;$] #math.equation(block: false, alt: "−9 open parenthesis −6 close parenthesis .")[$−9 ( −6 ) .$] #linebreak() If you missed this problem, review . #solutionbox[ −54, 54 ] ] === Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] You have already learned how to multiply binomials using #strong[FOIL]. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors. #figure(figph[Figure shows the equation open parentheses x plus 2 close parentheses open parentheses x plus 3 close parentheses equals x squared plus 5 x plus 6. The left side of the equation is labeled factors and the right is labeled product. An arrow pointing right is labeled multiply. An arrow pointing left is labeled factor.], alt: "Figure shows the equation open parentheses x plus 2 close parentheses open parentheses x plus 3 close parentheses equals x squared plus 5 x plus 6. The left side of the equation is labeled factors and the right is labeled product. An arrow pointing right is labeled multiply. An arrow pointing left is labeled factor.", caption: none) To figure out how we would factor a #strong[trinomial] of the form #math.equation(block: false, alt: "x squared plus b x plus c ,")[$x^(2) + b x + c ,$] such as #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$] and factor it to #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis ,")[$( x + 2 ) ( x + 3 ) ,$] let’s start with two general binomials of the form #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis")[$( x + m )$] and #math.equation(block: false, alt: "open parenthesis x plus n close parenthesis .")[$( x + n ) .$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[The image displays the algebraic expression (x + m)(x + n), which represents the product of two binomials. This is a common form in algebra, often encountered when expanding or factoring quadratic equations.], alt: "The image displays the algebraic expression (x + m)(x + n), which represents the product of two binomials. This is a common form in algebra, often encountered when expanding or factoring quadratic equations.", caption: none)]), [Foil to find the product.], [#figure(figph[A mathematical expression showing x squared plus mx plus nx plus mn.], alt: "A mathematical expression showing x squared plus mx plus nx plus mn.", caption: none)], [Factor the GCF from the middle terms.], [#figure(figph[An algebraic expression is shown: x² + (m + n)x + mn. This is a quadratic expression with variables x, m, and n, demonstrating the expanded form of (x+m)(x+n).], alt: "An algebraic expression is shown: x² + (m + n)x + mn. This is a quadratic expression with variables x, m, and n, demonstrating the expanded form of (x+m)(x+n).", caption: none)], [Our trinomial is of the form #math.equation(block: false, alt: "x squared plus b x plus c .")[$x^(2) + b x + c .$]], [#figure(figph[This image illustrates the relationship between the standard quadratic form x^2 + bx + c and its factored form components x^2 + (m + n)x + mn, highlighting b = m + n and c = mn.], alt: "This image illustrates the relationship between the standard quadratic form x^2 + bx + c and its factored form components x^2 + (m + n)x + mn, highlighting b = m + n and c = mn.", caption: none)], )) This tells us that to factor a trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c ,")[$x^(2) + b x + c ,$] we need two factors #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis")[$( x + m )$] and #math.equation(block: false, alt: "open parenthesis x plus n close parenthesis")[$( x + n )$] where the two numbers #emph[m] and #emph[n] multiply to #emph[c] and add to #emph[b]. #examplebox("Example 1")[How to Factor a Trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]][ Factor: #math.equation(block: false, alt: "x squared plus 11 x plus 24 .")[$x^(2) + 11 x + 24 .$] #solutionbox[ #figure(figph[Step 1 is to write the factors of x squared plus 11x plus 24 as two binomials with first terms x. Write two sets of parentheses and put x as the first term.], alt: "Step 1 is to write the factors of x squared plus 11x plus 24 as two binomials with first terms x. Write two sets of parentheses and put x as the first term.", caption: none) #figure(figph[Step 2 is to find two numbers m and n that multiply to c, m times n is c and add to b, m plus n is b. So, find two numbers that multiply to 24 and add to 11. Factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6. Sum of factors: 1 plus 24 is 25, 2 plus 12 is 14, 3 plus 8 is 11 and 4 plus 6 is 10.], alt: "Step 2 is to find two numbers m and n that multiply to c, m times n is c and add to b, m plus n is b. So, find two numbers that multiply to 24 and add to 11. Factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6. Sum of factors: 1 plus 24 is 25, 2 plus 12 is 14, 3 plus 8 is 11 and 4 plus 6 is 10.", caption: none) #figure(figph[Step 3 is to use m and n, in this case, 3 and 8, as the last terms of the binomials. So we get open parentheses x plus 3 close parentheses open parentheses x plus 8 close parentheses], alt: "Step 3 is to use m and n, in this case, 3 and 8, as the last terms of the binomials. So we get open parentheses x plus 3 close parentheses open parentheses x plus 8 close parentheses", caption: none) #figure(figph[Step 4 is to check by multiplying the factors to get the original polynomial.], alt: "Step 4 is to check by multiplying the factors to get the original polynomial.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "q squared plus 10 q plus 24 .")[$q^(2) + 10 q + 24 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis q plus 4 close parenthesis open parenthesis q plus 6 close parenthesis")[$( q + 4 ) ( q + 6 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "t squared plus 14 t plus 24 .")[$t^(2) + 14 t + 24 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis t plus 2 close parenthesis open parenthesis t plus 12 close parenthesis")[$( t + 2 ) ( t + 12 )$] ] ] Let’s summarize the steps we used to find the factors. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c .")[$x^(2) + b x + c .$]] + Write the factors as two binomials with first terms #emph[x]. #math.equation(block: false, alt: "x squared plus b x plus c; open parenthesis x close parenthesis open parenthesis x close parenthesis")[$#h(4em) x^(2) + b x + c \ ( x #h(1.2em) ) ( x #h(1.2em) )$] + Find two numbers #emph[m] and #emph[n] that - multiply to #math.equation(block: false, alt: "c , m times n equals c")[$c , m · n = c$] - add to #math.equation(block: false, alt: "b , m plus n equals b")[$b , m + n = b$] + Use #emph[m] and #emph[n] as the last terms of the factors. #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$#h(7em) ( x + m ) ( x + n )$] + Check by multiplying the factors. ] In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative. How do you get a #emph[positive product] and a #emph[negative sum]? We use two negative numbers. #examplebox("Example 2")[][ Factor: #math.equation(block: false, alt: "y squared minus 11 y plus 28.")[$y^(2) − 11 y + 28.$] #solutionbox[ Again, with the positive last term, 28, and the negative middle term, #math.equation(block: false, alt: "−11 y ,")[$−11 y ,$] we need two negative factors. Find two numbers that multiply 28 and add to #math.equation(block: false, alt: "−11 .")[$−11 .$] #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "y squared minus 11 y plus 28")[$y^(2) − 11 y + 28$]]), [Write the factors as two binomials with first terms #math.equation(block: false, alt: "y .")[$y .$]], [#math.equation(block: false, alt: "open parenthesis y close parenthesis open parenthesis y close parenthesis")[$( y #h(1.2em) ) ( y #h(1.2em) )$]], [Find two numbers that: multiply to 28 and add to −11.], [], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "28")[$28$]], [Sum of factors]), [#math.equation(block: false, alt: "− 1 , −28")[$"−" 1 , −28$] #linebreak() #linebreak() #math.equation(block: false, alt: "− 2 , −14")[$"−" 2 , −14$] #linebreak() #linebreak() #math.equation(block: false, alt: "− 4 , −7")[$"−" 4 , −7$]], [#math.equation(block: false, alt: "− 1 plus open parenthesis − 28 close parenthesis equals −29")[$"−" 1 + ( "−" 28 ) = −29$] #linebreak() #linebreak() #math.equation(block: false, alt: "− 2 plus open parenthesis − 14 close parenthesis equals −16")[$"−" 2 + ( "−" 14 ) = −16$] #linebreak() #linebreak() #math.equation(block: false, alt: "− 4 plus open parenthesis − 7 close parenthesis equals −11 to the power *")[$#h(0.5em) "−" 4 + ( "−" 7 ) = −11^(*)$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Use #math.equation(block: false, alt: "−4 , −7")[$−4 , −7$] as the last terms of the binomials.], [#math.equation(block: false, alt: "open parenthesis y minus 4 close parenthesis open parenthesis y minus 7 close parenthesis")[$( y − 4 ) ( y − 7 )$]]), [Check: #linebreak() #math.equation(block: false, alt: "open parenthesis y minus 4 close parenthesis open parenthesis y minus 7 close parenthesis; y squared minus 7 y minus 4 y plus 28; y squared minus 11 y plus 28 ✓")[$#h(4em) ( y − 4 ) ( y − 7 ) \ y^(2) − 7 y − 4 y + 28 \ y^(2) − 11 y + 28 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "u squared minus 9 u plus 18 .")[$u^(2) − 9 u + 18 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis u minus 3 close parenthesis open parenthesis u minus 6 close parenthesis")[$( u − 3 ) ( u − 6 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "y squared minus 16 y plus 63 .")[$y^(2) − 16 y + 63 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis y minus 7 close parenthesis open parenthesis y minus 9 close parenthesis")[$( y − 7 ) ( y − 9 )$] ] ] Now, what if the last term in the trinomial is negative? Think about #strong[FOIL]. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too. How do you get a #emph[negative product] and a #emph[positive sum]? We use one positive and one negative number. When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree. #examplebox("Example 3")[][ Factor: #math.equation(block: false, alt: "2 x plus x squared minus 48 .")[$2 x + x^(2) − 48 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "2 x plus x squared minus 48")[$2 x + x^(2) − 48$]]), [First we put the terms in decreasing degree order.], [#math.equation(block: false, alt: "x squared plus 2 x minus 48")[$x^(2) + 2 x − 48$]], [Factors will be two binomials with first terms #math.equation(block: false, alt: "x .")