#set document(title: "7.3 Factor Trinomials of the Form ax2+bx+c", 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")) == 7.3#h(0.6em)Factor Trinomials of the Form ax#super[2]+bx+c #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Find the GCF of #math.equation(block: false, alt: "45 p squared and 30 p to the power 6")[$45 p^(2) #h(0.2em) "and" #h(0.2em) 30 p^(6)$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "15 p squared")[$15 p^(2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Multiply #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"))[ Combine like terms #math.equation(block: false, alt: "12 x squared plus 3 x plus 5 x plus 9")[$12 x^(2) + 3 x + 5 x + 9$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "12 x squared plus 8 x plus 9")[$12 x^(2) + 8 x + 9$] ] ] === Recognize a Preliminary Strategy for Factoring Let’s summarize where we are so far with factoring polynomials. In the first two sections of this chapter, we used three methods of factoring: factoring the GCF, factoring by grouping, and factoring a trinomial by “undoing” FOIL. More methods will follow as you continue in this chapter, as well as later in your studies of algebra. How will you know when to use each factoring method? As you learn more methods of factoring, how will you know when to apply each method and not get them confused? It will help to organize the factoring methods into a strategy that can guide you to use the correct method. As you start to factor a polynomial, always ask first, “Is there a greatest common factor?” If there is, factor it first. The next thing to consider is the type of polynomial. How many terms does it have? Is it a binomial? A trinomial? Or does it have more than three terms? If it is a trinomial where the leading coefficient is one, #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$], use the “undo FOIL” method. If it has more than three terms, try the grouping method. This is the only method to use for polynomials of more than three terms. Some polynomials cannot be factored. They are called “prime.” Below we summarize the methods we have so far. These are detailed in Choose a strategy to factor polynomials completely. #figure(figph[This figure lists strategies for factoring polynomials. At the top of the figure is G C F, where factoring always starts. From there, the figure has three branches. The first is binomial, the second is trinomial with the form x ^ 2 + b x +c, and the third is “more than three terms”, which is labeled with grouping.], alt: "This figure lists strategies for factoring polynomials. At the top of the figure is G C F, where factoring always starts. From there, the figure has three branches. The first is binomial, the second is trinomial with the form x ^ 2 + b x +c, and the third is “more than three terms”, which is labeled with grouping.", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Choose a strategy to factor polynomials completely.] + Is there a greatest common factor? - Factor it out. + Is the polynomial a binomial, trinomial, or are there more than three terms? - If it is a binomial, right now we have no method to factor it. - If it is a trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]: Undo FOIL #math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1em) ) ( x #h(1em) )$] - If it has more than three terms: Use the grouping method. + Check by multiplying the factors. ] Use the preliminary strategy to completely factor a polynomial. A polynomial is factored completely if, other than monomials, all of its factors are prime. #examplebox("Example 1")[][ Identify the best method to use to factor each polynomial. + ⓐ #math.equation(block: false, alt: "6 y squared minus 72")[$6 y^(2) − 72$] + ⓑ #math.equation(block: false, alt: "r squared minus 10 r minus 24")[$r^(2) − 10 r − 24$] + ⓒ #math.equation(block: false, alt: "p squared plus 5 p plus p q plus 5 q")[$p^(2) + 5 p + p q + 5 q$] #solutionbox[ ⓐ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "6 y squared minus 72")[$6 y^(2) − 72$]]), [Is there a greatest common factor?], [Yes, 6.], [Factor out the 6.], [#math.equation(block: false, alt: "6 open parenthesis y squared minus 12 close parenthesis")[$6 ( y^(2) − 12 )$]], [Is it a binomial, trinomial, or are there #linebreak() more than 3 terms?], [Binomial, we have no method to factor binomials yet.], )) ⓑ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "r squared minus 10 r minus 24")[$r^(2) − 10 r − 24$]]), [Is there a greatest common factor?], [No, there is no common factor.], [Is it a binomial, trinomial, or are there #linebreak() more than three terms?], [Trinomial, with leading coefficient 1, so “undo” FOIL.], )) ⓒ #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "p squared plus 5 p plus p q plus 5 q")[$p^(2) + 5 p + p q + 5 q$]]), [Is there a greatest common factor?], [No, there is no common factor.], [Is it a binomial, trinomial, or are there #linebreak() more than three terms?], [More than three terms, so factor using grouping.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Identify the best method to use to factor each polynomial: + ⓐ #math.equation(block: false, alt: "4 y squared plus 32")[$4 y^(2) + 32$] + ⓑ #math.equation(block: false, alt: "y squared plus 10 y plus 21")[$y^(2) + 10 y + 21$] + ⓒ #math.equation(block: false, alt: "y z plus 2 y plus 3 z plus 6")[$y z + 2 y + 3 z + 6$] #solutionbox[ ⓐ no method ⓑ undo using FOIL ⓒ factor with grouping ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Identify the best method to use to factor each polynomial: + ⓐ #math.equation(block: false, alt: "a b plus a plus 4 b plus 4")[$a b + a + 4 b + 4$] + ⓑ #math.equation(block: false, alt: "3 k squared plus 15")[$3 k^(2) + 15$] + ⓒ #math.equation(block: false, alt: "p squared plus 9 p plus 8")[$p^(2) + 9 p + 8$] #solutionbox[ ⓐ factor using grouping ⓑ no method ⓒ undo using FOIL ] ] === Factor Trinomials of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] with a GCF Now that we have organized what we’ve covered so far, we are ready 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 GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods in the last section. Let’s do a few examples to see how this works. Watch out for the signs in the next two examples. #examplebox("Example 2")[][ Factor completely: #math.equation(block: false, alt: "2 n squared minus 8 n minus 42")[$2 n^(2) − 8 n − 42$]. #solutionbox[ Use the preliminary strategy. