Factoring Polynomials


 Nigel Nicholson
 5 years ago
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1 Factoring Polynomials Any Any Any natural number that that that greater greater than than than 1 1can can 1 be can be be factored into into into a a a product of of of prime prime numbers. For For For example 0 0 = 0 = = ()()(5) and and30 and 30 = 30 = ()(3)(5) /\ /\ Lf..5. IS /\ /\ 35 In Inth th th chapter chapter we ll we ll we ll learn learn an an analogous way way way to to factor factor polynomials. Fundamental Theorem of of of Algebra AA Amonic monic polynomial a a a polynomial whose whose whose leading leading coe coecoefficient cient cient equals equals equals So1. So So x 4 x 4 x x 3 +5x 3 +5x 7monic,andx monic,but3x 4notmonic. x3 + 5x 7 x 3x 4 monic. The Carl Carl following Friedrich Friedrich result Gauss Gauss tells was was usthe how the boy to boy who factor who polynomials. dcovered a really a really It essentially quick quick way way tells to to see see usthat what that 1 + the prime =5050. polynomials + = are: In In1799, 1799, a a grownup Gauss Gauss proved proved the the following following theorem: theorem: Any polynomial the product of a real number, Any and a Any polynomial collection polynomial the of monic the product quadratic product of polynomials of a real a real number, number, that and a collection of monic quadratic polynomials that do not and have a collection roots, and of monic of monic quadratic linear polynomials. that do do not not have have roots, roots, and and of of monic monic linear linear polynomials. polynomials. Th Th result called the Fundamental Theorem of Algebra. It one of the most Thresult important result called calledthe resultsthe Fundamental in Fundamental Theorem all of mathematics, Theoremof though of Algebra. Algebra. It from the It one form one of of the it s themost written most important inimportant results above, it sresults in probably inallall of di of mathematics, cult mathematics, though to immediately thoughfrom understand fromthetheform itsform it s importance. it swritten writtenin inabove, it s it sprobably probablydi difficult to toimmediately understand understanditsits importance. The importance. The explanation explanation for for why th th theorem theorem true true somewhat somewhat di di cult, and and it it beyond The beyond the explanation the scope scope of for of thwhy th course. th course. We ll theorem We ll have have to true to accept somewhat accept it it on on faith. difficult, and it beyond the scope of th course. We ll have to accept it on faith. faith. Examples. Examples. Examples. 4x 4x 1x 1x 4x +8canbefactoredintoaproductofanumber,4,and 1x a of a number, and two twomonic moniclinear linear polynomials, x x 11and andx x.. That That,, 4x 4x two 1x monic 1x linear polynomials, 1)(x 1)(x ). x 1 and x. That, 4~_ 1x+8=4(x 1)(x ). )
2 x 5 +x 4 7x 3 +14x 10x +0canbefactoredintoaproduct of a number, 1, a monic linear polynomial, x, and two monic quadratic polynomials that don t have roots, x + and x +5. That x 5 +x 4 7x 3 +14x 10x +0= (x )(x +)(x +5). (We can check the dcriminants of x +andx +5toseethat these two quadratics don t have roots.) x 4 x 3 +14x 6x +4=(x +3)(x x + 4). Again, x +3 and x x +4donothaveroots. Notice that in each of the above examples, the real number that appears in the product of polynomials 4 in the first example, 1inthesecond, and in the third the same as the leading coe cient for the original polynomial. Th always happens, so the Fundamental Theorem of Algebra can be more precely stated as follows: If p(x) =a n x n + a n 1 x n a 0, then p(x) theproductoftherealnumbera n, and a collection of monic quadratic polynomials that do not have roots, and of monic linear polynomials. Completely factored Apolynomialcompletely factored if it written as a product of a real number (which will be the same number as the leading coe cient of the polynomial), and a collection of monic quadratic polynomials that do not have roots, and of monic linear polynomials. Looking at the examples above, 4(x 1)(x ) and (x )(x +)(x +5) and (x +3)(x x +4)arecompletelyfactored. One reason it s nice to completely factor a polynomial because if you do, then it s easy to read o what the roots of the polynomial are. Example. Suppose p(x) = x 5 +10x 4 +x 3 38x +4x 48. Written in th form, its di cult to see what the roots of p(x) are. Butafterbeing completely factored, p(x) = (x +)(x 3)(x 4)(x +1). Therootsof 158