[$x .$]], [#math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1.2em) ) ( x #h(1.2em) )$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "−48")[$−48$]], [Sum of factors]), [#math.equation(block: false, alt: "−1 , 48")[$−1 , 48$] #linebreak() #math.equation(block: false, alt: "−2 , 24")[$−2 , 24$] #linebreak() #math.equation(block: false, alt: "−3 , 16")[$−3 , 16$] #linebreak() #math.equation(block: false, alt: "−4 , 12")[$−4 , 12$] #linebreak() #math.equation(block: false, alt: "−6 , 8")[$−6 , 8$]], [#math.equation(block: false, alt: "−1 plus 48 equals 47")[$−1 + 48 = 47$] #linebreak() #math.equation(block: false, alt: "−2 plus 24 equals 22")[$−2 + 24 = 22$] #linebreak() #math.equation(block: false, alt: "−3 plus 16 equals 13")[$−3 + 16 = 13$] #linebreak() #math.equation(block: false, alt: "−4 plus 12 equals 8")[$−4 + 12 = 8$] #linebreak() #math.equation(block: false, alt: "−6 plus 8 equals 2 to the power *")[$#h(0.5em) −6 + 8 = 2^(*)$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Use −6 , 8 as the last terms of the binomials.")[$"Use" #h(0.2em) −6 , 8 #h(0.2em) "as the last terms of the binomials."$]], [#math.equation(block: false, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 8 close parenthesis")[$( x − 6 ) ( x + 8 )$]]), [Check: #linebreak() #math.equation(block: false, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 8 close parenthesis; x squared minus 6 q plus 8 q minus 48; x squared plus 2 x minus 48 ✓")[$#h(4em) ( x − 6 ) ( x + 8 ) \ x^(2) − 6 q + 8 q − 48 \ x^(2) + 2 x − 48 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "9 m plus m squared plus 18 .")[$9 m + m^(2) + 18 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m plus 3 close parenthesis open parenthesis m plus 6 close parenthesis")[$( m + 3 ) ( m + 6 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "−7 n plus 12 plus n squared .")[$−7 n + 12 + n^(2) .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis n minus 3 close parenthesis open parenthesis n minus 4 close parenthesis")[$( n − 3 ) ( n − 4 )$] ] ] Sometimes you’ll need to factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x y plus c y squared")[$x^(2) + b x y + c y^(2)$] with two variables, such as #math.equation(block: false, alt: "x squared plus 12 x y plus 36 y squared .")[$x^(2) + 12 x y + 36 y^(2) .$] The first term, #math.equation(block: false, alt: "x squared ,")[$x^(2) ,$] is the product of the first terms of the binomial factors, #math.equation(block: false, alt: "x times x .")[$x · x .$] The #math.equation(block: false, alt: "y squared")[$y^(2)$] in the last term means that the second terms of the binomial factors must each contain #emph[y]. To get the coefficients #emph[b] and #emph[c], you use the same process summarized in How To Factor trinomials. #examplebox("Example 4")[][ Factor: #math.equation(block: false, alt: "r squared minus 8 r s minus 9 s squared .")[$r^(2) − 8 r s − 9 s^(2) .$] #solutionbox[ We need #emph[r] in the first term of each binomial and #emph[s] in the second term. The last term of the trinomial is negative, so the factors must have opposite signs. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "r squared minus 8 r s minus 9 s squared")[$r^(2) − 8 r s − 9 s^(2)$]]), [Note that the first terms are #math.equation(block: false, alt: "r ,")[$r ,$] last terms contain #math.equation(block: false, alt: "s .")[$s .$]], [#math.equation(block: false, alt: "open parenthesis r s close parenthesis open parenthesis r s close parenthesis")[$( r #h(1.2em) s ) ( r #h(1.2em) s )$]], [Find the numbers that multiply to −9 and add to −8.], [], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "−9")[$−9$]], [Sum of factors]), [#math.equation(block: false, alt: "1 , −9")[$#h(0.65em) 1 , −9$]], [#math.equation(block: false, alt: "−1 plus 9 equals 8")[$#h(0.65em) −1 + 9 = 8$]], [#math.equation(block: false, alt: "−1 , 9")[$−1 , 9$]], [#math.equation(block: false, alt: "1 plus open parenthesis −9 close parenthesis equals − 8 to the power *")[$1 + ( −9 ) = "−" 8^(*)$]], [#math.equation(block: false, alt: "3 , −3")[$#h(0.65em) 3 , −3$]], [#math.equation(block: false, alt: "3 plus open parenthesis −3 close parenthesis equals 0")[$3 + ( −3 ) = 0$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Use 1 , −9 as coefficients of the last terms.")[$"Use" #h(0.2em) 1 , −9 #h(0.2em) "as coefficients of the last terms."$]], [#math.equation(block: false, alt: "open parenthesis r plus s close parenthesis open parenthesis r minus 9 s close parenthesis")[$( r + s ) ( r − 9 s )$]]), [Check: #linebreak() #math.equation(block: false, alt: "open parenthesis r minus 9 s close parenthesis open parenthesis r plus s close parenthesis; r squared plus r s minus 9 r s minus 9 s squared; r squared minus 8 r s minus 9 s squared ✓")[$#h(4em) ( r − 9 s ) ( r + s ) \ r^(2) + r s − 9 r s − 9 s^(2) \ r^(2) − 8 r s − 9 s^(2) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "a squared minus 11 a b plus 10 b squared .")[$a^(2) − 11 a b + 10 b^(2) .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a minus b close parenthesis open parenthesis a minus 10 b close parenthesis")[$( a − b ) ( a − 10 b )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "m squared minus 13 m n plus 12 n squared .")[$m^(2) − 13 m n + 12 n^(2) .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m minus n close parenthesis open parenthesis m minus 12 n close parenthesis")[$( m − n ) ( m − 12 n )$] ] ] Some trinomials are prime. The only way to be certain a trinomial is #strong[prime] is to list all the possibilities and show that none of them work. #examplebox("Example 5")[][ Factor: #math.equation(block: false, alt: "u squared minus 9 u v minus 12 v squared .")[$u^(2) − 9 u v − 12 v^(2) .$] #solutionbox[ We need #emph[u] in the first term of each binomial and #emph[v] in the second term. The last term of the trinomial is negative, so the factors must have opposite signs. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "u squared minus 9 u v minus 12 v squared")[$u^(2) − 9 u v − 12 v^(2)$]]), [Note that the first terms are #math.equation(block: false, alt: "u ,")[$u ,$] last terms contain #math.equation(block: false, alt: "v .")[$v .$]], [#math.equation(block: false, alt: "open parenthesis u v close parenthesis open parenthesis u v close parenthesis")[$( u #h(1.2em) v ) ( u #h(1.2em) v )$]], [Find the numbers that multiply to −12 and add to −9.], [], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "− 12")[$"−" 12$]], [Sum of factors]), [#math.equation(block: false, alt: "1 , −12")[$#h(0.65em) 1 , −12$] #linebreak() #math.equation(block: false, alt: "−1 , 12")[$−1 , 12$] #linebreak() #math.equation(block: false, alt: "2 , −6")[$#h(0.65em) 2 , −6$] #linebreak() #math.equation(block: false, alt: "−2 , 6")[$−2 , 6$] #linebreak() #math.equation(block: false, alt: "3 , −4")[$#h(0.65em) 3 , −4$] #linebreak() #math.equation(block: false, alt: "−3 , 4")[$−3 , 4$]], [#math.equation(block: false, alt: "1 plus open parenthesis −12 close parenthesis equals −11")[$1 + ( −12 ) = −11$] #linebreak() #math.equation(block: false, alt: "−1 plus 12 equals 11")[$#h(0.65em) −1 + 12 = 11$] #linebreak() #math.equation(block: false, alt: "2 plus open parenthesis −6 close parenthesis equals −4")[$#h(0.5em) 2 + ( −6 ) = −4$] #linebreak() #math.equation(block: false, alt: "−2 plus 6 equals 4")[$#h(1.2em) −2 + 6 = 4$] #linebreak() #math.equation(block: false, alt: "3 plus open parenthesis −4 close parenthesis equals −1")[$#h(0.6em) 3 + ( −4 ) = −1$] #linebreak() #math.equation(block: false, alt: "−3 plus 4 equals 1")[$#h(1.2em) −3 + 4 = 1$]], )) Note there are no factor pairs that give us #math.equation(block: false, alt: "−9")[$−9$] as a sum. The trinomial is prime. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "x squared minus 7 x y minus 10 y squared .")[$x^(2) − 7 x y − 10 y^(2) .$] #solutionbox[ prime ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "p squared plus 15 p q plus 20 q squared .")[$p^(2) + 15 p q + 20 q^(2) .$] #solutionbox[ prime ] ] Let’s summarize the method we just developed to factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c .")[$x^(2) + b x + c .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Strategy for Factoring Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]] When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors. #math.equation(block: true, alt: "x squared plus b x plus c; open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis; When c is positive, m and n have the same sign.; b positive b negative; m , n positive m , n negative; x squared plus 5 x plus 6 x squared minus 6 x plus 8; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis open parenthesis x minus 4 close parenthesis open parenthesis x minus 2 close parenthesis; same signs same signs; When c is negative, m and n have opposite signs.; x squared plus x minus 12 x squared minus 2 x minus 15; open parenthesis x plus 4 close parenthesis open parenthesis x minus 3 close parenthesis open parenthesis x minus 5 close parenthesis open parenthesis x plus 3 close parenthesis; opposite signs opposite signs")[$x^(2) + b x + c \ ( x + m ) ( x + n ) \ "When" #h(0.2em) "c" #h(0.2em) "is positive," #h(0.2em) "m" #h(0.2em) "and" #h(0.2em) "n" #h(0.2em) "have the same sign." \ b #h(0.2em) "positive" #h(16.5em) b #h(0.2em) "negative" \ m , n #h(0.2em) "positive" #h(15em) m , n #h(0.2em) "negative" \ x^(2) + 5 x + 6 #h(16em) x^(2) − 6 x + 8 \ ( x + 2 ) ( x + 3 ) #h(15em) ( x − 4 ) ( x − 2 ) \ "same signs" #h(16em) "same signs" \ "When" #h(0.2em) "c" #h(0.2em) "is negative," #h(0.2em) "m" #h(0.2em) "and" #h(0.2em) "n" #h(0.2em) "have opposite signs." \ x^(2) + x − 12 #h(16em) x^(2) − 2 x − 15 \ ( x + 4 ) ( x − 3 ) #h(15em) ( x − 5 ) ( x + 3 ) \ "opposite signs" #h(15em) "opposite signs"$]Notice that, in the case when #emph[m] and #emph[n] have opposite signs, the sign of the one with the larger absolute value matches the sign of #emph[b]. ] === Factor Trinomials of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] using Trial and Error Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form #math.equation(block: false, alt: "a x squared plus b x plus c .")