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor?], [#math.equation(block: false, alt: "2 n squared minus 8 n minus 42")[$2 n^(2) − 8 n − 42$]]), [Yes, GCF = 2. Factor it out.], [#math.equation(block: false, alt: "2 open parenthesis n squared minus 4 n minus 21 close parenthesis")[$2 ( n^(2) − 4 n − 21 )$]], )) Inside the parentheses, is it a binomial, trinomial, or are there more than three terms? #figure(table( columns: 2, align: left, inset: 6pt, table.header([It is a trinomial whose coefficient is 1, so undo FOIL.], [#math.equation(block: false, alt: "2 open parenthesis n close parenthesis open parenthesis n close parenthesis")[$2 ( n #h(1em) ) ( n #h(1em) )$]]), [Use 3 and −7 as the last terms of the binomials.], [#math.equation(block: false, alt: "2 open parenthesis n plus 3 close parenthesis open parenthesis n minus 7 close parenthesis")[$2 ( n + 3 ) ( n − 7 )$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "−21")[$−21$]], [Sum of factors]), [#math.equation(block: false, alt: "1 , −21")[$1 , −21$]], [#math.equation(block: false, alt: "1 plus open parenthesis −21 close parenthesis equals −20")[$1 + ( −21 ) = −20$]], [#math.equation(block: false, alt: "3 , −7")[$3 , −7$]], [#math.equation(block: false, alt: "3 plus open parenthesis −7 close parenthesis equals −4 *")[$3 + ( −7 ) = −4 "*"$]], )) Check. #math.equation(block: true, alt: "2 open parenthesis n plus 3 close parenthesis open parenthesis n minus 7 close parenthesis")[$#h(2.5em) 2 ( n + 3 ) ( n − 7 )$] #math.equation(block: true, alt: "2 open parenthesis n squared minus 7 n plus 3 n minus 21 close parenthesis")[$#h(2.5em) 2 ( n^(2) − 7 n + 3 n − 21 )$] #math.equation(block: true, alt: "2 open parenthesis n squared minus 4 n minus 21 close parenthesis")[$#h(2.5em) 2 ( n^(2) − 4 n − 21 )$] #math.equation(block: true, alt: "2 n squared minus 8 n minus 42 ✓")[$#h(2.5em) 2 n^(2) − 8 n − 42 #h(0.2em) ✓$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "4 m squared minus 4 m minus 8")[$4 m^(2) − 4 m − 8$]. #solutionbox[ #math.equation(block: true, alt: "4 open parenthesis m plus 1 close parenthesis open parenthesis m minus 2 close parenthesis")[$4 ( m + 1 ) ( m − 2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "5 k squared minus 15 k minus 50")[$5 k^(2) − 15 k − 50$]. #solutionbox[ #math.equation(block: true, alt: "5 open parenthesis k plus 2 close parenthesis open parenthesis k minus 5 close parenthesis")[$5 ( k + 2 ) ( k − 5 )$] ] ] #examplebox("Example 3")[][ Factor completely: #math.equation(block: false, alt: "4 y squared minus 36 y plus 56")[$4 y^(2) − 36 y + 56$]. #solutionbox[ Use the preliminary strategy. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor?], [#math.equation(block: false, alt: "4 y squared minus 36 y plus 56")[$4 y^(2) − 36 y + 56$]]), [Yes, GCF = 4. Factor it.], [#math.equation(block: false, alt: "4 open parenthesis y squared minus 9 y plus 14 close parenthesis")[$4 ( y^(2) − 9 y + 14 )$]], [Inside the parentheses, is it a binomial, trinomial, or are #linebreak() there more than three terms?], [], [It is a trinomial whose coefficient is 1. So undo FOIL.], [#math.equation(block: false, alt: "4 open parenthesis y close parenthesis open parenthesis y close parenthesis")[$4 ( y #h(1.5em) ) ( y #h(1.5em) )$]], [Use a table like the one below to find two numbers that multiply to #linebreak() 14 and add to −9.], [], [Both factors of 14 must be negative.], [#math.equation(block: false, alt: "4 open parenthesis y minus 2 close parenthesis open parenthesis y minus 7 close parenthesis")[$4 ( y − 2 ) ( y − 7 )$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "14")[$14$]], [Sum of factors]), [#math.equation(block: false, alt: "−1 , −14")[$−1 , −14$]], [#math.equation(block: false, alt: "−1 plus open parenthesis −14 close parenthesis equals −15")[$−1 + ( −14 ) = −15$]], [#math.equation(block: false, alt: "−2 , −7")[$−2 , −7$]], [#math.equation(block: false, alt: "−2 plus open parenthesis −7 close parenthesis equals −9 *")[$−2 + ( −7 ) = −9 "*"$]], )) Check. #math.equation(block: true, alt: "4 open parenthesis y minus 2 close parenthesis open parenthesis y minus 7 close parenthesis")[$#h(2.5em) 4 ( y − 2 ) ( y − 7 )$] #math.equation(block: true, alt: "4 open parenthesis y squared minus 7 y minus 2 y plus 14 close parenthesis")[$#h(2.5em) 4 ( y^(2) − 7 y − 2 y + 14 )$] #math.equation(block: true, alt: "4 open parenthesis y squared minus 9 y plus 14 close parenthesis")[$#h(2.5em) 4 ( y^(2) − 9 y + 14 )$] #math.equation(block: true, alt: "4 y squared minus 36 y plus 56 ✓")[$#h(2.5em) 4 y^(2) − 36 y + 56 #h(0.2em) ✓$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "3 r squared minus 9 r plus 6")[$3 r^(2) − 9 r + 6$]. #solutionbox[ #math.equation(block: true, alt: "3 open parenthesis r minus 1 close parenthesis open parenthesis r minus 2 close parenthesis")[$3 ( r − 1 ) ( r − 2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "2 t squared minus 10 t plus 12")[$2 t^(2) − 10 t + 12$]. #solutionbox[ #math.equation(block: true, alt: "2 open parenthesis t minus 2 close parenthesis open parenthesis t minus 3 close parenthesis")[$2 ( t − 2 ) ( t − 3 )$] ] ] In the next example the GCF will include a variable. #examplebox("Example 4")[][ Factor completely: #math.equation(block: false, alt: "4 u cubed plus 16 u squared minus 20 u")[$4 u^(3) + 16 u^(2) − 20 u$]. #solutionbox[ Use the preliminary strategy. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor?], [#math.equation(block: false, alt: "4 u cubed plus 16 u squared minus 20 u")[$4 u^(3) + 16 u^(2) − 20 u$]]), [Yes, GCF = 4#emph[u]. Factor it.], [#math.equation(block: false, alt: "4 u open parenthesis u squared plus 4 u minus 5 close parenthesis")[$4 u ( u^(2) + 4 u − 5 )$]], [Binomial, trinomial, or more than three terms?], [], [It is a trinomial. So “undo FOIL.”], [#math.equation(block: false, alt: "4 u open parenthesis u close parenthesis open parenthesis u close parenthesis")[$4 u ( u #h(1em) ) ( u #h(1em) )$]], [Use a table like the table below to find two numbers that #linebreak() multiply to −5 and add to 4.], [#math.equation(block: false, alt: "4 u open parenthesis u minus 1 close parenthesis open parenthesis u plus 5 close parenthesis")[$4 u ( u − 1 ) ( u + 5 )$]], )) #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")[$−1 , 5$]], [#math.equation(block: false, alt: "−1 plus 5 equals 4 *")[$−1 + 5 = 4 "*"$]], [#math.equation(block: false, alt: "1 , −5")[$1 , −5$]], [#math.equation(block: false, alt: "1 plus open parenthesis −5 close parenthesis equals −4")[$1 + ( −5 ) = −4$]], )) Check. #math.equation(block: true, alt: "4 u open parenthesis u minus 1 close parenthesis open parenthesis u plus 5 close parenthesis")[$#h(2.5em) 4 u ( u − 1 ) ( u + 5 )$] #math.equation(block: true, alt: "4 u open parenthesis u squared plus 5 u minus u minus 5 close parenthesis")[$#h(2.5em) 4 u ( u^(2) + 5 u − u − 5 )$] #math.equation(block: true, alt: "4 u open parenthesis u squared plus 4 u minus 5 close parenthesis")[$#h(2.5em) 4 u ( u^(2) + 4 u − 5 )$] #math.equation(block: true, alt: "4 u cubed plus 16 u squared minus 20 u ✓")[$#h(2.5em) 4 u^(3) + 16 u^(2) − 20 u #h(0.2em) ✓$] ] ] #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")[$6 y ( y − 2 ) ( y + 5 )$] ] ] === Factor Trinomials using Trial and Error 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")[$3 x^(2) + 5 x + 2 \ ( #h(1em) ) ( #h(1em) )$]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[This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied.], alt: "This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied.", 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 and 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[This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both, there are the terms 1x, 3x under the 3x^2. Also, there are the factors 1,2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3x + 2) and the next is (x + 2)(3x + 1).], alt: "This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both, there are the terms 1x, 3x under the 3x^2. Also, there are the factors 1,2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3x + 2) and the next is (x + 2)(3x + 1).", caption: none) Which factors are correct? To decide that, we multiply the inner and outer terms. #figure(figph[This figure demonstrates the possible factors of the polynomial 3 x^ 2 + 5 x +2. The polynomial is written twice. Underneath both, there are the terms 1 x, 3 x under the 3 x ^ 2. Also, there are the factors 1, 2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3 x + 2). Underneath this factorization are the products 3 x from multiplying the middle terms 1 and 3 x. Also there is the product of 2 x from multiplying the outer terms x and 2. These products of 3 x and 2 x add to 5 x. Underneath the second factorization are the products 6 x from multiplying the middle terms 2 and 3 x. Also there is the product of 1 x from multiplying the outer terms x and 1. These two products of 6 x and 1 x add to 7 x.], alt: "This figure demonstrates the possible factors of the polynomial 3 x^ 2 + 5 x +2. The polynomial is written twice. Underneath both, there are the terms 1 x, 3 x under the 3 x ^ 2. Also, there are the factors 1, 2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3 x + 2). Underneath this factorization are the products 3 x from multiplying the middle terms 1 and 3 x. Also there is the product of 2 x from multiplying the outer terms x and 2. These products of 3 x and 2 x add to 5 x. Underneath the second factorization are the products 6 x from multiplying the middle terms 2 and 3 x. Also there is the product of 1 x from multiplying the outer terms x and 1. These two products of 6 x and 1 x add to 7 x.", caption: none) Since the middle term of the trinomial is 5#emph[x], the factors in the first case will work. Let’s 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 ✓")[$( x + 1 ) ( 3 x + 2 ) \ 3 x^(2) + 2 x + 3 x + 2 \ 3 x^(2) + 5 x + 2 #h(0.2em) ✓$]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 5")[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][ Factor completely: #math.equation(block: false, alt: "3 y squared plus 22 y plus 7")[$3 y^(2) + 22 y + 7$]. #solutionbox[ #figure(figph[This table summarizes the steps for factoring 3 y ^ 2 + 22 y + 7. The first row states write the trinomial in descending order. The polynomial is written 3 y ^ 2 +22 y + 7.], alt: "This table summarizes the steps for factoring 3 y ^ 2 + 22 y + 7. The first row states write the trinomial in descending order. The polynomial is written 3 y ^ 2 +22 y + 7.", caption: none) #figure(figph[The second row states find all the factor pairs of the first term. The only pairs listed are 1 y, 3 y. Then, since there is only one pair, they are in the parentheses written (1 y ) and (3 y ).], alt: "The second row states find all the factor pairs of the first term. The only pairs listed are 1 y, 3 y. Then, since there is only one pair, they are in the parentheses written (1 y ) and (3 y ).", caption: none) #figure(figph[The third row states “find all the factored pairs of the third term”. It also states the only factors of 7 are 1 and 7.], alt: "The third row states “find all the factored pairs of the third term”. It also states the only factors of 7 are 1 and 7.", caption: none) #figure(figph[The fourth row states test all the possible combinations of the factors until the correct product is found. The possible factors are shown (y + 1)(3 y + 7) and (y + 7)(3y + 1). Under each factor is the products of the outer terms and the inner terms. For the first it is 7y and 3y. For the second it is 21 y and y. The combination (y + 7)(3 y + 1) is the correct factoring.], alt: "The fourth row states test all the possible combinations of the factors until the correct product is found. The possible factors are shown (y + 1)(3 y + 7) and (y + 7)(3y + 1). Under each factor is the products of the outer terms and the inner terms. For the first it is 7y and 3y. For the second it is 21 y and y. The combination (y + 7)(3 y + 1) is the correct factoring.", caption: none) #figure(figph[The last row states to check by multiplying. The product of (y + 7)(3 y + 1) is shown as 3 y ^ 2 + 22 y + 7.], alt: "The last row states to check by multiplying. The product of (y + 7)(3 y + 1) is shown as 3 y ^ 2 + 22 y + 7.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #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: #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. + 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. ] When the middle term is negative and the last term is positive, the signs in the binomials must both be negative. #examplebox("Example 6")[][ Factor completely: #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 shows the quadratic expression 6b^2 - 13b + 5.], alt: "The image shows the quadratic expression 6b^2 - 13b + 5.", caption: none)]), [Find the factors of the first term.], [#figure(figph[Quadratic expression 6b^2 - 13b + 5, with factor options for 6b^2: 1b \* 6b and 2b \* 3b. This is the setup for factoring a trinomial, exploring combinations for the first term.], alt: "Quadratic expression 6b^2 - 13b + 5, with factor options for 6b^2: 1b * 6b and 2b * 3b. This is the setup for factoring a trinomial, exploring combinations for the first term.", caption: none)], [Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.], [#figure(figph[Factoring the quadratic expression 6b^2 - 13b + 5, showing possible factors for 6b^2 (1b \* 6b, 2b \* 3b) and the constant term +5 (-1, -5).], alt: "Factoring the quadratic