P.6rrr FactoringPolynomials 571

FactoringPolynomiats

In SectionP.5 we studiedmultiplication of polynomials.In this sectionwe will fac- tor polynomials.Factoring "reverses" multiplication. By factoring,we canexpress a complicatedpolynomial as a productof severalsimpler expressionsn which areoften easierto study.

Factoring Out the GreatestCommon Factor To factor 6x2 - 3x, notice that 3.r is a monomial that can be divided evenly into eachterm. We canuse the distributiveproperty to wdte

6x2- 3x :3x(2x - l). - We call this processfactoring out 3x. Both 3r and2x - I arefactors of 6x2 3x. Since3 is a factorof 6x2 and3r, 3 is a commonfactor of the termsof the polyno- mial. The greatestcommon factor (GCF) is a monomialthat includesevery num- ber andvariable that is a factorof all termsof the polynomial.The monomial3x is the greatestcofilmon factar of 6x2 - 3x. Usually the commonfactor has a positive coefficient,but at times it is useful to factor out a commonfactor with a negativeco- efficient.For example,we couldfactor out -3x from 6x2 - 3x:

6x2- 3x: -3x(-2x + l)

fua*t*/a ! nattorturgout the greatest commonfa$or Factor out the greatestcornmon factor from eachpolynomial, first using the GCF with a positive coefficient and then using a negativecoefficient. a. 9xa - 6x3+ l2x2 b. ,2y'l l\xy + 25y c. -z - w 58 ChapterPrrr Prerequisites

Sotution a. 9xa- 6x3+ tz"* : z*(z - zx + 4) : -l*(-l* + 2.x- 4) b. xzy-t llxy + 25y: y(f + llx + 25) : _y(_f _ lox _ 25) c.-z-w:l(-z-w) : -l(z + w)

77V 7fu1. FactorouttheGCFin-12a3 + 8a2- l6a.

Factoringby Grouping Somefour-term polynomials can be factoredby groupingthe terms in pairsand fac- toring out a commonfactor from eachpair of terms.We showthis techniquein the next example.As we will seelater, grouping can alsobe usedto factortrinomials.

fuan/a l| ractoringfour-term polynomials Factoreach polynomial by grouping. a. x3 + x2 + 3x + 3 b. aw * bc - bw - ac

Solution a. Factorthe common factorx2 out of the first two terms and the commonfactor 3 out of the lasttwo terms: x3+* * 3x* 3:f(x + 1) + 3(x+ l) Factoroutcommonfactors.

:(x+lxl+3) Factor out the common factor (x f l), b. We must first arrangethe polynomial so that the first group of two terms has a commonfactor and the last group of two terms also hasa commonfactor: aw I bc - bw - ac: ctw- bw - ac I bc Rearranse.

: w(a - b) * c(a - b) Factor out common factors

:(w-c)(a-b) Factorout (a - b). Try Tfus. Factorw3+ wz - 3w - 3.

Factoringax2 * bx * c Factoring by grouping was used in Example 2 to factor polynomials with four terms. Factoring by grouping can be used also to factor a trinomial that is the product of two binomials, becausethere are four terms in the product of two binomials (FOIL). If we examine the multiplication of two binomials using the distributive property, we can develop a strategy for factoring trinomials. (r + 3)(x + 5): (x + 3)r + (x + 3)5 Distributiveproperty : f + 3x * 5x * 15 Distributiveproperty

:f + 8x*15 Combineliketerms. P.5r rn FactoringPolynomials 59

To factor x' j 8x * 15 we simply reverse the steps in this multiplication. First write 8x as 3x * 5x, and then factor by grouping. We could write 8x as x *'7x,2x -f 6x,and so on, butwe choose3r * 5xbecause3 and 5 haveaproduct of15. Sotofactorx2 * 8x * 15proceedasfollows:

x2 + 8x+ 15 : xz + 3x* 5r * 15 ReplaceSxbv3xr5-r.

: (r * 3)x + (x + 3)5 Factorout commonfactors

:(x+3)(x+5) Factoroutr + 3.

