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.