[$a x^(2) + b x + c .$] Remember to always check for a #strong[GCF] first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works. #examplebox("Example 6")[][ Factor completely: #math.equation(block: false, alt: "4 x cubed plus 16 x squared minus 20 x .")[$4 x^(3) + 16 x^(2) − 20 x .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor?], [#math.equation(block: false, alt: "4 x cubed plus 16 x squared minus 20 x")[$4 x^(3) + 16 x^(2) − 20 x$]]), [Yes, #math.equation(block: false, alt: "GCF equals 4 x .")[$"GCF" = 4 x .$] Factor it.], [#math.equation(block: false, alt: "4 x open parenthesis x squared plus 4 x minus 5 close parenthesis")[$4 x ( x^(2) + 4 x − 5 )$]], [Binomial, trinomial, or more than three terms?], [], [It is a trinomial. So “undo FOIL.”], [#math.equation(block: false, alt: "4 x open parenthesis x close parenthesis open parenthesis x close parenthesis")[$4 x ( x #h(1.8em) ) ( x #h(1.5em) )$]], [Use a table like the one shown to find two numbers that #linebreak() multiply to −5 and add to 4.], [#math.equation(block: false, alt: "4 x open parenthesis x minus 1 close parenthesis open parenthesis x plus 5 close parenthesis")[$4 x ( x − 1 ) ( x + 5 )$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "− 5")[$"−" 5$]], [Sum of factors]), [#math.equation(block: false, alt: "− 1 , 5")[$"−" 1 , 5$] #linebreak() #math.equation(block: false, alt: "1 , −5")[$#h(0.6em) 1 , −5$]], [#math.equation(block: false, alt: "− 1 plus 5 equals 4 to the power *")[$#h(0.6em) "−" 1 + 5 = 4^(*)$] #linebreak() #math.equation(block: false, alt: "1 plus open parenthesis − 5 close parenthesis equals −4")[$1 + ( "−" 5 ) = −4$]], )) #figure(table( columns: 1, align: left, inset: 6pt, [Check: #linebreak() #math.equation(block: false, alt: "4 x open parenthesis x minus 1 close parenthesis open parenthesis x plus 5 close parenthesis; 4 x open parenthesis x squared plus 5 x minus x minus 5 close parenthesis; 4 x open parenthesis x squared plus 4 x minus 5 close parenthesis; 4 x cubed plus 16 x squared minus 20 x ✓")[$#h(4em) 4 x ( x − 1 ) ( x + 5 ) \ 4 x ( x^(2) + 5 x − x − 5 ) \ 4 x ( x^(2) + 4 x − 5 ) \ 4 x^(3) + 16 x^(2) − 20 x ✓$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "5 x cubed plus 15 x squared minus 20 x .")[$5 x^(3) + 15 x^(2) − 20 x .$] #solutionbox[ #math.equation(block: true, alt: "5 x open parenthesis x minus 1 close parenthesis open parenthesis x plus 4 close parenthesis")[$5 x ( x − 1 ) ( x + 4 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "6 y cubed plus 18 y squared minus 60 y .")[$6 y^(3) + 18 y^(2) − 60 y .$] #solutionbox[ #math.equation(block: true, alt: "6 y open parenthesis y minus 2 close parenthesis open parenthesis y plus 5 close parenthesis")[$6 y ( y − 2 ) ( y + 5 )$] ] ] What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method. Let’s factor the trinomial #math.equation(block: false, alt: "3 x squared plus 5 x plus 2 .")[$3 x^(2) + 5 x + 2 .$] From our earlier work, we expect this will factor into two binomials. #math.equation(block: true, alt: "3 x squared plus 5 x plus 2; open parenthesis close parenthesis open parenthesis close parenthesis")[$3 x^(2) + 5 x + 2 \ ( #h(2em) ) ( #h(2em) )$]We know the first terms of the binomial factors will multiply to give us #math.equation(block: false, alt: "3 x squared .")[$3 x^(2) .$] The only factors of #math.equation(block: false, alt: "3 x squared")[$3 x^(2)$] are #math.equation(block: false, alt: "1 x , 3 x .")[$1 x , 3 x .$] We can place them in the binomials. #figure(figph[The polynomial is 3x squared plus 5x plus 2. There are two pairs of parentheses, with the first terms in them being x and 3x.], alt: "The polynomial is 3x squared plus 5x plus 2. There are two pairs of parentheses, with the first terms in them being x and 3x.", caption: none) Check: Does #math.equation(block: false, alt: "1 x times 3 x equals 3 x squared ?")[$1 x · 3 x = 3 x^(2) ?$] We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1. #figure(figph[Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses.], alt: "Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses.", caption: none) Which factors are correct? To decide that, we multiply the inner and outer terms. #figure(figph[Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses. In each case, arrows are shown pairing the first term of the first factor with the last term of the second factor and the first term of the second factor with the last term of the first factor.], alt: "Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses. In each case, arrows are shown pairing the first term of the first factor with the last term of the second factor and the first term of the second factor with the last term of the first factor.", caption: none) Since the middle term of the trinomial is #math.equation(block: false, alt: "5 x ,")[$5 x ,$] the factors in the first case will work. Let’s use FOIL to check. #math.equation(block: true, alt: "open parenthesis x plus 1 close parenthesis open parenthesis 3 x plus 2 close parenthesis; 3 x squared plus 2 x plus 3 x plus 2; 3 x squared plus 5 x plus 2 ✓")[$( x + 1 ) ( 3 x + 2 ) \ 3 x^(2) + 2 x + 3 x + 2 \ 3 x^(2) + 5 x + 2 ✓$]Our result of the factoring is: #math.equation(block: true, alt: "3 x squared plus 5 x plus 2; open parenthesis x plus 1 close parenthesis open parenthesis 3 x plus 2 close parenthesis")[$3 x^(2) + 5 x + 2 \ ( x + 1 ) ( 3 x + 2 )$]#examplebox("Example 7")[How to Factor a Trinomial Using Trial and Error][ Factor completely using trial and error: #math.equation(block: false, alt: "3 y squared plus 22 y plus 7 .")[$3 y^(2) + 22 y + 7 .$] #solutionbox[ #figure(figph[Step 1 is to write the trinomial in descending order. The trinomial 3 y squared plus 22y plus 7 is already in descending order.], alt: "Step 1 is to write the trinomial in descending order. The trinomial 3 y squared plus 22y plus 7 is already in descending order.", caption: none) #figure(figph[Step 2 is to factor the GCF. Here, there is none.], alt: "Step 2 is to factor the GCF. Here, there is none.", caption: none) #figure(figph[Step 3 is Find all the factor pairs of the first term. The only factors here are 1y and 3y. Since there is only one pair, we can put each as the first term in the parentheses.], alt: "Step 3 is Find all the factor pairs of the first term. The only factors here are 1y and 3y. Since there is only one pair, we can put each as the first term in the parentheses.", caption: none) #figure(figph[Step 4 is to find all the factor pairs of the third term. Here, the only pair is 1 and 7.], alt: "Step 4 is to find all the factor pairs of the third term. Here, the only pair is 1 and 7.", caption: none) #figure(figph[Step 5 is to test all the possible combinations of the factors until the correct product is found. For possible factors open parentheses y plus 1 close parentheses open parentheses 37 plus 7 close parentheses, the product is 3 y squared plus 10y plus 7. For the possible factors open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses, the product is 3 y squared plus 22y plus 7, which is the correct product. Hence, the correct factors are open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses.], alt: "Step 5 is to test all the possible combinations of the factors until the correct product is found. For possible factors open parentheses y plus 1 close parentheses open parentheses 37 plus 7 close parentheses, the product is 3 y squared plus 10y plus 7. For the possible factors open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses, the product is 3 y squared plus 22y plus 7, which is the correct product. Hence, the correct factors are open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses.", caption: none) #figure(figph[Step 6 is to check by multiplying.], alt: "Step 6 is to check by multiplying.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "2 a squared plus 5 a plus 3 .")[$2 a^(2) + 5 a + 3 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 1 close parenthesis open parenthesis 2 a plus 3 close parenthesis")[$( a + 1 ) ( 2 a + 3 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "4 b squared plus 5 b plus 1 .")[$4 b^(2) + 5 b + 1 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis b plus 1 close parenthesis open parenthesis 4 b plus 1 close parenthesis")[$( b + 1 ) ( 4 b + 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[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.] + Write the trinomial in descending order of degrees as needed. + Factor any GCF. + Find all the factor pairs of the first term. + Find all the factor pairs of the third term. + Test all the possible combinations of the factors until the correct product is found. + Check by multiplying. ] Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative. #examplebox("Example 8")[][ Factor completely using trial and error: #math.equation(block: false, alt: "6 b squared minus 13 b plus 5 .")