The key to factoring axz + bx 1 c with a : 1 is to find two numbers that have a product equal to c and a sum equal to D, then proceed as above. After you have prac- ticed this procedure,you should skip the middle two stepsshown here and just write the answer. fuawla p ractoringax2 * bx * c with a = ! Factoreach trinomial. a.x2-5x-14 b.x2+4x-2l c,x2-sx-l 6

Solutlon a. Twonumbers that havea productof - 14and a sumof -5 are -7 and2.

xz - 5x - 14: xz - 7x * 2x - 14 Reolace-5.rbv-7x*2x.

: (r - 7)x + (x - 7)2 Factoroutcommonfactors.

:(x-7)(x+2) Factorout (x - 7). Checkby usingFOIL. b. Two numbersthat havea product of -21 and a sum of 4 are7 and -3. x2+4x-2t:(x+7)(x-3) Checkby usingFOIL. c. Twonumbers that have a productof 6 anda sumof -5 are -2 and-3.

x2-Sx *6: (x-2)(x-3)

Checkby usingFOIL. fty Tirt. Factorx2 - 3x - 70.

Beforetrying to factoraxz I bx * c with a * 1, considerthe productof two linearfactors:

(Mx + N)(Px + Q): MP* + (MQ + NP)x + NQ

Observethat the product of MQ and NP (from the coefficient of x) is MNPQ. The product of MP (the leading coefficient) and,NQ (the constantterm) is alsoMNPQ. So if the trinomial axz + bx I c canbe factored there are two numbersthat havea sum equalto b (the coefficientofx) and a productequal to ac (lhe productofthe leading coefficient and the constantterm). This fact is the lqeyto the ac-method for factoring ax2 + bx + c. 60 ChapterP r rr Prerequisites

ffireffiffiffiffiffiffiffi The ac-Method for Factoring Tofactoraxz + bx + cwith a+ 1i 1. Findtwo numberswhose sum isb and whoseproduct is ac. 2. Replaceb by the sumof thesetwo numbers. 3. Factorthe resultingfour-term polynomial by grouping.

fuawla I ractoringax2 + bx* c with a + t Factor each trinomial.

a.2x2 *5r*2 b.6x2-x-12 c. l5x2-l4x-l 3

Solution a. Sinceac : 2 . 2 : 4 andb : 5, we needtwo numbersthathaveaproduct of4 and a sum of5. The numbersare 4 and 1.

2xz + 5x -l 2 : 2xz + 4x * x -l 2 Replace 5n by 4;r -l- ;r.

: (x + 2)2x + (x + 2)l Factorbygrouping

: (r + 2)(2x+ l) Chcckusing FOIL.

b. Sinceac:6(-12): -72 andb: -1, we needtwo numbersthat havea productof -72 and a sum of - l. The numbersare 8 and -9.

6x2 - x - 12 : 6x2 - 9x t 8x - 12 Replace :r by -9:r f 8"r

: (2x - 3)3x + (2x - 3)4 Factorby grouping.

: (2x - 3)(3x+ 4) Checkusing FOIL. c. Sinceac : 15 . 3 : 45 and b = - 14, we needtwo numbersthat have a prod- uct of 45 and a sum of - 14.The numbersare -5 and -9.

75x2- l4x * 3 : l5x2 - 5x - 9x -l 3 Rcpface- l4x by -5x - 9r

: (3.r - 1)5x + (3x - l)(-3) Factorbygrouping

: (3x - 1)(5x - 3) checkusingFolL.

TrV 1hs. Factorl2x2 * 5x - 3.

After factoring some polynomials by the grouping method of Example 3 and the ac-method of Example 4, you will find that you can often guess the factors without going through all of the steps in those methods. Guessing the factors is an acceptablemethod that is called the trial-and-error method. For this method you simply write down any factors that you think might work, use FOIL to check if they are correct, then try again if they are wrong.

fua.*V/e fl Factoringby trial and error Use trial and error to factor 2x' + llx - 6. P.6rrr Factoringpolynomials61

Sotution

since2xzis2x'x and 6 is either2 . 3 or I . 6, thereareseveralpossibilitiestotry for the factors:

(2x l)(x 6) (2x 2)(x 3) (2x 6)(x l) (2* 3)(" 2)

The factors must have opposite signs to get the negative sign in the original polyno- mial. You can rule out a binomial that has a common factor such as 2x + 6 because there is no common factor in the originar trinomial. we now use FOIL to multiply some possible factors until we find the correct factors:

(2x + l)(x - 6): 2x2- llx - 6 Error. - (2x l)(x + 6) : 2x2 + llx - 6 Correct. - So2x2 * 11x 6 is factoredas (2x - 1)(x + 6).