expression 6b^2 - 13b + 5, showing possible factors for 6b^2 (1b * 6b, 2b * 3b) and the constant term +5 (-1, -5).", 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")[$6 b^(2) − 13 b + 5$] \*], [#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 ✓")[$#h(2.5em) ( 2 b − 1 ) ( 3 b − 5 ) \ #h(2.5em) 6 b^(2) − 10 b − 3 b + 5 \ #h(2.5em) 6 b^(2) − 13 b + 5 #h(0.2em) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #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: #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 7")[][ Factor completely: #math.equation(block: false, alt: "14 x squared minus 47 x minus 7")[$14 x^(2) − 47 x − 7$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The trinomial is already in descending order.], [#figure(figph[The image shows the mathematical expression 14x^2 - 47x - 7.], alt: "The image shows the mathematical expression 14x^2 - 47x - 7.", caption: none)]), [Find the factors of the first term.], [#figure(figph[A math problem presenting the quadratic expression 14x^2 - 47x - 7, and the factor pairs of 1x \* 14x and 2x \* 7x for the 14x^2 term, used to facilitate factoring the trinomial.], alt: "A math problem presenting the quadratic expression 14x^2 - 47x - 7, and the factor pairs of 1x * 14x and 2x * 7x for the 14x^2 term, used to facilitate factoring the trinomial.", caption: none)], [Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative.], [#figure(figph[The quadratic expression 14x^2 - 47x - 7 is presented with its first term factors (1x \* 14x, 2x \* 7x) and constant term factors (1, -7; -1, 7) for factorization.], alt: "The quadratic expression 14x^2 - 47x - 7 is presented with its first term factors (1x * 14x, 2x * 7x) and constant term factors (1, -7; -1, 7) for factorization.", caption: none)], )) Consider all the combinations of factors. We use each pair of the factors of #math.equation(block: false, alt: "14 x squared")[$14 x^(2)$] with each pair of factors of #math.equation(block: false, alt: "−7 .")[$−7 .$] #figure(table( columns: 3, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "14 x squared")[$14 x^(2)$]], [Pair with], [Factors of #math.equation(block: false, alt: "−7")[$−7$]]), [#math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "14 x")[$14 x$]], [], [#math.equation(block: false, alt: "1")[$1$], #math.equation(block: false, alt: "−7")[$−7$] #linebreak() #math.equation(block: false, alt: "−7")[$−7$], #math.equation(block: false, alt: "1")[$1$] #linebreak() (reverse order)], [#math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "14 x")[$14 x$]], [], [#math.equation(block: false, alt: "−1")[$−1$], #math.equation(block: false, alt: "7")[$7$] #linebreak() #math.equation(block: false, alt: "7")[$7$], #math.equation(block: false, alt: "−1")[$−1$] #linebreak() (reverse order)], [#math.equation(block: false, alt: "2 x , 7 x")[$2 x , 7 x$]], [], [#math.equation(block: false, alt: "1")[$1$], #math.equation(block: false, alt: "−7")[$−7$] #linebreak() #math.equation(block: false, alt: "−7")[$−7$], #math.equation(block: false, alt: "1")[$1$] #linebreak() (reverse order)], [#math.equation(block: false, alt: "2 x , 7 x")[$2 x , 7 x$]], [], [#math.equation(block: false, alt: "−1")[$−1$], #math.equation(block: false, alt: "7")[$7$] #linebreak() #math.equation(block: false, alt: "7")[$7$], #math.equation(block: false, alt: "−1")[$−1$] #linebreak() (reverse order)], )) These pairings lead to the following eight combinations. #figure(figph[This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”], alt: "This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”", caption: none) #figure(table( columns: 2, align: left, inset: 6pt, table.header([The correct factors are those whose product is the #linebreak() original trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 x minus 7 close parenthesis open parenthesis 7 x plus 1 close parenthesis")[$( 2 x − 7 ) ( 7 x + 1 )$]]), [Check by multiplying. #linebreak() #math.equation(block: false, alt: "open parenthesis 2 x minus 7 close parenthesis open parenthesis 7 x plus 1 close parenthesis; 14 x squared plus 2 x minus 49 x minus 7; 14 x squared minus 47 x minus 7 ✓")[$#h(2.5em) ( 2 x − 7 ) ( 7 x + 1 ) \ #h(2.5em) 14 x^(2) + 2 x − 49 x − 7 \ #h(2.5em) 14 x^(2) − 47 x − 7 #h(0.2em) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "8 a squared minus 3 a minus 5")[$8 a^(2) − 3 a − 5$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis a minus 1 close parenthesis open parenthesis 8 a plus 5 close parenthesis")[$( a − 1 ) ( 8 a + 5 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "6 b squared minus b minus 15")[$6 b^(2) − b − 15$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 b plus 3 close parenthesis open parenthesis 3 b minus 5 close parenthesis")[$( 2 b + 3 ) ( 3 b − 5 )$] ] ] #examplebox("Example 8")[][ Factor completely: #math.equation(block: false, alt: "18 n squared minus 37 n plus 15")[$18 n^(2) − 37 n + 15$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([The trinomial is already in descending order.], [#math.equation(block: false, alt: "18 n squared minus 37 n plus 15")[$18 n^(2) − 37 n + 15$]]), [Find the factors of the first term.], [#figure(figph[The quadratic expression 18n^2 - 37n + 15 is displayed, with potential factor pairs for the 18n^2 term listed below in red.], alt: "The quadratic expression 18n^2 - 37n + 15 is displayed, with potential factor pairs for the 18n^2 term listed below in red.", caption: none)], [Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative facotrs.], [#figure(figph[This image illustrates the initial step in factoring the quadratic trinomial 18n^2 - 37n + 15, showing possible factor pairs for the 18n^2 term and the constant term +15.], alt: "This image illustrates the initial step in factoring the quadratic trinomial 18n^2 - 37n + 15, showing possible factor pairs for the 18n^2 term and the constant term +15.", caption: none)], )) Consider all the combinations of factors. #figure(figph[This table has the heading 18 n ^ 2 – 37n + 15. This table has two columns. The first column is labeled possible factors and the second column is labeled product. The first column lists all the combinations of possible factors and the second column has the products. Eight rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (2n – 3)(9n – 5) with the product 18n^2 minus 37 n + 15 has an asterisk.], alt: "This table has the heading 18 n ^ 2 – 37n + 15. This table has two columns. The first column is labeled possible factors and the second column is labeled product. The first column lists all the combinations of possible factors and the second column has the products. Eight rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (2n – 3)(9n – 5) with the product 18n^2 minus 37 n + 15 has an asterisk.", caption: none) #figure(table( columns: 2, align: left, inset: 6pt, table.header([The correct factors are those whose product is #linebreak() the original trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 n minus 3 close parenthesis open parenthesis 9 n minus 5 close parenthesis")[$( 2 n − 3 ) ( 9 n − 5 )$]]), [Check by multiplying. #linebreak() #math.equation(block: false, alt: "open parenthesis 2 n minus 3 close parenthesis open parenthesis 9 n minus 5 close parenthesis; 18 n squared minus 10 n minus 27 n plus 15; 18 n squared minus 37 n plus 15 ✓")[$#h(2.5em) ( 2 n − 3 ) ( 9 n − 5 ) \ #h(2.5em) 18 n^(2) − 10 n − 27 n + 15 \ #h(2.5em) 18 n^(2) − 37 n + 15 #h(0.2em) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "18 x squared minus 3 x minus 10")[$18 x^(2) − 3 x − 10$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 x plus 2 close parenthesis open parenthesis 6 x minus 5 close parenthesis")[$( 3 x + 2 ) ( 6 x − 5 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #math.equation(block: false, alt: "30 y squared minus 53 y minus 21")[$30 y^(2) − 53 y − 21$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 y plus 1 close parenthesis open parenthesis 10 y minus 21 close parenthesis")[$( 3 y + 1 ) ( 10 y − 21 )$] ] ] Don’t forget to look for a GCF first. #examplebox("Example 9")[][ Factor completely: #math.equation(block: false, alt: "10 y to the power 4 plus 55 y cubed plus 60 y squared")[$10 y^(4) + 55 y^(3) + 60 y^(2)$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "10 y to the power 4 plus 55 y cubed plus 60 y squared")[$10 y^(4) + 55 y^(3) + 60 y^(2)$]]), [Notice the greatest common factor, and factor it first.], [#math.equation(block: false, alt: "15 y squared open parenthesis 2 y squared plus 11 y plus 12 close parenthesis")[$15 y^(2) ( 2 y^(2) + 11 y + 12 )$]], [Factor the trinomial.], [#figure(figph[An algebraic expression 5y^2(2y^2 + 11y + 12) is displayed with red text illustrating factorization hints. These include factors of 2y^2 (y\*2y) and factors of 12 (1\*12, 2\*6, 3\*4), aiding the factoring of the quadratic.], alt: "An algebraic expression 5y^2(2y^2 + 11y + 12) is displayed with red text illustrating factorization hints. These include factors of 2y^2 (y*2y) and factors of 12 (1*12, 2*6, 3*4), aiding the factoring of the quadratic.", caption: none)], )) Consider all the combinations. #figure(figph[This table has the heading 2 y squared + 11 y + 12 This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. Four rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (y + 4)(2y + 3) with the product 2 y squared + 11 y + 12 has an asterisk.], alt: "This table has the heading 2 y squared + 11 y + 12 This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. Four rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (y + 4)(2y + 3) with the product 2 y squared + 11 y + 12 has an asterisk.", 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 5#emph[y]#super[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 plus 55 y cubed plus 60 y squared ✓")[$#h(2.5em) 5 y^(2) ( y + 4 ) ( 2 y + 3 ) \ #h(2.5em) 5 y^(2) ( 2 y^(2) + 8 y + 3 y + 12 ) \ #h(2.5em) 10 y^(4) + 55 y^(3) + 60 y^(2) #h(0.2em) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor completely: #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: #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 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 10")[How to Factor Trinomials Using the “ac” Method][ Factor: #math.equation(block: false, alt: "6 x squared plus 7 x plus 2")[$6 x^(2) + 7 x + 2$]. #solutionbox[ #figure(figph[This table lists the steps for factoring 6 x ^ 2 + 7 x + 2. The first step is to factor the GCF. This polynomial has none.], alt: "This table lists the steps for factoring 6 x ^ 2 + 7 x + 2. The first step is to factor the GCF. This polynomial has none.", caption: none) #figure(figph[The second row states to find the product a c. Then, it lists a c as 6 times 2 = 12.], alt: "The second row states to find the product a c. Then, it lists a c as 6 times 2 = 12.", caption: none) #figure(figph[The third step is to find two numbers m and n in which m times n = a c and m + n = b. The middle column reads, “find two numbers that add to 7. Both factors must be positive”. The numbers are 3 and 4. 3 times 4 is 12 and 3 + 4 is 7.], alt: "The third step is to find two numbers m and n in which m times n = a c and m + n = b. The middle column reads, “find two numbers that add to 7. Both factors must be positive”. The numbers are 3 and 4. 3 times 4 is 12 and 3 + 4 is 7.", caption: none) #figure(figph[The next step is to split the middle term using m and n. That is, to write 7 x as 3 x + 4 x. Therefore, 6 x ^ 2 + 7 x + 2 is rewritten as 6 x ^ 2 +3 x + 4 x + 2.], alt: "The next step is to split the middle term using m and n. That is, to write 7 x as 3 x + 4 x. Therefore, 6 x ^ 2 + 7 x + 2 is rewritten as 6 x ^ 2 +3 x + 4 x + 2.", caption: none) #figure(figph[The next step is to factor by grouping. 3 x(2 x + 1) + 2(2 x + 1) then factor again (2 x + 1)(3 x + 2).], alt: "The next step is to factor by grouping. 3 x(2 x + 1) + 2(2 x + 1) then factor again (2 x + 1)(3 x + 2).", caption: none) #figure(figph[The last step is to check by multiplying. Multiply the factors (2 x + 1)(3 x + 2) to get 6 x ^ 2 + 7 x + 2.], alt: "The last step is to check by multiplying. Multiply the factors (2 x + 1)(3 x + 2) to get 6 x ^ 2 + 7 x + 2.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #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: #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")[$( 2 y + 1 ) ( 2 y + 3 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Factor trinomials of the form using the “ac” method.] + Factor any GCF. + Find the product 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")[$"Multiply to" #h(0.2em) a c & & & m · n = a · c \ "Add to" #h(0.2em) b & & & m + n = b$] + Split the middle term using #emph[m] and #emph[n]: #figure(figph[This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.], alt: "This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.", caption: none) + Factor by grouping. + Check by multiplying the factors. ] When the third term of the trinomial is negative, the factors of the third term will have opposite signs. #examplebox("Example 11")[][ Factor: #math.equation(block: false, alt: "8 u squared minus 17 u minus 21")[$8 u^(2) − 17 u − 21$]. #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([Is there a greatest common factor? No.], [], [#figure(figph[Two lines of algebraic expressions, with the top line showing 'ax^2 + bx + c' in red, and the bottom line showing '8u^2 - 17u - 21' in black.], alt: "Two lines of algebraic expressions, with the top line showing 'ax^2 + bx + c' in red, and the bottom line showing '8u^2 - 17u - 21' in black.", caption: none)]), [Find #math.equation(block: false, alt: "a times c")[$a ⋅ c$].], [#math.equation(block: false, alt: "a times c")[$a ⋅ c$]], [], [], [#math.equation(block: false, alt: "8 open parenthesis −21 close parenthesis")[$8 ( −21 )$]], [], [], [#math.equation(block: false, alt: "−168")[$−168$]], [], )) Find two numbers that multiply to #math.equation(block: false, alt: "−168")[$−168$] and add to #math.equation(block: false, alt: "−17")[$−17$]. The larger factor must be negative. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "−168")[$−168$]], [Sum of factors]), [#math.equation(block: false, alt: "1 , −168")[$1 , −168$]], [#math.equation(block: false, alt: "1 plus open parenthesis −168 close parenthesis equals −167")[$1 + ( −168 ) = −167$]], [#math.equation(block: false, alt: "2 , −84")[$2 , −84$]], [#math.equation(block: false, alt: "2 plus open parenthesis −84 close parenthesis equals −82")[$2 + ( −84 ) = −82$]], [#math.equation(block: false, alt: "3 , −56")[$3 , −56$]], [#math.equation(block: false, alt: "3 plus open parenthesis −56 close parenthesis equals −53")[$3 + ( −56 ) = −53$]], [#math.equation(block: false, alt: "4 , −42")[$4 , −42$]], [#math.equation(block: false, alt: "4 plus open parenthesis −42 close parenthesis equals −38")[$4 + ( −42 ) = −38$]], [#math.equation(block: false, alt: "6 , −28")[$6 , −28$]], [#math.equation(block: false, alt: "6 plus open parenthesis −28 close parenthesis equals −22")[$6 + ( −28 ) = −22$]], [#math.equation(block: false, alt: "7 , −24")[$7 , −24$]], [#math.equation(block: false, alt: "7 plus open parenthesis −24 close parenthesis equals −17 *")[$7 + ( −24 ) = −17 "*"$]], [#math.equation(block: false, alt: "8 , −21")[$8 , −21$]], [#math.equation(block: false, alt: "8 plus open parenthesis −21 close parenthesis equals −13")[$8 + ( −21 ) = −13$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([Split the middle term using 7#emph[u] and −24#emph[u].], [#math.equation(block: false, alt: "8 u squared minus 17 ↙ u ↘ minus 21; 8 u squared plus 7 u ⎵ −24 u minus 21 ⎵")[$#h(2em) 8 u^(2) − limits(17)_("↙") limits(u)_("↘") − 21 \ #h(2em) underbrace(8 u^(2) + 7 u) #h(0.2em) underbrace(−24 u − 21)$]]), [Factor by grouping.], [#math.equation(block: false, alt: "u open parenthesis 8 u plus 7 close parenthesis minus 3 open parenthesis 8 u plus 7 close parenthesis; open parenthesis 8 u plus 7 close parenthesis open parenthesis u minus 3 close parenthesis")[$#h(2em) u ( 8 u + 7 ) − 3 ( 8 u + 7 ) \ #h(2em) ( 8 u + 7 ) ( u − 3 )$]], [Check by multiplying. #linebreak() #math.equation(block: false, alt: "open parenthesis 8 u plus 7 close parenthesis open parenthesis u minus 3 close parenthesis; 8 u squared minus 24 u plus 7 u minus 21; 8 u squared minus 17 u minus 21 ✓")[$#h(2.5em) ( 8 u + 7 ) ( u − 3 ) \ #h(2.5em) 8 u^(2) − 24 u + 7 u − 21 \ #h(2.5em) 8 u^(2) − 17 u − 21 #h(0.2em) ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "20 h squared plus 13 h minus 15")[$20 h^(2) + 13 h − 15$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 h plus 5 close parenthesis open parenthesis 5 h minus 3 close parenthesis")[$( 4 h + 5 ) ( 5 h − 3 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "6 g squared plus 19 g minus 20")[$6 g^(2) + 19 g − 20$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis g plus 4 close parenthesis open parenthesis 6 g minus 5 close parenthesis")[$( g + 4 ) ( 6 g − 5 )$] ] ] #examplebox("Example 12")[][ Factor: #math.equation(block: false, alt: "2 x squared plus 6 x plus 5")[$2 x^(2) + 6 x + 5$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor? No.], [#figure(figph[A general quadratic equation ax^2 + bx + c, shown in red, with a specific example 2x^2 + 6x + 5, shown in black, illustrating the standard form of a quadratic polynomial.], alt: "A general quadratic equation ax^2 + bx + c, shown in red, with a specific example 2x^2 + 6x + 5, shown in black, illustrating the standard form of a quadratic polynomial.", caption: none)]), [Find #math.equation(block: false, alt: "a times c")[$a ⋅ c$].], [#math.equation(block: false, alt: "a times c")[$a ⋅ c$]], [], [#math.equation(block: false, alt: "2 open parenthesis 5 close parenthesis")[$2 ( 5 )$]], [], [#math.equation(block: false, alt: "10")[$10$]], )) Find two numbers that multiply to 10 and add to 6. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of #math.equation(block: false, alt: "10")[$10$]], [Sum of factors]), [#math.equation(block: false, alt: "1 , 10")[$1 , 10$]], [#math.equation(block: false, alt: "1 plus 10 equals 11")[$1 + 10 = 11$]], [2, 5], [#math.equation(block: false, alt: "2 plus 5 equals 7")[$2 + 5 = 7$]], )) There are no factors that multiply to 10 and add to 6. The polynomial is prime. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "10 t squared plus 19 t minus 15")[$10 t^(2) + 19 t − 15$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 t plus 5 close parenthesis open parenthesis 5 t minus 3 close parenthesis")[$( 2 t + 5 ) ( 5 t − 3 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #math.equation(block: false, alt: "3 u squared plus 8 u plus 5")[$3 u^(2) + 8 u + 5$]. #solutionbox[ #math.equation(block: true, alt: "open parenthesis u plus 1 close parenthesis open parenthesis 3 u plus 5 close parenthesis")[$( u + 1 ) ( 3 u + 5 )$] ] ] Don’t forget to look for a common factor! #examplebox("Example 13")[][ Factor: #math.equation(block: false, alt: "10 y squared minus 55 y plus 70")[$10 y^(2) − 55 y + 70$]. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Is there a greatest common factor? Yes. The GCF is 5.], [#figure(figph[The image displays the quadratic expression 10y^2 - 55y + 70, potentially for factoring or solving.], alt: "The image displays the quadratic expression 10y^2 - 55y + 70, potentially for factoring or solving.", caption: none)]), [Factor it. Be careful to keep the factor of 5 all the way through the solution!], [#figure(figph[The image displays the mathematical expression 5(2y^2 - 11y + 14) in white text on a plain white background.], alt: "The image displays the mathematical expression 5(2y^2 - 11y + 14) in white text on a plain white background.", caption: none)], [The trinomial inside the parentheses has a leading coefficient that is not 1.], [#figure(figph[Two mathematical expressions are shown: 'ax^2 + bx + c' in red, followed by '5(2y^2 - 11y + 14)' in black, illustrating a quadratic formula alongside a factored quadratic expression.], alt: "Two mathematical expressions are shown: 'ax^2 + bx + c' in red, followed by '5(2y^2 - 11y + 14)' in black, illustrating a quadratic formula alongside a factored quadratic expression.", caption: none)], [Factor the trinomial.], [#figure(figph[A mathematical expression showing the product of 5, (y-2), and (2y-7).], alt: "A mathematical expression showing the product of 5, (y-2), and (2y-7).", caption: none)], [Check by mulitplying all three factors.], [], [#math.equation(block: false, alt: "5 open parenthesis 2 y squared minus 2 y minus 4 y plus 14 close parenthesis")[$5 ( 2 y^(2) − 2 y − 4 y + 14 )$]], [], [#math.equation(block: false, alt: "5 open parenthesis 2 y squared minus 11 y plus 14 close parenthesis")[$5 ( 2 y^(2) − 11 y + 14 )$]], [], [#math.equation(block: false, alt: "10 y squared minus 55 y plus 70 ✓")[$10 y^(2) − 55 y + 70 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Factor: #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: #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")[$3 ( 3 w − 2 ) ( 2 w − 3 )$] ] ] We can now update the Preliminary Factoring Strategy, as shown in and detailed in Choose a strategy to factor polynomials completely (updated), to include trinomials of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$]. Remember, some polynomials are prime and so they cannot be factored. #figure(figph[This figure has the strategy for factoring polynomials. At the top of the figure is GCF. Below this, there are three options. The first is binomial. The second is trinomial. Under trinomial there are x squared + b x + c and a x squared + b x +c. The two methods here are trial and error and the “a c” method. The third option is for more than three terms. It is grouping.], alt: "This figure has the strategy for factoring polynomials. At the top of the figure is GCF. Below this, there are three options. The first is binomial. The second is trinomial. Under trinomial there are x squared + b x + c and a x squared + b x +c. The two methods here are trial and error and the “a c” method. The third option is for more than three terms. It is grouping.", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Choose a strategy to factor polynomials completely (updated).] + Is there a greatest common factor? - Factor it. + Is the polynomial a binomial, trinomial, or are there more than three terms? - If it is a binomial, right now we have no method to factor it. - If it is a trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] #linebreak() Undo FOIL #math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1em) ) ( x #h(1em) )$]. - If it is a trinomial of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] #linebreak() Use Trial and Error or the “ac” method. - If it has more than three terms #linebreak() Use the grouping method. + Check by multiplying the factors. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with factoring trinomials of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$]. - #link("https://openstax.org/l/25FactorTrinom")[Factoring Trinomials, a is not 1] ] === Key Concepts - #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:] + Write the trinomial in descending order of degrees. + 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[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 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")[$"Multiply to" #h(0.2em) a c & & & m · n = a · c \ "Add to" #h(0.2em) b & & & m + n = b$] + Split the middle term using #emph[m] and #emph[n]: #linebreak() #figure(figph[This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.], alt: "This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.", caption: none) + Factor by grouping. + Check by multiplying the factors. - #strong[Choose a strategy to factor polynomials completely (updated):] + Is there a greatest common factor? Factor it. + Is the polynomial a binomial, trinomial, or are there more than three terms? #linebreak() If it is a binomial, right now we have no method to factor it. #linebreak() If it is a trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] #linebreak()    Undo FOIL #math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1em) ) ( x #h(1em) )$]. #linebreak() If it is a trinomial of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] #linebreak()    Use Trial and Error or the “ac” method. #linebreak() If it has more than three terms #linebreak()    Use the grouping method. + Check by multiplying the factors. ==== Practice Makes Perfect #strong[Recognize a Preliminary Strategy to Factor Polynomials Completely] In the following exercises, identify the best method to use to factor each polynomial. + ⓐ #math.equation(block: false, alt: "10 q squared plus 50")[$10 q^(2) + 50$] + ⓑ #math.equation(block: false, alt: "a squared minus 5 a minus 14")[$a^(2) − 5 a − 14$] + ⓒ #math.equation(block: false, alt: "u v plus 2 u plus 3 v plus 6")[$u v + 2 u + 3 v + 6$] #solutionbox[ ⓐ factor the GCF, binomial ⓑ Undo FOIL ⓒ factor by grouping ] + ⓐ #math.equation(block: false, alt: "n squared plus 10 n plus 24")[$n^(2) + 10 n + 24$] + ⓑ #math.equation(block: false, alt: "8 u squared plus 16")[$8 u^(2) + 16$] + ⓒ #math.equation(block: false, alt: "p q plus 5 p plus 2 q plus 10")[$p q + 5 p + 2 q + 10$] + ⓐ #math.equation(block: false, alt: "x squared plus 4 x minus 21")[$x^(2) + 4 x − 21$] + ⓑ #math.equation(block: false, alt: "a b plus 10 b plus 4 a plus 40")[$a b + 10 b + 4 a + 40$] + ⓒ #math.equation(block: false, alt: "6 c squared plus 24")[$6 c^(2) + 24$] #solutionbox[ ⓐ undo FOIL ⓑ factor by grouping ⓒ factor the GCF, binomial ] + ⓐ #math.equation(block: false, alt: "20 x squared plus 100")[$20 x^(2) + 100$] + ⓑ #math.equation(block: false, alt: "u v plus 6 u plus 4 v plus 24")[$u v + 6 u + 4 v + 24$] + ⓒ #math.equation(block: false, alt: "y squared minus 8 y plus 15")[$y^(2) − 8 y + 15$] #strong[Factor Trinomials of the form #math.equation(block: false, alt: "a x squared plus b x plus c")[$a x^(2) + b x + c$] with a GCF] In the following exercises, factor completely. #math.equation(block: true, alt: "5 x squared plus 35 x plus 30")[$5 x^(2) + 35 x + 30$] #solutionbox[ #math.equation(block: true, alt: "5 open parenthesis x plus 1 close parenthesis open parenthesis x plus 6 close parenthesis")[$5 ( x + 1 ) ( x + 6 )$] ] #math.equation(block: true, alt: "12 s squared plus 24 s plus 12")[$12 s^(2) + 24 s + 12$] #math.equation(block: true, alt: "2 z squared minus 2 z minus 24")[$2 z^(2) − 2 z − 24$] #solutionbox[ #math.equation(block: true, alt: "2 open parenthesis z minus 4 close parenthesis open parenthesis z plus 3 close parenthesis")[$2 ( z − 4 ) ( z + 3 )$] ] #math.equation(block: true, alt: "3 u squared minus 12 u minus 36")[$3 u^(2) − 12 u − 36$] #math.equation(block: true, alt: "7 v squared minus 63 v plus 56")[$7 v^(2) − 63 v + 56$] #solutionbox[ #math.equation(block: true, alt: "7 open parenthesis v minus 1 close parenthesis open parenthesis v minus 8 close parenthesis")[$7 ( v − 1 ) ( v − 8 )$] ] #math.equation(block: true, alt: "5 w squared minus 30 w plus 45")[$5 w^(2) − 30 w + 45$] #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")[$6 y^(4) + 12 y^(3) − 48 y^(2)$] #strong[Factor Trinomials Using Trial and Error] In the following exercises, factor. #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: "6 p squared minus 19 p plus 10")[$6 p^(2) − 19 p + 10$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 p minus 2 close parenthesis open parenthesis 2 p minus 5 close parenthesis")[$( 3 p − 2 ) ( 2 p − 5 )$] ] #math.equation(block: true, alt: "21 m squared minus 29 m plus 10")[$21 m^(2) − 29 m + 10$] #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: "4 p squared plus 17 p minus 15")[$4 p^(2) + 17 p − 15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 p minus 3 close parenthesis open parenthesis p plus 5 close parenthesis")[$( 4 p − 3 ) ( p + 5 )$] ] #math.equation(block: true, alt: "6 u squared plus 5 u minus 14")[$6 u^(2) + 5 u − 14$] #math.equation(block: true, alt: "16 x squared minus 32 x plus 16")[$16 x^(2) − 32 x + 16$] #solutionbox[ #math.equation(block: true, alt: "16 open parenthesis x minus 1 close parenthesis open parenthesis x minus 1 close parenthesis")[$16 ( x − 1 ) ( x − 1 )$] ] #math.equation(block: true, alt: "81 a squared plus 153 a minus 18")[$81 a^(2) + 153 a − 18$] #math.equation(block: true, alt: "30 q cubed plus 140 q squared plus 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 plus 30 y squared minus 35 y")[$5 y^(3) + 30 y^(2) − 35 y$] #strong[Factor Trinomials using the ‘ac’ Method] In the following exercises, factor. #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: "9 z squared plus 15 z plus 4")[$9 z^(2) + 15 z + 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 z plus 1 close parenthesis open parenthesis 3 z plus 4 close parenthesis")[$( 3 z + 1 ) ( 3 z + 4 )$] ] #math.equation(block: true, alt: "3 m squared plus 26 m plus 48")[$3 m^(2) + 26 m + 48$] #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: "4 q squared minus 9 q plus 5")[$4 q^(2) − 9 q + 5$] #math.equation(block: true, alt: "5 s squared minus 9 s plus 4")[$5 s^(2) − 9 s + 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 s minus 4 close parenthesis open parenthesis s minus 1 close parenthesis")[$( 5 s − 4 ) ( s − 1 )$] ] #math.equation(block: true, alt: "4 r squared minus 20 r plus 25")[$4 r^(2) − 20 r + 25$] #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: "3 x squared plus 5 x plus 4")[$3 x^(2) + 5 x + 4$] #solutionbox[ prime ] #math.equation(block: true, alt: "4 y squared plus 15 y plus 6")[$4 y^(2) + 15 y + 6$] #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[Mixed Practice] In the following exercises, factor. #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: "a squared minus a minus 20")[$a^(2) − a − 20$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis a minus 5 close parenthesis open parenthesis a plus 4 close parenthesis")[$( a − 5 ) ( a + 4 )$] ] #math.equation(block: true, alt: "m squared minus m minus 12")[$m^(2) − m − 12$] #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: "12 y squared minus 37 y plus 21")[$12 y^(2) − 37 y + 21$] #math.equation(block: true, alt: "2 p squared plus 4 p plus 3")[$2 p^(2) + 4 p + 3$] #solutionbox[ prime ] #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: "x squared plus 3 x minus 28")[$x^(2) + 3 x − 28$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 7 close parenthesis open parenthesis x minus 4 close parenthesis")[$( x + 7 ) ( x − 4 )$] ] #math.equation(block: true, alt: "6 u squared plus 7 u minus 5")[$6 u^(2) + 7 u − 5$] #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 squared plus 2 x minus 24")[$x^(2) + 2 x − 24$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis x plus 6 close parenthesis open parenthesis x minus 4 close parenthesis")[$( x + 6 ) ( x − 4 )$] ] #math.equation(block: true, alt: "2 b squared minus 7 b plus 4")[$2 b^(2) − 7 b + 4$] ==== Everyday Math #strong[Height of a toy rocket] The height of a toy rocket launched with an initial speed of 80 feet per second from the balcony of an apartment building is related to the number of seconds, #emph[t], since it is launched by the trinomial #math.equation(block: false, alt: "−16 t squared plus 80 t plus 96")[$−16 t^(2) + 80 t + 96$]. Completely factor the trinomial. #solutionbox[ #math.equation(block: true, alt: "−16 open parenthesis t minus 6 close parenthesis open parenthesis t plus 1 close parenthesis")[$−16 ( t − 6 ) ( t + 1 )$] ] #strong[Height of a beach ball] The height of a beach ball tossed up with an initial speed of 12 feet per second from a height of 4 feet is related to the number of seconds, #emph[t], since it is tossed by the trinomial #math.equation(block: false, alt: "−16 t squared plus 12 t plus 4")[$−16 t^(2) + 12 t + 4$]. Completely factor the trinomial. ==== Writing Exercises 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 may vary. ] How is the “ac” method similar to the “undo FOIL” method? How is it different? What are the questions, in order, that you ask yourself as you start to factor a polynomial? What do you need to do as a result of the answer to each question? #solutionbox[ Answers may vary. ] On your paper draw the chart that summarizes the factoring strategy. Try to do it without looking at the book. When you are done, look back at the book to finish it or verify it. ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has the following statements all to be preceded by “I can…”. The first row is “recognize a preliminary strategy to factor polynomials completely”. The second row is “factor trinomials of the form a x ^ 2 + b x + c with a GCF”. The third row is “factor trinomials using trial and error”. And the fourth row is “factor trinomials using the “ac” method”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.], alt: "This table has the following statements all to be preceded by “I can…”. The first row is “recognize a preliminary strategy to factor polynomials completely”. The second row is “factor trinomials of the form a x ^ 2 + b x + c with a GCF”. The third row is “factor trinomials using trial and error”. And the fourth row is “factor trinomials using the “ac” method”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.", caption: none) ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?