[$6 b^(2) − 13 b + 5 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The trinomial is already in descending order.], [#figure(figph[The image displays the quadratic expression 6b^2 - 13b + 5.], alt: "The image displays the quadratic expression 6b^2 - 13b + 5.", caption: none)]), [Find the factors of the first term.], [#figure(figph[A quadratic expression, 6b^2 - 13b + 5, is shown with potential factors for the 6b^2 term listed below it: 1b \* 6b and 2b \* 3b, in red text.], alt: "A quadratic expression, 6b^2 - 13b + 5, is shown with potential factors for the 6b^2 term listed below it: 1b * 6b and 2b * 3b, in red text.", caption: none)], [Find the factors of the last term. Consider the signs. #linebreak() Since the last term, 5, is positive its factors must both be #linebreak() positive or both be negative. The coefficient of the #linebreak() middle term is negative, so we use the negative factors.], [#figure(figph[The image shows the quadratic expression 6b^2 - 13b + 5, with potential factors for the first term (1b\*6b, 2b\*3b) and the constant term (-1, -5) listed below, indicating the process of factoring.], alt: "The image shows the quadratic expression 6b^2 - 13b + 5, with potential factors for the first term (1b*6b, 2b*3b) and the constant term (-1, -5) listed below, indicating the process of factoring.", caption: none)], )) Consider all the combinations of factors. #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "6 b squared minus 13 b plus 5")[$6 b^(2) − 13 b + 5$]], [], [Possible factors], [Product], [#math.equation(block: false, alt: "open parenthesis b minus 1 close parenthesis open parenthesis 6 b minus 5 close parenthesis")[$( b − 1 ) ( 6 b − 5 )$]], [#math.equation(block: false, alt: "6 b squared minus 11 b plus 5")[$6 b^(2) − 11 b + 5$]], [#math.equation(block: false, alt: "open parenthesis b minus 5 close parenthesis open parenthesis 6 b minus 1 close parenthesis")[$( b − 5 ) ( 6 b − 1 )$]], [#math.equation(block: false, alt: "6 b squared minus 31 b plus 5")[$6 b^(2) − 31 b + 5$]], [#math.equation(block: false, alt: "open parenthesis 2 b minus 1 close parenthesis open parenthesis 3 b minus 5 close parenthesis")[$( 2 b − 1 ) ( 3 b − 5 )$]], [#math.equation(block: false, alt: "6 b squared minus 13 b plus 5 to the power *")[$6 b^(2) − 13 b + 5^(*)$]], [#math.equation(block: false, alt: "open parenthesis 2 b minus 5 close parenthesis open parenthesis 3 b minus 1 close parenthesis")[$( 2 b − 5 ) ( 3 b − 1 )$]], [#math.equation(block: false, alt: "6 b squared minus 17 b plus 5")[$6 b^(2) − 17 b + 5$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([The correct factors are those whose product #linebreak() is the original trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 b minus 1 close parenthesis open parenthesis 3 b minus 5 close parenthesis")[$( 2 b − 1 ) ( 3 b − 5 )$]]), [Check by multiplying: #linebreak() #math.equation(block: false, alt: "open parenthesis 2 b minus 1 close parenthesis open parenthesis 3 b minus 5 close parenthesis; 6 b squared minus 10 b minus 3 b plus 5; 6 b squared minus 13 b plus 5 ✓")[$#h(4em) ( 2 b − 1 ) ( 3 b − 5 ) \ 6 b^(2) − 10 b − 3 b + 5 \ 6 b^(2) − 13 b + 5 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "8 x squared minus 14 x plus 3 .")[$8 x^(2) − 14 x + 3 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 x minus 3 close parenthesis open parenthesis 4 x minus 1 close parenthesis")[$( 2 x − 3 ) ( 4 x − 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "10 y squared minus 37 y plus 7 .")[$10 y^(2) − 37 y + 7 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 y minus 7 close parenthesis open parenthesis 5 y minus 1 close parenthesis")[$( 2 y − 7 ) ( 5 y − 1 )$] ] ] When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations. #examplebox("Example 9")[][ Factor completely using trial and error: #math.equation(block: false, alt: "18 x squared minus 37 x y plus 15 y squared .")[$18 x^(2) − 37 x y + 15 y^(2) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The trinomial is already in descending order.], [#figure(figph[The image shows the algebraic expression 18x^2 - 37xy + 15y^2.], alt: "The image shows the algebraic expression 18x^2 - 37xy + 15y^2.", caption: none)]), [Find the factors of the first term.], [#figure(figph[Steps to factor the trinomial 18x^2 - 37xy + 15y^2 are shown, with initial factor pairs for the 18x^2 term listed as 1x\*18x, 2x\*9x, and 3x\*6x.], alt: "Steps to factor the trinomial 18x^2 - 37xy + 15y^2 are shown, with initial factor pairs for the 18x^2 term listed as 1x*18x, 2x*9x, and 3x*6x.", caption: none)], [Find the factors of the last term. Consider the signs. #linebreak() Since 15 is positive and the coefficient of the middle #linebreak() term is negative, we use the negative factors.], [#figure(figph[An image displaying the algebraic expression 18x^2 - 37xy + 15y^2, with potential factors for the first and last terms shown in red below it, indicating steps for factoring the trinomial.], alt: "An image displaying the algebraic expression 18x^2 - 37xy + 15y^2, with potential factors for the first and last terms shown in red below it, indicating steps for factoring the trinomial.", caption: none)], )) Consider all the combinations of factors. #figure(figph[This table shows the possible factors and corresponding products of the trinomial 18 x squared minus 37xy plus 15 y squared. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: open parentheses x minus 1y close parentheses open parentheses 18x minus 15y close parentheses, highlighted. Factor, open parentheses x minus 15y close parentheses open parentheses 18x minus 1y close parentheses; product: 18 x squared minus 271xy plus 15 y squared. Factor open parentheses x minus 3y close parentheses open parentheses 18x minus 5 y close parentheses; product: 18 x squared minus 59xy plus 15 y squared. Factor: open parentheses x minus 5y close parentheses open parentheses 18x minus 3y close parentheses highlighted. Factor: open parentheses 2x minus 1y close parentheses open parentheses 9x minus 15y close parentheses highlighted. Factor: open parentheses 2x minus 15y close parentheses open parentheses 9x minus 1y close parentheses; product 18 x squared minus 137 xy plus 15y squared. Factor: open parentheses 2x minus 3y close parentheses open parentheses 9x minus 5y close parentheses; product: 18 x squared minus 37xy plus 15 y squared, which is the original trinomial. Factor: open parentheses 2x minus 57 close parentheses open parentheses 9x minus 3y close parentheses highlighted. Factor: open parentheses 3x minus 1y close parentheses open parentheses 6x minus 15y close parentheses highlighted. Factor: open parentheses 3x minus 15y close parentheses highlighted open parentheses 6x minus 1y close parentheses. Factor: open parentheses 3x minus 3y close parentheses highlighted open parentheses 6x minus 5y.], alt: "This table shows the possible factors and corresponding products of the trinomial 18 x squared minus 37xy plus 15 y squared. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: open parentheses x minus 1y close parentheses open parentheses 18x minus 15y close parentheses, highlighted. Factor, open parentheses x minus 15y close parentheses open parentheses 18x minus 1y close parentheses; product: 18 x squared minus 271xy plus 15 y squared. Factor open parentheses x minus 3y close parentheses open parentheses 18x minus 5 y close parentheses; product: 18 x squared minus 59xy plus 15 y squared. Factor: open parentheses x minus 5y close parentheses open parentheses 18x minus 3y close parentheses highlighted. Factor: open parentheses 2x minus 1y close parentheses open parentheses 9x minus 15y close parentheses highlighted. Factor: open parentheses 2x minus 15y close parentheses open parentheses 9x minus 1y close parentheses; product 18 x squared minus 137 xy plus 15y squared. Factor: open parentheses 2x minus 3y close parentheses open parentheses 9x minus 5y close parentheses; product: 18 x squared minus 37xy plus 15 y squared, which is the original trinomial. Factor: open parentheses 2x minus 57 close parentheses open parentheses 9x minus 3y close parentheses highlighted. Factor: open parentheses 3x minus 1y close parentheses open parentheses 6x minus 15y close parentheses highlighted. Factor: open parentheses 3x minus 15y close parentheses highlighted open parentheses 6x minus 1y close parentheses. Factor: open parentheses 3x minus 3y close parentheses highlighted open parentheses 6x minus 5y.", caption: none) #figure(table( columns: 2, align: left, inset: 6pt, table.header([The correct factors are those whose product is the original trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 x minus 3 y close parenthesis open parenthesis 9 x minus 5 y close parenthesis")[$( 2 x − 3 y ) ( 9 x − 5 y )$]]), [Check by multiplying: #linebreak() #math.equation(block: false, alt: "open parenthesis 2 x minus 3 y close parenthesis open parenthesis 9 x minus 5 y close parenthesis; 18 x squared minus 10 x y minus 27 x y plus 15 y squared; 18 x squared minus 37 x y plus 15 y squared ✓")[$#h(4em) ( 2 x − 3 y ) ( 9 x − 5 y ) \ 18 x^(2) − 10 x y − 27 x y + 15 y^(2) \ 18 x^(2) − 37 x y + 15 y^(2) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error #math.equation(block: false, alt: "18 x squared minus 3 x y minus 10 y squared .")[$18 x^(2) − 3 x y − 10 y^(2) .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 x plus 2 y close parenthesis open parenthesis 6 x minus 5 y close parenthesis")[$( 3 x + 2 y ) ( 6 x − 5 y )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "30 x squared minus 53 x y minus 21 y squared .")[$30 x^(2) − 53 x y − 21 y^(2) .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 x plus y close parenthesis open parenthesis 10 x minus 21 y close parenthesis")[$( 3 x + y ) ( 10 x − 21 y )$] ] ] Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF. #examplebox("Example 10")[][ Factor completely using trial and error: #math.equation(block: false, alt: "−10 y to the power 4 minus 55 y cubed minus 60 y squared .")[$−10 y^(4) − 55 y^(3) − 60 y^(2) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression: negative ten y to the fourth power, minus fifty-five y cubed, minus sixty y squared.], alt: "A mathematical expression: negative ten y to the fourth power, minus fifty-five y cubed, minus sixty y squared.", caption: none)]), [Notice the greatest common factor, so factor it first.], [#figure(figph[negative five y squared times open parenthesis two y squared plus eleven y plus twelve close parenthesis], alt: "negative five y squared times open parenthesis two y squared plus eleven y plus twelve close parenthesis", caption: none)], [Factor the trinomial.], [#figure(figph[A mathematical expression -5y²(2y² + 11y + 12) illustrating steps to factor the quadratic trinomial. Red text shows factors for 2y² (y\*2y) and factors for 12 (1\*12, 2\*6, 3\*4).], alt: "A mathematical expression -5y²(2y² + 11y + 12) illustrating steps to factor the quadratic trinomial. Red text shows factors for 2y² (y*2y) and factors for 12 (1*12, 2*6, 3*4).", caption: none)], )) Consider all the combinations. #figure(figph[This table shows the possible factors and product of the trinomial 2 y squared plus 11y plus 12. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: y plus 1, 2y plus 12 highlighted. Factor: y plus 12, 2y plus 1; product: 2 y squared plus 25y plus 12. Factor: y plus 2, 2y plus 6 highlighted. Factor: y plus 6, 2y plus 2 highlighted. Factor: y plus 3, 2y plus 4 highlighted. Factor: y plus 4, 2y plus 3; product: 2 y squared plus 11y plus 12. This is the original trinomial.], alt: "This table shows the possible factors and product of the trinomial 2 y squared plus 11y plus 12. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: y plus 1, 2y plus 12 highlighted. Factor: y plus 12, 2y plus 1; product: 2 y squared plus 25y plus 12. Factor: y plus 2, 2y plus 6 highlighted. Factor: y plus 6, 2y plus 2 highlighted. Factor: y plus 3, 2y plus 4 highlighted. Factor: y plus 4, 2y plus 3; product: 2 y squared plus 11y plus 12. This is the original trinomial.", caption: none) #figure(table( columns: 2, align: left, inset: 6pt, table.header([The correct factors are those whose product #linebreak() is the original trinomial. Remember to include #linebreak() the factor #math.equation(block: false, alt: "− 5 y squared .")[$"−" 5 y^(2) .$]], [#math.equation(block: false, alt: "− 5 y squared open parenthesis y plus 4 close parenthesis open parenthesis 2 y plus 3 close parenthesis")[$"−" 5 y^(2) ( y + 4 ) ( 2 y + 3 )$]]), [Check by multiplying: #linebreak() #math.equation(block: false, alt: "− 5 y squared open parenthesis y plus 4 close parenthesis open parenthesis 2 y plus 3 close parenthesis; − 5 y squared open parenthesis 2 y squared plus 8 y plus 3 y plus 12 close parenthesis; − 10 y to the power 4 minus 55 y cubed minus 60 y squared ✓")[$#h(4em) "−" 5 y^(2) ( y + 4 ) ( 2 y + 3 ) \ "−" 5 y^(2) ( 2 y^(2) + 8 y + 3 y + 12 ) \ "−" 10 y^(4) − 55 y^(3) − 60 y^(2) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "15 n cubed minus 85 n squared plus 100 n .")[$15 n^(3) − 85 n^(2) + 100 n .$] #solutionbox[ #math.equation(block: true, alt: "5 n open parenthesis n minus 4 close parenthesis open parenthesis 3 n minus 5 close parenthesis")[$5 n ( n − 4 ) ( 3 n − 5 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely using trial and error: #math.equation(block: false, alt: "56 q cubed plus 320 q squared minus 96 q .")[$56 q^(3) + 320 q^(2) − 96 q .$] #solutionbox[ #math.equation(block: true, alt: "8 q open parenthesis q plus 6 close parenthesis open parenthesis 7 q minus 2 close parenthesis")[$8 q ( q + 6 ) ( 7 q − 2 )$] ] ] === 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 Another way to factor trinomials of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works! #examplebox("Example 11")[How to Factor Trinomials using the “ac” Method][ Factor using the #emph[‘ac’] method: #math.equation(block: false, alt: "6 x squared plus 7 x plus 2 .")[$6 x^(2) + 7 x + 2 .$] #solutionbox[ #figure(figph[Step 1 is to factor the GCF. There is none in 6 x squared plus 7x plus 2.], alt: "Step 1 is to factor the GCF. There is none in 6 x squared plus 7x plus 2.", caption: none) #figure(figph[Step 2 is to find the product of a and c. The product of 6 and 2 is 12.], alt: "Step 2 is to find the product of a and c. The product of 6 and 2 is 12.", caption: none) #figure(figph[Step 3 is to find 2 numbers m and n such that mn is ac and m plus n is b. So we need to numbers that multiply to 12 and add to 7. Both factors must be positive. 3 times 4 is 12 and 3 plus 4 is 7.], alt: "Step 3 is to find 2 numbers m and n such that mn is ac and m plus n is b. So we need to numbers that multiply to 12 and add to 7. Both factors must be positive. 3 times 4 is 12 and 3 plus 4 is 7.", caption: none) #figure(figph[Step 4 is to split the middle term using m and n. So we rewrite 7 x as 3x plus 4x. It would give the same result if we used 4x plus 3x. Rewriting, we get 6 x squared plus 3x plus 4x plus 2. Notice that this is the same as the original polynomial. We just split the middle term to get a more useful form], alt: "Step 4 is to split the middle term using m and n. So we rewrite 7 x as 3x plus 4x. It would give the same result if we used 4x plus 3x. Rewriting, we get 6 x squared plus 3x plus 4x plus 2. Notice that this is the same as the original polynomial. We just split the middle term to get a more useful form", caption: none) #figure(figph[Step 5 is to factor by grouping. So, we get, 3x open parentheses 2x plus 1 close parentheses plus 2 open parentheses 2x plus 1 close parentheses. This is equal to 2x plus 1, 3x plus 2.], alt: "Step 5 is to factor by grouping. So, we get, 3x open parentheses 2x plus 1 close parentheses plus 2 open parentheses 2x plus 1 close parentheses. This is equal to 2x plus 1, 3x plus 2.", caption: none) #figure(figph[Step 6 is to check by multiplying the factors.], alt: "Step 6 is to check by multiplying the factors.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor using the ‘ac’ method: #math.equation(block: false, alt: "6 x squared plus 13 x plus 2 .")[$6 x^(2) + 13 x + 2 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis 6 x plus 1 close parenthesis")[$( x + 2 ) ( 6 x + 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor using the ‘ac’ method: #math.equation(block: false, alt: "4 y squared plus 8 y plus 3 .")[$4 y^(2) + 8 y + 3 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 y plus 1 close parenthesis open parenthesis 2 y plus 3 close parenthesis")[$( 2 y + 1 ) ( 2 y + 3 )$] ] ] The “ac” method is summarized here. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[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.] + Factor any GCF. + Find the product #emph[ac]. + Find two numbers #emph[m] and #emph[n] that: #linebreak() #math.equation(block: false, alt: "Multiply to a c, m times n equals a times c; Add to b, m plus n equals b; a x squared plus b x plus c")[$"Multiply to" #h(0.2em) a c & & & & & m · n = a · c \ "Add to" #h(0.2em) b & & & & & m + n = b \ & & & & & a x^(2) + b x + c$] + Split the middle term using #emph[m] and #emph[n]. #math.equation(block: false, alt: "a x squared plus m x plus n x plus c")[$#h(5em) a x^(2) + m x + n x + c$] + Factor by grouping. + Check by multiplying the factors. ] Don’t forget to look for a common factor! #examplebox("Example 12")[][ Factor using the #emph[‘ac’] method: #math.equation(block: false, alt: "10 y squared minus 55 y plus 70 .")[$10 y^(2) − 55 y + 70 .$] #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([Is there a greatest common factor?], [], []), [Yes. The GCF is 5.], [], [#figure(figph[The image shows the quadratic expression 10y^2 - 55y + 70.], alt: "The image shows the quadratic expression 10y^2 - 55y + 70.", caption: none)], [Factor it.], [], [#figure(figph[The image shows the mathematical expression 5(2y^2 - 11y + 14).], alt: "The image shows the mathematical expression 5(2y^2 - 11y + 14).", caption: none)], [The trinomial inside the parentheses has a #linebreak() leading coefficient that is not 1.], [], [#figure(figph[The image displays two algebraic expressions: 'ax^2 + bx + c' in red, representing a general quadratic equation, and '5(2y^2 - 11y + 14)' in black, a factored quadratic expression.], alt: "The image displays two algebraic expressions: 'ax^2 + bx + c' in red, representing a general quadratic equation, and '5(2y^2 - 11y + 14)' in black, a factored quadratic expression.", caption: none)], [Find the product #math.equation(block: false, alt: "a c .")[$a c .$]], [#math.equation(block: false, alt: "a c equals 28")[$#h(3.3em) a c = 28$]], [], [Find two numbers that multiply to #math.equation(block: false, alt: "a c")[$a c$]], [#math.equation(block: false, alt: "open parenthesis −4 close parenthesis open parenthesis −7 close parenthesis equals 28")[$#h(0.5em) ( −4 ) ( −7 ) = 28$]], [], [and add to #emph[b].], [#math.equation(block: false, alt: "−4 plus open parenthesis −7 close parenthesis equals −11")[$−4 + ( −7 ) = −11$]], [], [Split the middle term.], [], [#figure(figph[Mathematical expression 5(2y^2 - 11y + 14) with arrows highlighting the -11y term, likely for breaking it down during factorization.], alt: "Mathematical expression 5(2y^2 - 11y + 14) with arrows highlighting the -11y term, likely for breaking it down during factorization.", caption: none)], [], [], [#figure(figph[The algebraic expression 5(2y^2 - 7y - 4y + 14) is shown, with blue brackets indicating the grouping of terms (2y^2 - 7y) and (-4y + 14), typically a step in factorization.], alt: "The algebraic expression 5(2y^2 - 7y - 4y + 14) is shown, with blue brackets indicating the grouping of terms (2y^2 - 7y) and (-4y + 14), typically a step in factorization.", caption: none)], [Factor the trinomial by grouping.], [], [#figure(figph[A mathematical expression featuring the difference of two products, 5(y(2y - 7)) - 2(2y - 7)), suitable for algebraic manipulation or solving.], alt: "A mathematical expression featuring the difference of two products, 5(y(2y - 7)) - 2(2y - 7)), suitable for algebraic manipulation or solving.", caption: none)], [], [], [#figure(figph[A mathematical expression displays 5 multiplied by the quantity (y minus 2), which is then multiplied by the quantity (2y minus 7), representing a factored quadratic expression.], alt: "A mathematical expression displays 5 multiplied by the quantity (y minus 2), which is then multiplied by the quantity (2y minus 7), representing a factored quadratic expression.", caption: none)], [Check by multiplying all three factors. #linebreak() #linebreak() #math.equation(block: false, alt: "5 open parenthesis y minus 2 close parenthesis open parenthesis 2 y minus 7 close parenthesis; 5 open parenthesis 2 y squared minus 7 y minus 4 y plus 14 close parenthesis; 5 open parenthesis 2 y squared minus 11 y plus 14 close parenthesis; 10 y squared minus 55 y plus 70 ✓")[$#h(4em) 5 ( y − 2 ) ( 2 y − 7 ) \ 5 ( 2 y^(2) − 7 y − 4 y + 14 ) \ 5 ( 2 y^(2) − 11 y + 14 ) \ 10 y^(2) − 55 y + 70 ✓$]], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor using the ‘ac’ method: #math.equation(block: false, alt: "16 x squared minus 32 x plus 12 .")[$16 x^(2) − 32 x + 12 .$] #solutionbox[ #math.equation(block: true, alt: "4 open parenthesis 2 x minus 3 close parenthesis open parenthesis 2 x minus 1 close parenthesis")[$4 ( 2 x − 3 ) ( 2 x − 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor using the ‘ac’ method: #math.equation(block: false, alt: "18 w squared minus 39 w plus 18 .")[$18 w^(2) − 39 w + 18 .