TrV 77at. Factor3x2 - 2ox- 7. r

Factoring the Specialproducts

In Section P.5we learned how to find the special products: the squareof a sum, the squareofa difference, and the product ofa sum and a difference. The trinomial that results from squaring a sum or a difference is called a perfect square trinomial. we can write a factoring rule for each special product.

Factoringthe a2-b2:(a*b)(a-b) Differenceof two squares SpecialProducts a2+2ab+b2:(a+b)2 Pcrfectsquare trinomial a2_2ab+b2:(a_b)2 Perfectsquare trinomial

erctfla @ Factoringspecial products Factoreach polynomial. - - a. 4x2 7 b, x2 6x t 9 c. 9y2 + 30y + 25 d. x2t - g Sotution - - a.4xz 1: (Zx)2 12 Recognizethedifferenceoftwosquares. : (2x * t)(2x _ l) - : - b. x2 6x * 9 x2 2(3x) + 32 Recognizethe perfect square trinomiar. :(x-12

: c. 9y2 + 30y + 25 (3y)2 + 2(3y)(5) + 52 Recognizethe perfect square trinomial. : (3y + 5)2

- g : d. xzt (x'- 3)(x/ + :) Recognizethedifferenceoftwosquares.

77V Thlt. Factor4x2 - 28x+ 49. r 62 ChapterP r rr Prerequisites

Factoringthe Difference and Sum of Two Cubes The following formulas are used to factor the difference of two cubes and the sum of two cubes.You should verif' these formulas using multiplication.

Factoringthe Difference o3 - b3 : (a - b)(a2 + ab + bz) and Sumof Two Cubes a3 + b3 : (a + b)(o2 - ab + bz)

ha.*V/e I Factoringdifferences and sumsof cubes Factor each polynomial. a. x3 - 27 b. 8w6 I 12523 c. yt^ - |

Solution a. Sincex3 - 27 : x3 - 33,we use a : x andb : 3 inthe formula forfactoring the difference of two cubes:

x3-27:(r-3)(x2+3x+9)

b. Since 8w6 + 12523: (Zr2)3 + (52)3,we use a : 2wz andb :52 in the for- mula for factoring the sum of two cubes: gw6 + 12523: (2w2 + 5z)(4wa - l0w2z + 25zz)

c. Since y3^ : (y^)3, we can use the formula for the difference of two cubes with e: v^ andb: l: y3^ - | : (y. - t)(y2* + ym + l)

/FV 77ur. Factor8x3 + r2s.

Factoring by Substitution When a polynomial involves a complicated expression,we can use two substitutions to help us factor. First we replace the complicated expression by a single variable and factor the simpler-looking polynomial. Then we replace the single variable by the complicated expression,This method is called substitution.

ha*Vle lt Factoringhigher-degree potynomials Usesubstitution to factoreach polynomial. a. w4- 6wz- t6 b. (o' - l)' + 7(a2- 3) + tz Sotution a. Replace wz by x andwa by x2. wa-6vt?-16:f-ex-rc

: (x - 8)(r + 2) Factorthetrinomial

: (w2 - 8)(w2 + Z) Replacexbyw2. P.5rrm FactoringPolynomials 63

b. Replacea2 - 3bybinthepolynomial. - (o' z)' + 7(a2- 3) + 12: b2+ 7b + 12 : (b + 3)(b+ 4) Factorthe trinornial. :(a2_3+ - 3)(o2 3 + 4) Replacebbyu2 - 3 : a2(a2+ l) Siniplily.

Try Tbl. Factor(wz + D2 - z@2 + t) - 18.

Factoring Gompletety A polynomial is factored when it is written as a product. Unless noted otherwise, our discussionof factoring will continue to be limited to polynomials whose factors have integral coefficients. polynomials that cannot be factored using integral coeffi_ called.prim_e or irreducible over :r;n:r,u." the integers. fo, oi-*^i, D- + b + l' and x * 5 areprime "iu_pI"",- polynomialsbecause they cannotbe expressedas a product (in a nontrivial manner). A polynomial is said to be factored comptetely when it is written as a product of prime polynomials. when factoring polynomials, we usually do not factor integers that are common factors. For exampli'a*;Q* _ ii is factored completely even though the coefficient 4 could be factored.

fua.nF/e p Factoringcomptetely Factoreach polynomial completely. a. 2wa- 32 b. -6x7 + 6x Solution a. 2wa- 32 : z(w4 - rc) Factofout thc greatcstcon.lrnon lactor : - q@2 2(w2 + 4) Difl'clcnccof two squarcs : - 2(w 2)(. + Z)(w2 + 4) Dillbrcnceot.two squarcs The polynomial is now factored completely because w2 + 4is prime. b. -6x7 * 6x : -6x(x6 - l) Gl'Cltest e()t)llilr)n l:tcl(rl : -6x(x3 - l)(x3 + t) Differoncc of two squares :-6x(x-l)(*2*xf l)(x + l)(x2- -r + l) Diff-ercnccol'lwo cubcs;sum of two cubes The polynomial is factored completely becauseall of the factors are prime. 4fV Thl. Factor2ya- 162completely.

If 7 is a factorof r47, thenTis a divisorof r47.If we divide r47by 7 to get2l (with no remainder), :7 .2r. thenwe have r47 By factoring2r,we factorr47 completelyas r47 : '72. 3 The sameideas hord foipolynom[ts. irwe knowone factorof a polynomial, we canuse division to find the otherfactor and thenfactor thepolynomial completely. ChapterPrrr Prerequisites

,ta*V/e @ using division to factor cornptetety Factorx3- 7x * 6completely,giventhatx - 2isafactot.

$alution Uselong divisionto find the otherfactor: x2+2x -3 x- 28\of-7x+6 x3-2f ,t4*uzf-t,

3x*6 0 Becausetheremainderis0,wehave13- 7x * 6: (x -2)(r2 +2x - 3).Fac- toring completelygives x3 _ 7x * 6 : (x _ 2)(x + 3)(r _ t).

7rV 77A1. Factory3- 6y'+ lly - 6giveny - l isafactor.

For Thought Trueor False?Mark true if the polynomialis factoredcorrectly and falseif it is not. Explain.

l. xz + 6x - 16 : x(x t 6) - 16r 6. x3 - y3: (x - y)(xz+ 2xy+ y2)r

2, 2xa - 5x2= -x215 - 2x21r 7. 8a3+ 27b3: (2a + 3b)(4a2+ 6ab + gb2)F

3. 2xa- Sxz- 3 = (x2 - 3)(2x2+ l)r B. xz + 8x - t2: (r * 2)(x- 6)r

4. az_ l: (a _ l)2r 9. (a + 3)6- (a -r 3)x: (a + 3)(6-.r)T

5. a3- t: (a * t)(o2+ l)F 10.a - b: -l(b - a) r

IEl Exercises

Factorout thegreatest common factor from eachpolynomial, first 7. m- nl(m-n),-I(n-m) 8. y*xl(y-x),-1(x-y) usinga positivecofficient on the GCF and thenusing a negative cofficient. (Example1) Factor eachpolynomial by grouping. (Example2)