$] #solutionbox[ #math.equation(block: true, alt: "3 open parenthesis 3 w minus 2 close parenthesis open parenthesis 2 w minus 3 close parenthesis")[$3 ( 3 w − 2 ) ( 2 w − 3 )$] ] ] === Factor Using Substitution Sometimes a trinomial does not appear to be in the #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] form. However, we can often make a thoughtful substitution that will allow us to make it fit the #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] form. This is called #strong[factoring by substitution]. It is standard to use #emph[u] for the substitution. In the #math.equation(block: false, alt: "a x squared plus b x plus c ,")[$a x^(2) + b x + c ,$] the middle term has a variable, #emph[x], and its square, #math.equation(block: false, alt: "x squared ,")[$x^(2) ,$] is the variable part of the first term. Look for this relationship as you try to find a substitution. #examplebox("Example 13")[][ Factor by substitution: #math.equation(block: false, alt: "x to the power 4 minus 4 x squared minus 5.")[$x^(4) − 4 x^(2) − 5.$] #solutionbox[ The variable part of the middle term is #math.equation(block: false, alt: "x squared")[$x^(2)$] and its square, #math.equation(block: false, alt: "x to the power 4 ,")[$x^(4) ,$] is the variable part of the first term. (We know #math.equation(block: false, alt: "open parenthesis x squared close parenthesis squared equals x to the power 4 close parenthesis .")[$attach(\( x^(2) \), t: 2) = x^(4) \) .$] If we let #math.equation(block: false, alt: "u equals x squared ,")[$u = x^(2) ,$] we can put our trinomial in the #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] form we need to factor it. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression reads 'x to the power of 4 minus 4x squared minus 5'.], alt: "A mathematical expression reads 'x to the power of 4 minus 4x squared minus 5'.", caption: none)]), [Rewrite the trinomial to prepare for the substitution.], [#figure(figph[A mathematical expression showing a quadratic in terms of x squared: (x^2)^2 - 4(x^2) - 5. The 'x^2' terms are highlighted in red.], alt: "A mathematical expression showing a quadratic in terms of x squared: (x^2)^2 - 4(x^2) - 5. The 'x^2' terms are highlighted in red.", caption: none)], [Let #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] and substitute.], [#figure(figph[The image shows the mathematical expression u squared minus 4u minus 5, with the 'u' characters in a reddish hue and the numbers and minus signs in grey/black.], alt: "The image shows the mathematical expression u squared minus 4u minus 5, with the 'u' characters in a reddish hue and the numbers and minus signs in grey/black.", caption: none)], [Factor the trinomial.], [#figure(figph[The mathematical expression (u+1)(u-5) is shown, representing the product of two binomials.], alt: "The mathematical expression (u+1)(u-5) is shown, representing the product of two binomials.", caption: none)], [Replace #emph[u] with #math.equation(block: false, alt: "x squared .")[$x^(2) .$]], [#figure(figph[The image displays the mathematical expression (x^2 + 1)(x^2 - 5), which represents the product of two binomials involving x squared.], alt: "The image displays the mathematical expression (x^2 + 1)(x^2 - 5), which represents the product of two binomials involving x squared.", caption: none)], [Check: #linebreak() #linebreak() #math.equation(block: false, alt: "open parenthesis x squared plus 1 close parenthesis open parenthesis x squared minus 5 close parenthesis; x to the power 4 minus 5 x squared plus x squared minus 5; x to the power 4 minus 4 x squared minus 5 ✓")[$#h(3em) ( x^(2) + 1 ) ( x^(2) − 5 ) \ x^(4) − 5 x^(2) + x^(2) − 5 \ x^(4) − 4 x^(2) − 5 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor by substitution: #math.equation(block: false, alt: "h to the power 4 plus 4 h squared minus 12 .")[$h^(4) + 4 h^(2) − 12 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis h squared minus 2 close parenthesis open parenthesis h squared plus 6 close parenthesis")[$( h^(2) − 2 ) ( h^(2) + 6 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor by substitution: #math.equation(block: false, alt: "y to the power 4 minus y squared minus 20 .")[$y^(4) − y^(2) − 20 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis y squared plus 4 close parenthesis open parenthesis y squared minus 5 close parenthesis")[$( y^(2) + 4 ) ( y^(2) − 5 )$] ] ] Sometimes the expression to be substituted is not a monomial. #examplebox("Example 14")[][ Factor by substitution: #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12")[$attach(( x − 2 ), t: 2) + 7 ( x − 2 ) + 12$] #solutionbox[ The binomial in the middle term, #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis")[$( x − 2 )$] is squared in the first term. If we let #math.equation(block: false, alt: "u equals x minus 2")[$u = x − 2$] and substitute, our trinomial will be in #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] form. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression is displayed, which reads as quantity x minus 2 squared, plus 7 times quantity x minus 2, plus 12. This is a quadratic expression in terms of (x-2).], alt: "A mathematical expression is displayed, which reads as quantity x minus 2 squared, plus 7 times quantity x minus 2, plus 12. This is a quadratic expression in terms of (x-2).", caption: none)]), [Rewrite the trinomial to prepare for the substitution.], [#figure(figph[A mathematical expression reads as 'open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12'.], alt: "A mathematical expression reads as 'open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12'.", caption: none)], [Let #math.equation(block: false, alt: "u equals x minus 2")[$u = x − 2$] and substitute.], [#figure(figph[A quadratic expression is displayed against a white background, reading 'u^2 + 7u + 12' with the 'u' characters in red.], alt: "A quadratic expression is displayed against a white background, reading 'u^2 + 7u + 12' with the 'u' characters in red.", caption: none)], [Factor the trinomial.], [#figure(figph[An algebraic expression showing the product of two binomials, (u+3) and (u+4).], alt: "An algebraic expression showing the product of two binomials, (u+3) and (u+4).", caption: none)], [Replace #emph[u] with #math.equation(block: false, alt: "x minus 2 .")[$x − 2 .$]], [#figure(figph[A mathematical expression featuring (x-2) plus 3 times (x-2) plus 4. The 'x-2' terms are highlighted in red, indicating a common factor within the polynomial expression.], alt: "A mathematical expression featuring (x-2) plus 3 times (x-2) plus 4. The 'x-2' terms are highlighted in red, indicating a common factor within the polynomial expression.", caption: none)], [Simplify inside the parentheses.], [#figure(figph[The mathematical expression (x+1)(x+2) is displayed in black text on a white background.], alt: "The mathematical expression (x+1)(x+2) is displayed in black text on a white background.", caption: none)], )) This could also be factored by first multiplying out the #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared")[$attach(( x − 2 ), t: 2)$] and the #math.equation(block: false, alt: "7 open parenthesis x minus 2 close parenthesis")[$7 ( x − 2 )$] and then combining like terms and then factoring. Most students prefer the substitution method. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor by substitution: #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis squared plus 6 open parenthesis x minus 5 close parenthesis plus 8 .")[$attach(( x − 5 ), t: 2) + 6 ( x − 5 ) + 8 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x minus 1 close parenthesis")[$( x − 3 ) ( x − 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor by substitution: #math.equation(block: false, alt: "open parenthesis y minus 4 close parenthesis squared plus 8 open parenthesis y minus 4 close parenthesis plus 15 .")[$attach(( y − 4 ), t: 2) + 8 ( y − 4 ) + 15 .$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis y minus 1 close parenthesis open parenthesis y plus 1 close parenthesis")[$( y − 1 ) ( y + 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access this online resource for additional instruction and practice with factoring. - #link("https://openstax.org/l/37ACmethod")[Factor a trinomial using the AC method] ] === Key Concepts - #strong[How to factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c .")[$x^(2) + b x + c .$]] + Write the factors as two binomials with first terms #emph[x]. #math.equation(block: false, alt: "x squared plus b x plus c; open parenthesis x close parenthesis open parenthesis x close parenthesis")[$#h(4em) x^(2) + b x + c \ ( x #h(1.2em) ) ( x #h(1.2em) )$] + Find two numbers #emph[m] and #emph[n] that #linebreak() #math.equation(block: false, alt: "multiply to, c , m times n equals c; add to, b , m plus n equals b")[$"multiply to" & & c , m · n = c \ "add to" & & b , m + n = b$] + Use #emph[m] and #emph[n] as the last terms of the factors. #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$#h(8em) ( x + m ) ( x + n )$] #linebreak() + Check by multiplying the factors. - #strong[Strategy for Factoring Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]:] When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors. #linebreak() #math.equation(block: false, alt: "x squared plus b x plus c; open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis; When c is positive, m and n have the same sign.; b positive b negative; m , n positive m , n negative; x squared plus 5 x plus 6 x squared minus 6 x plus 8; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis open parenthesis x minus 4 close parenthesis open parenthesis x minus 2 close parenthesis; same signs same signs; When c is negative, m and n have opposite signs.; x squared plus x minus 12 x squared minus 2 x minus 15; open parenthesis x plus 4 close parenthesis open parenthesis x minus 3 close parenthesis open parenthesis x minus 5 close parenthesis open parenthesis x plus 3 close parenthesis; opposite signs opposite signs")[$x^(2) + b x + c \ ( x + m ) ( x + n ) \ "When" #h(0.2em) "c" #h(0.2em) "is positive," #h(0.2em) "m" #h(0.2em) "and" #h(0.2em) "n" #h(0.2em) "have the same sign." \ b #h(0.2em) "positive" #h(16.5em) b #h(0.2em) "negative" \ m , n #h(0.2em) "positive" #h(15em) m , n #h(0.2em) "negative" \ x^(2) + 5 x + 6 #h(16em) x^(2) − 6 x + 8 \ ( x + 2 ) ( x + 3 ) #h(15em) ( x − 4 ) ( x − 2 ) \ "same signs" #h(16em) "same signs" \ "When" #h(0.2em) "c" #h(0.2em) "is negative," #h(0.2em) "m" #h(0.2em) "and" #h(0.2em) "n" #h(0.2em) "have opposite signs." \ x^(2) + x − 12 #h(15em) x^(2) − 2 x − 15 \ ( x + 4 ) ( x − 3 ) #h(15em) ( x − 5 ) ( x + 3 ) \ "opposite signs" #h(15em) "opposite signs"$] #linebreak() Notice that, in the case when #emph[m] and #emph[n] have opposite signs, the sign of the one with the larger absolute value matches the sign of #emph[b]. - #strong[How to 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.] + Write the trinomial in descending order of degrees as needed. + Factor any GCF. + Find all the factor pairs of the first term. + Find all the factor pairs of the third term. + Test all the possible combinations of the factors until the correct product is found. + Check by multiplying. - #strong[How to 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.] + Factor any GCF. + Find the product #emph[ac]. + Find two numbers #emph[m] and #emph[n] that: #linebreak() #math.equation(block: false, alt: "Multiply to a c ., m times n equals a times c; Add to b ., m plus n equals b; a x squared plus b x plus c")[$"Multiply to" #h(0.2em) a c . & & & & & m · n = a · c \ "Add to" #h(0.2em) b . & & & & & m + n = b \ & & & & & a x^(2) + b x + c$] + Split the middle term using #emph[m] and #emph[n]. #math.equation(block: false, alt: "a x squared plus m x plus n x plus c")[$#h(4em) a x^(2) + m x + n x + c$] + Factor by grouping. + Check by multiplying the factors. ==== Practice Makes Perfect #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: "p squared plus 11 p plus 30")[$p^(2) + 11 p + 30$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis p plus 5 close parenthesis open parenthesis p plus 6 close parenthesis")[$( p + 5 ) ( p + 6 )$] ] #math.equation(block: true, alt: "w squared plus 10 w plus 21")[$w^(2) + 10 w + 21$] #math.equation(block: true, alt: "n squared plus 19 n plus 48")[$n^(2) + 19 n + 48$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis n plus 3 close parenthesis open parenthesis n plus 16 close parenthesis")[$( n + 3 ) ( n + 16 )$] ] #math.equation(block: true, alt: "b squared plus 14 b plus 48")[$b^(2) + 14 b + 48$] #math.equation(block: true, alt: "a squared plus 25 a plus 100")[$a^(2) + 25 a + 100$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 5 close parenthesis open parenthesis a plus 20 close parenthesis")[$( a + 5 ) ( a + 20 )$] ] #math.equation(block: true, alt: "u squared plus 101 u plus 100")[$u^(2) + 101 u + 100$] #math.equation(block: true, alt: "x squared minus 8 x plus 12")[$x^(2) − 8 x + 12$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x minus 6 close parenthesis")[$( x − 2 ) ( x − 6 )$] ] #math.equation(block: true, alt: "q squared minus 13 q plus 36")[$q^(2) − 13 q + 36$] #math.equation(block: true, alt: "y squared minus 18 y plus 45")[$y^(2) − 18 y + 45$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis y minus 3 close parenthesis open parenthesis y minus 15 close parenthesis")[$( y − 3 ) ( y − 15 )$] ] #math.equation(block: true, alt: "m squared minus 13 m plus 30")[$m^(2) − 13 m + 30$] #math.equation(block: true, alt: "x squared minus 8 x plus 7")[$x^(2) − 8 x + 7$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis open parenthesis x minus 7 close parenthesis")[$( x − 1 ) ( x − 7 )$] ] #math.equation(block: true, alt: "y squared minus 5 y plus 6")[$y^(2) − 5 y + 6$] #math.equation(block: true, alt: "5 p minus 6 plus p squared")[$5 p − 6 + p^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis p minus 1 close parenthesis open parenthesis p plus 6 close parenthesis")[$( p − 1 ) ( p + 6 )$] ] #math.equation(block: true, alt: "6 n minus 7 plus n squared")[$6 n − 7 + n^(2)$] #math.equation(block: true, alt: "8 minus 6 x plus x squared")[$8 − 6 x + x^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 4 close parenthesis open parenthesis x minus 2 close parenthesis")[$( x − 4 ) ( x − 2 )$] ] #math.equation(block: true, alt: "7 x plus x squared plus 6")[$7 x + x^(2) + 6$] #math.equation(block: true, alt: "x squared minus 12 minus 11 x")[$x^(2) − 12 − 11 x$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 12 close parenthesis open parenthesis x plus 1 close parenthesis")[$( x − 12 ) ( x + 1 )$] ] #math.equation(block: true, alt: "−11 minus 10 x plus x squared")[$−11 − 10 x + x^(2)$] In the following exercises, factor each trinomial of the form #math.equation(block: false, alt: "x squared plus b x y plus c y squared .")[$x^(2) + b x y + c y^(2) .$] If the trinomial cannot be factored, answer “Prime.” #math.equation(block: true, alt: "x squared minus 2 x y minus 80 y squared")[$x^(2) − 2 x y − 80 y^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 8 y close parenthesis open parenthesis x minus 10 y close parenthesis")[$( x + 8 y ) ( x − 10 y )$] ] #math.equation(block: true, alt: "p squared minus 8 p q minus 65 q squared")[$p^(2) − 8 p q − 65 q^(2)$] #math.equation(block: true, alt: "m squared minus 64 m n minus 65 n squared")[$m^(2) − 64 m n − 65 n^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis m plus n close parenthesis open parenthesis m minus 65 n close parenthesis")[$( m + n ) ( m − 65 n )$] ] #math.equation(block: true, alt: "p squared minus 2 p q minus 35 q squared")[$p^(2) − 2 p q − 35 q^(2)$] #math.equation(block: true, alt: "a squared plus 5 a b minus 24 b squared")[$a^(2) + 5 a b − 24 b^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus 8 b close parenthesis open parenthesis a minus 3 b close parenthesis")[$( a + 8 b ) ( a − 3 b )$] ] #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: "x squared minus 3 x y minus 14 y squared")[$x^(2) − 3 x y − 14 y^(2)$] #solutionbox[ Prime ] #math.equation(block: true, alt: "u squared minus 8 u v minus 24 v squared")[$u^(2) − 8 u v − 24 v^(2)$] #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 ] #math.equation(block: true, alt: "c squared minus 7 c d plus 18 d squared")[$c^(2) − 7 c d + 18 d^(2)$] #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 Trial and Error] In the following exercises, factor completely using trial and error. #math.equation(block: true, alt: "p cubed minus 8 p squared minus 20 p")[$p^(3) − 8 p^(2) − 20 p$] #solutionbox[ #math.equation(block: true, alt: "p open parenthesis p minus 10 close parenthesis open parenthesis p plus 2 close parenthesis")[$p ( p − 10 ) ( p + 2 )$] ] #math.equation(block: true, alt: "q cubed minus 5 q squared minus 24 q")[$q^(3) − 5 q^(2) − 24 q$] #math.equation(block: true, alt: "3 m cubed minus 21 m squared plus 30 m")[$3 m^(3) − 21 m^(2) + 30 m$] #solutionbox[ #math.equation(block: true, alt: "3 m open parenthesis m minus 5 close parenthesis open parenthesis m minus 2 close parenthesis")[$3 m ( m − 5 ) ( m − 2 )$] ] #math.equation(block: true, alt: "11 n cubed minus 55 n squared plus 44 n")[$11 n^(3) − 55 n^(2) + 44 n$] #math.equation(block: true, alt: "5 x to the power 4 plus 10 x cubed minus 75 x squared")[$5 x^(4) + 10 x^(3) − 75 x^(2)$] #solutionbox[ #math.equation(block: true, alt: "5 x squared open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis")[$5 x^(2) ( x − 3 ) ( x + 5 )$] ] #math.equation(block: true, alt: "6 y to the power 4 plus 12 y cubed minus 48 y squared")[$6 y^(4) + 12 y^(3) − 48 y^(2)$] #math.equation(block: true, alt: "2 t squared plus 7 t plus 5")[$2 t^(2) + 7 t + 5$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 t plus 5 close parenthesis open parenthesis t plus 1 close parenthesis")[$( 2 t + 5 ) ( t + 1 )$] ] #math.equation(block: true, alt: "5 y squared plus 16 y plus 11")[$5 y^(2) + 16 y + 11$] #math.equation(block: true, alt: "11 x squared plus 34 x plus 3")[$11 x^(2) + 34 x + 3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 11 x plus 1 close parenthesis open parenthesis x plus 3 close parenthesis")[$( 11 x + 1 ) ( x + 3 )$] ] #math.equation(block: true, alt: "7 b squared plus 50 b plus 7")[$7 b^(2) + 50 b + 7$] #math.equation(block: true, alt: "4 w squared minus 5 w plus 1")[$4 w^(2) − 5 w + 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 w minus 1 close parenthesis open parenthesis w minus 1 close parenthesis")[$( 4 w − 1 ) ( w − 1 )$] ] #math.equation(block: true, alt: "5 x squared minus 17 x plus 6")[$5 x^(2) − 17 x + 6$] #math.equation(block: true, alt: "4 q squared minus 7 q minus 2")[$4 q^(2) − 7 q − 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 q plus 1 close parenthesis open parenthesis q minus 2 close parenthesis")[$( 4 q + 1 ) ( q − 2 )$] ] #math.equation(block: true, alt: "10 y squared minus 53 y minus 11")[$10 y^(2) − 53 y − 11$] #math.equation(block: true, alt: "6 p squared minus 19 p q plus 10 q squared")[$6 p^(2) − 19 p q + 10 q^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 p minus 5 q close parenthesis open parenthesis 3 p minus 2 q close parenthesis")[$( 2 p − 5 q ) ( 3 p − 2 q )$] ] #math.equation(block: true, alt: "21 m squared minus 29 m n plus 10 n squared")[$21 m^(2) − 29 m n + 10 n^(2)$] #math.equation(block: true, alt: "4 a squared plus 17 a b minus 15 b squared")[$4 a^(2) + 17 a b − 15 b^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 a minus 3 b close parenthesis open parenthesis a plus 5 b close parenthesis")[$( 4 a − 3 b ) ( a + 5 b )$] ] #math.equation(block: true, alt: "6 u squared plus 5 u v minus 14 v squared")[$6 u^(2) + 5 u v − 14 v^(2)$] #math.equation(block: true, alt: "−16 x squared minus 32 x minus 16")[$−16 x^(2) − 32 x − 16$] #solutionbox[ #math.equation(block: true, alt: "−16 open parenthesis x plus 1 close parenthesis open parenthesis x plus 1 close parenthesis")[$−16 ( x + 1 ) ( x + 1 )$] ] #math.equation(block: true, alt: "−81 a squared plus 153 a plus 18")[$−81 a^(2) + 153 a + 18$] #math.equation(block: true, alt: "−30 q cubed minus 140 q squared minus 80 q")[$−30 q^(3) − 140 q^(2) − 80 q$] #solutionbox[ #math.equation(block: true, alt: "−10 q open parenthesis 3 q plus 2 close parenthesis open parenthesis q plus 4 close parenthesis")[$−10 q ( 3 q + 2 ) ( q + 4 )$] ] #math.equation(block: true, alt: "−5 y cubed minus 30 y squared plus 35 y")[$−5 y^(3) − 30 y^(2) + 35 y$] #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 using the ‘ac’ method. #math.equation(block: true, alt: "5 n