l, 6x3 - l2x2 2. nxz * 18x3 9. x3 + 2x2+ 5x.+-!0 10. 2# -1w2 + 3w - 3 6*(x - 2), -6f (-x + 2) 6x2(2+ 3x),-6x2(-z - 3x) (2w2'+3)(w - 1) - 3.4a 8ab 4.3wm i l5wm2 -y2 -3y+3 -7x-7 4a(1 - 2b),-4a(2b - l) 3wm(1+ 5m),-3wm(-1 - 5m) ll.y3 12,x3 tx2 (f - t)(y - t) ,- (x2- 7)(x+ r) 5, -ax3 + 5ax2 - 5ax 6, -sa3 * sb3 - sb - - - -f ax(-x2 + 5.r 5), s(-o3 + b3 D, 13. ady w * d- aW .... 14. xy * ab by 4- ax -*(* - 5.r + 5) -s(a3 - b3 + b) (d-w)(ay+r) (x+b)b,+a) P.6rrr Exercises / 15. x2y2 t ab - ay2 - bx2 16. 6yz - 3y - l0z-l 5 58. (t3 + s)z+ 5(/3+ 5) + 4(f + e)(l + 6) - - - - b? b)(f a) (3y s)(22 r) Factor eachpolynomial completely.(E4nmple 9) Factor each trinomial. See the procedurefor factoring by the ac- / 59. -3x3 + 27x 60.a"b'- l6b" method on page 60. @xamples3-5) -3x(x-3)(r+3) b2(a-2)(a+2)(a2+4) 17.x2 + lox + 16 18. .x2+ 7x + 12 61. l6t4 * 54w3t 62. 8a6 - a3b3 ("r+2)(x+8) (x+3)(x+4) 2t(2t + 3w)(4?- 6tu + 9w2) a3(2a- b)(4a2+ 2ab+ b2) 19.x2-4x-12 20.y2+3y-18 63.a3 + a2- 4a- 4 64.2b3+3b2 - ]8b-27 (x-6)(x+2) (y+6)(y-3) (a - 2)(a+ 2)(a+ 1) (b - 3)(b+ 3)Qb+ 3) - 21. m2 12m+ 20 22.n2-8n-t7 65.xa -2x3 -8x+ 16 66.aa - a3+ a- 1. (m-2)(m-10) (n-7)(n-t) (x-2)2(x2 +2x+ 4) (a+ l)(a- 1)(a2- a+ l) 23.t2+5t-84 24.s2-6s-27 -36x3 -6aa - (t-7)(t+12) (s+3)(s-9) 67. + l8x2 + 4x 68. a3 + l5a2 -2x(6x+ l)(3x- 2) -o2(3a+ 5)(2a- 3) 25.2x2-7x-4 26.3x2+5x-2 (2x+I)@-a) (3x-1)(x+2) 69.a1-a6-64a+64 - - - 27. 8x2- 10x - 3 28. 18x2- 15x + 2 (a 2)(a2+ 2a + 4)(a+ 2)(a2 2a+ 4)(a 1) (4x+t)(2x-3) (6x- 1)(3x- 2) 7o.as-4a4-4a+16 ,,/n3o. 29.6y2+7y-5 r5r2 - r4x - 8 (a2-2)(a2+2)(a-4) (3y+s)(2y-r) (3x-4)(5x+2) 71. -6x2 - x * 15 72. -6x2 - 9x + 42 -(3.r + s)(2x- 3) -3(2x + 7)(x- 2) Factor each special product. (Example6) 73. (a2 + 2)2- 4Q2 + 2) + 3 (a - 1)(a+ t)(a2+ t)

31,t2 - u2(t- u)(t+ u) 32.gf - f Gt-v)(3t+v) 74. (23+ 5)2- fiQ3 + 5) + 24 - - 33.t2 + 2t + l(t + t)2 34. m2 * l\m * 25 (m + 5)2 (z 1)(* + z + 1)(z+ l)(/ z + t)

35. 4w2 - 4w * | (zw - t)2 36. 9x2- l?-xy+ 4y2 For eachpair of polynonials,use long divisionto determine - (3x 2y)z whetherthe Jirst polynomial is afactor of thesecond. (Example I 0) 37.ya' - 25 1rz'- 5)(12'+5) 38. 1.21w2t- L (llwt - 1)(1lwr+ 1) 75.x * 3,x3+ 4x2+ 4x * 3yes 39.922x2* 24zx-+16(3zx + 4)240.25t2- 20tw3+ 4w6 (5t - 2w3\2 76.x - 2,x3- 3x2+ 5x- 6Yes

Factor each sum or difference oftvvo cubes. (Example7) 77. x - 1,3x3+ 5x2- lzx- 9No

41. t3 - u3 (t - u)(f - tu - u2) 42, m3 + n3 78. x * 1,2x3+ 4x2- 9x * 2No \m+n)(mz-mn+n2) 43.a3 - 8(a _ 2)(az+ 2a+ 4) ^). ri I r\o, _ t + D y'46.1-Bou In eachpair of polryomials,the second polynomial is afactor of 45.27v3+8 theJirst. Factorthe Jirst polynomial completely. (Example 10) (li+2)(gy2-6y+4) 0 -2a\(t +2a2+ 4a4) - - 47.27x3y6- 8ze 48. st3h:+^n: 79. x3 + 4x2+ x 6,x 1 ("r- l)(.v+ 2)(x+ 3) (lxyz - U3)(gx2y4+ 6ry2i + 4261 Qth + n3)(4fh2- 2thn3+ n6) 80. x3 - 73x- l2,x* I (x + l)(x - 4)(.r+ 3) 49.x3', - 8 50. y3',+ 8 (t - 2)(f' + 2/ + 4) + z)(yr* - 2y. + 4) U; 81. x3 - x2 - 4x - 6,x - 3 (x - 3)(x2+ 2x + 2) Usesibstitution to eachpolyn'omial. (Example 8) factor 82. 2x3t 7x2+ 5x I l,2x + 1 (h+ 1)(x2+ 3:r+ 1) 51. y6 + l0y3 + 25 (y3+ s)2 52. y8 + 8y4+ 12 83. .xa+ 5x3+ 5x2- 5x - 6,x * 2 (ya+2)(ya+6) (x+2)(x+3)(x-l)(x+l) s3. 4a4bg- 8a2b4- 5 54. 3c2nra- 22cm7+ 7 84. 4xa + 4x3- 25xz- x * 6,x - 2 (2a2b4+ t)(2a2b4- 5) (3cm7- l)(cm1- l) (x-2)(x+3)(2x+ 1)(2x-1) 55. (2a + 1)2+ 2(2a + l) - 24 (2a+ 7)(2a- 3)

56.2(w- 3)'+ 3(w- 3) - 2(2w- 7)(w- r)

57. (b2 + 2)2- 5(b2+ 2) + 4@2- D(b2+ D 66 ChapterP I ffiriu Prerequisites

Solve eachproblem. < 6 in.----- >

85, VolumeoJ a Rectangular Box lf the volume of a box is x3 - I ft3 and the height is x - t ft, then what polynomial I representsthe area of the bottom? r: + ,r + 1 t'li;..'i,,tiV: LWH. I 7 in.

T T x x , i v v < r>l * Figurefor Exercises87 and 88

88. Area and Volume Find a polynomial that representsthe area ofthe bottom ofthe box in the accompanying figure. Find a polynomial that representsthe volume of the box 4.rr 2(r.r* 42. :l-rj 26.12+ 42.r n Figurefor Exercise85 For Writing/Discussion

89. Which of the following is not a perfect squaretrinomial? Ex- 86. Area ofaTrapezoid Ifthe areaofa trapezoidis plain your reasoning in writing. b 2x2 + 5x * 2 squaremeters and the two parallel sides are x a. 4x8 - 20x4y3+ 25y6 b. 1000a2- 200ab + b2 meters and x * I meters, then what polynomial representsthe height ofthe trapezoid?2: + 4 c.400wa-40w2+1 d. 36ata- 36a1+ 9 , t_._ A:1h(b1 + b2). 90. Which of the following is not a difference of two squares?Ex- plain your reasoningin writing. c a. 196a6ba- 2ggw8 b. 100x16- 161ttoo

c. 25we - gy2s d. | - 9y36

91. Another Lost Rule? Is it true that the sum of two cubesfac- tors as the cube of a sum?Explain. Give examples.No

92. Cooperative Learning Work in a small group to write a sum- mary of the techniquesused for factoring polynomials. When .r+ I given a polynomial to factor completely, what should you look e Figurefor Exercise86 for first, second,third and so on? Evaluate your summary by using it as a guide to factor somepolynomials selected from the exercises. 87. Making a Box An open-top puzzlebox is to be constructed from a 6 in. by 7 in. piece of cardboardby cutting out squares /,( from each corner and folding up the sides as shown in the fig- fu*tg 9"r?t(rtk Pot ure. Find the volume of the box for x : 0.5 in.. I in.. l4/rongDivisior Is it possible to divide the integers from 1 and 2 in. Which of these values for x produces the largest through 9 into three sets of any size so that the product of the inte- volume?l5 in.r.20in.r. l2 in.-r.I in. gers in eachofthe three setsis lessthan 72? No