squared plus 21 n plus 4")[$5 n^(2) + 21 n + 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 n plus 1 close parenthesis open parenthesis n plus 4 close parenthesis")[$( 5 n + 1 ) ( n + 4 )$] ] #math.equation(block: true, alt: "8 w squared plus 25 w plus 3")[$8 w^(2) + 25 w + 3$] #math.equation(block: true, alt: "4 k squared minus 16 k plus 15")[$4 k^(2) − 16 k + 15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 k minus 3 close parenthesis open parenthesis 2 k minus 5 close parenthesis")[$( 2 k − 3 ) ( 2 k − 5 )$] ] #math.equation(block: true, alt: "5 s squared minus 9 s plus 4")[$5 s^(2) − 9 s + 4$] #math.equation(block: true, alt: "6 y squared plus y minus 15")[$6 y^(2) + y − 15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 y plus 5 close parenthesis open parenthesis 2 y minus 3 close parenthesis")[$( 3 y + 5 ) ( 2 y − 3 )$] ] #math.equation(block: true, alt: "6 p squared plus p minus 22")[$6 p^(2) + p − 22$] #math.equation(block: true, alt: "2 n squared minus 27 n minus 45")[$2 n^(2) − 27 n − 45$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 n plus 3 close parenthesis open parenthesis n minus 15 close parenthesis")[$( 2 n + 3 ) ( n − 15 )$] ] #math.equation(block: true, alt: "12 z squared minus 41 z minus 11")[$12 z^(2) − 41 z − 11$] #math.equation(block: true, alt: "60 y squared plus 290 y minus 50")[$60 y^(2) + 290 y − 50$] #solutionbox[ #math.equation(block: true, alt: "10 open parenthesis 6 y minus 1 close parenthesis open parenthesis y plus 5 close parenthesis")[$10 ( 6 y − 1 ) ( y + 5 )$] ] #math.equation(block: true, alt: "6 u squared minus 46 u minus 16")[$6 u^(2) − 46 u − 16$] #math.equation(block: true, alt: "48 z cubed minus 102 z squared minus 45 z")[$48 z^(3) − 102 z^(2) − 45 z$] #solutionbox[ #math.equation(block: true, alt: "3 z open parenthesis 8 z plus 3 close parenthesis open parenthesis 2 z minus 5 close parenthesis")[$3 z ( 8 z + 3 ) ( 2 z − 5 )$] ] #math.equation(block: true, alt: "90 n cubed plus 42 n squared minus 216 n")[$90 n^(3) + 42 n^(2) − 216 n$] #math.equation(block: true, alt: "16 s squared plus 40 s plus 24")[$16 s^(2) + 40 s + 24$] #solutionbox[ #math.equation(block: true, alt: "8 open parenthesis 2 s plus 3 close parenthesis open parenthesis s plus 1 close parenthesis")[$8 ( 2 s + 3 ) ( s + 1 )$] ] #math.equation(block: true, alt: "24 p squared plus 160 p plus 96")[$24 p^(2) + 160 p + 96$] #math.equation(block: true, alt: "48 y squared plus 12 y minus 36")[$48 y^(2) + 12 y − 36$] #solutionbox[ #math.equation(block: true, alt: "12 open parenthesis 4 y minus 3 close parenthesis open parenthesis y plus 1 close parenthesis")[$12 ( 4 y − 3 ) ( y + 1 )$] ] #math.equation(block: true, alt: "30 x squared plus 105 x minus 60")[$30 x^(2) + 105 x − 60$] #strong[Factor Using Substitution] In the following exercises, factor using substitution. #math.equation(block: true, alt: "x to the power 4 minus 6 x squared minus 7")[$x^(4) − 6 x^(2) − 7$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x squared plus 1 close parenthesis open parenthesis x squared minus 7 close parenthesis")[$( x^(2) + 1 ) ( x^(2) − 7 )$] ] #math.equation(block: true, alt: "x to the power 4 plus 2 x squared minus 8")[$x^(4) + 2 x^(2) − 8$] #math.equation(block: true, alt: "x to the power 4 minus 3 x squared minus 28")[$x^(4) − 3 x^(2) − 28$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x squared minus 7 close parenthesis open parenthesis x squared plus 4 close parenthesis")[$( x^(2) − 7 ) ( x^(2) + 4 )$] ] #math.equation(block: true, alt: "x to the power 4 minus 13 x squared minus 30")[$x^(4) − 13 x^(2) − 30$] #math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared minus 5 open parenthesis x minus 3 close parenthesis minus 36")[$attach(( x − 3 ), t: 2) − 5 ( x − 3 ) − 36$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 12 close parenthesis open parenthesis x plus 1 close parenthesis")[$( x − 12 ) ( x + 1 )$] ] #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared minus 3 open parenthesis x minus 2 close parenthesis minus 54")[$attach(( x − 2 ), t: 2) − 3 ( x − 2 ) − 54$] #math.equation(block: true, alt: "open parenthesis 3 y minus 2 close parenthesis squared minus open parenthesis 3 y minus 2 close parenthesis minus 2")[$attach(( 3 y − 2 ), t: 2) − ( 3 y − 2 ) − 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 y minus 4 close parenthesis open parenthesis 3 y minus 1 close parenthesis")[$( 3 y − 4 ) ( 3 y − 1 )$] ] #math.equation(block: true, alt: "open parenthesis 5 y minus 1 close parenthesis squared minus 3 open parenthesis 5 y minus 1 close parenthesis minus 18")[$attach(( 5 y − 1 ), t: 2) − 3 ( 5 y − 1 ) − 18$] #strong[Mixed Practice] In the following exercises, factor each expression using any method. #math.equation(block: true, alt: "u squared minus 12 u plus 36")[$u^(2) − 12 u + 36$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis u minus 6 close parenthesis open parenthesis u minus 6 close parenthesis")[$( u − 6 ) ( u − 6 )$] ] #math.equation(block: true, alt: "x squared minus 14 x minus 32")[$x^(2) − 14 x − 32$] #math.equation(block: true, alt: "r squared minus 20 r s plus 64 s squared")[$r^(2) − 20 r s + 64 s^(2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis r minus 4 s close parenthesis open parenthesis r minus 16 s close parenthesis")[$( r − 4 s ) ( r − 16 s )$] ] #math.equation(block: true, alt: "q squared minus 29 q r minus 96 r squared")[$q^(2) − 29 q r − 96 r^(2)$] #math.equation(block: true, alt: "12 y squared minus 29 y plus 14")[$12 y^(2) − 29 y + 14$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 y minus 7 close parenthesis open parenthesis 3 y minus 2 close parenthesis")[$( 4 y − 7 ) ( 3 y − 2 )$] ] #math.equation(block: true, alt: "12 x squared plus 36 y minus 24 z")[$12 x^(2) + 36 y − 24 z$] #math.equation(block: true, alt: "6 n squared plus 5 n minus 4")[$6 n^(2) + 5 n − 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 n minus 1 close parenthesis open parenthesis 3 n plus 4 close parenthesis")[$( 2 n − 1 ) ( 3 n + 4 )$] ] #math.equation(block: true, alt: "3 q squared plus 6 q plus 2")[$3 q^(2) + 6 q + 2$] #math.equation(block: true, alt: "13 z squared plus 39 z minus 26")[$13 z^(2) + 39 z − 26$] #solutionbox[ #math.equation(block: true, alt: "13 open parenthesis z squared plus 3 z minus 2 close parenthesis")[$13 ( z^(2) + 3 z − 2 )$] ] #math.equation(block: true, alt: "5 r squared plus 25 r plus 30")[$5 r^(2) + 25 r + 30$] #math.equation(block: true, alt: "3 p squared plus 21 p")[$3 p^(2) + 21 p$] #solutionbox[ #math.equation(block: true, alt: "3 p open parenthesis p plus 7 close parenthesis")[$3 p ( p + 7 )$] ] #math.equation(block: true, alt: "7 x squared minus 21 x")[$7 x^(2) − 21 x$] #math.equation(block: true, alt: "6 r squared plus 30 r plus 36")[$6 r^(2) + 30 r + 36$] #solutionbox[ #math.equation(block: true, alt: "6 open parenthesis r plus 2 close parenthesis open parenthesis r plus 3 close parenthesis")[$6 ( r + 2 ) ( r + 3 )$] ] #math.equation(block: true, alt: "18 m squared plus 15 m plus 3")[$18 m^(2) + 15 m + 3$] #math.equation(block: true, alt: "24 n squared plus 20 n plus 4")[$24 n^(2) + 20 n + 4$] #solutionbox[ #math.equation(block: true, alt: "4 open parenthesis 2 n plus 1 close parenthesis open parenthesis 3 n plus 1 close parenthesis")[$4 ( 2 n + 1 ) ( 3 n + 1 )$] ] #math.equation(block: true, alt: "4 a squared plus 5 a plus 2")[$4 a^(2) + 5 a + 2$] #math.equation(block: true, alt: "x to the power 4 minus 4 x squared minus 12")[$x^(4) − 4 x^(2) − 12$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x squared plus 2 close parenthesis open parenthesis x squared minus 6 close parenthesis")[$( x^(2) + 2 ) ( x^(2) − 6 )$] ] #math.equation(block: true, alt: "x to the power 4 minus 7 x squared minus 8")[$x^(4) − 7 x^(2) − 8$] #math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis squared minus 9 open parenthesis x plus 3 close parenthesis minus 36")[$attach(( x + 3 ), t: 2) − 9 ( x + 3 ) − 36$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x minus 9 close parenthesis open parenthesis x plus 6 close parenthesis")[$( x − 9 ) ( x + 6 )$] ] #math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis squared minus 25 open parenthesis x plus 2 close parenthesis minus 54")[$attach(( x + 2 ), t: 2) − 25 ( x + 2 ) − 54$] ==== Writing Exercises Many trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] factor into the product of two binomials #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis .")[$( x + m ) ( x + n ) .$] Explain how you find the values of #emph[m] and #emph[n]. #solutionbox[ Answers will vary. ] Tommy factored #math.equation(block: false, alt: "x squared minus x minus 20")[$x^(2) − x − 20$] as #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis open parenthesis x minus 4 close parenthesis .")[$( x + 5 ) ( x − 4 ) .$] Sara factored it as #math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x minus 5 close parenthesis .")[$( x + 4 ) ( x − 5 ) .$] Ernesto factored it as #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis open parenthesis x minus 4 close parenthesis .")[$( x − 5 ) ( x − 4 ) .$] Who is correct? Explain why the other two are wrong. List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form #math.equation(block: false, alt: "a x squared plus b x plus c .")[$a x^(2) + b x + c .$] #solutionbox[ Answers will vary. ] How is the “ac” method similar to the “undo FOIL” method? How is it different? ==== 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, 4 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: factor trinomials of the form x squared plus bx plus c, factor trinomials of the form a x squared plus b x plus c using trial and error, factor trinomials of the form a x squared plus bx plus c with using the “ac” method, factor using substitution.], alt: "This table has 4 columns, 4 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: factor trinomials of the form x squared plus bx plus c, factor trinomials of the form a x squared plus b x plus c using trial and error, factor trinomials of the form a x squared plus bx plus c with using the “ac” method, factor using substitution.", caption: none) ⓑ After reviewing this checklist, what will you do to become confident for all objectives?