3 th polynomial can be read from the monic linear factors. They are, 3, and 4. (Notice that p(x) = (x +)(x 3)(x 4)(x +1)completelyfactored because x +1hasnoroots.) * * * * * * * * * * * * * Factoring linears To completely factor a linear polynomial, just factor out its leading coe  cient: ax + b = a x + b a For example, to completely factor x +6,writeitastheproduct(x +3). Factoring quadratics What a completely factored quadratic polynomial looks like will depend on how many roots it has. 0Roots.If the quadratic polynomial ax + bx + c has 0 roots, then it can be completely factored by factoring out the leading coe cient: ax + bx + c = a x + b a x + c a (The graphs of ax +bx+c and x + b a x+ c a di er by a vertical stretch or shrink that depends on a. A vertical stretch or shrink of a graph won t change the number of xintercepts, so x + b a x + c a won t have any roots since ax + bx + c doesn t have any roots. Thus, x + b a x + c a completely factored.) Example. The dcriminant of 4x x+ equals ( ) 4(4)() = 4 3 = 8, a negative number. Therefore, 4x x + has no roots, and it completely factored as 4(x 1 x + 1 ). Roots. If the quadratic polynomial ax + bx + c has roots, we can name them 1 and. Roots give linear factors, so we know that (x 1 ) 159
4 and (x )arefactorsofax + bx + c. That means that there some polynomial q(x) suchthat ax + bx + c = q(x)(x 1 )(x ) The degree of ax + bx + c equals. Because the sum of the degrees of the factors equals the degree of the product, we know that the degree of q(x) plus the degree of (x 1 )plusthedegreeof(x ) equals. In other words, the degree of q(x) plus1plus1equals. Zero the only number that you can add to to get, so q(x) must have degree 0, which means that q(x) justaconstantnumber. Because the leading term of ax + bx + c namely ax theproductof the leading terms of q(x), (x 1 ), and (x ) namelythenumberq(x), x, andx itmustbethatq(x) =a. Therefore, ax + bx + c = a(x 1 )(x ) Example. The dcriminant of x +4x equals4 4()( ) = = 3, a positive number, so there are two roots. We can use the quadratic formula to find the two roots, but before we do, it s best to simplify the square root of the dcriminant: p 3 = p (4)(4)() = 4 p. Now we use the quadratic formula to find that the roots are and 4+4 p () 4 4 p () = 4+4p 4 = 4 4p 4 = 1+ p = 1 Therefore, x +4x completelyfactoredas x ( 1+ p p ) x ( 1 ) =(x +1 p p )(x +1+ p ) 1Root.If ax + bx + c has exactly 1 root (let s call it 1 )then(x afactorofax + bx + c. Hence, ax + bx + c = g(x)(x 1 ) for some polynomial g(x) )
5 Because the degree of a product the sum of the degrees of the factors, g(x) mustbeadegree1polynomial,anditcanbecompletelyfactoredinto something of the form (x ) where, R. Therefore, ax + bx + c = (x )(x 1 ) Notice that a root of (x )(x 1 ), so a root of ax + bx + c since they are the same polynomial. But we know that ax + bx + c has only one root, namely 1, so must equal 1. That means that ax + bx + c = (x 1 )(x 1 ) The leading term of ax +bx+c ax. The leading term of (x 1 )(x 1 ) x. Since ax + bx + c equals (x 1 )(x 1 ), they must have the same leading term. Therefore, ax = x. Hence, a =. Replace with a in the equation above, and we are left with ax + bx + c = a(x 1 )(x 1 ) Example. The dcriminant of 3x 6x+3 equals ( 6) 4(3)(3) = = 0, so there exactly one root. We find the root using the quadratic formula: ( 6) + p 0 (3) = 6 6 =1 Therefore, 3x 6x +3completelyfactoredas3(x 1)(x 1). Summary. The following chart summarizes the dcussion above. roots of ax + bx + c completely factored form of ax + bx + c no roots a(x + b a x + c a ) roots: 1 and a(x 1 )(x ) 1root: 1 a(x 1 )(x 1 ) 161
6 * * * * * * * * * * * * * Factors in Z Recall that the factors of an integer n are all of the integers k such that n = mk for some third integer m. Examples. 1 = 3 4, so 4 a factor of = 15, so 15 a factor of 30. 1, 1, n and n are all factors of an integer n. That s because n = n 1andn =( n)( 1). Important special case. If 1,,... n Z, then each of these numbers are factors of the product 1 n. For example,, 10, and 7 are each factors of 10 7=140. Check factors of degree 0 coe roots If k, 1,and are all integers, then the polynomial cient when searching for q(x) =k(x 1 )(x )=kx k( 1 + )x + k 1 has 1 and as roots, and each of these roots are factors of the degree 0 coe cient of q(x). (The degree 0 coe cient k 1.) More generally, if k, 1,,..., n Z, then the degree 0 coe polynomial g(x) =k(x 1 )(x ) (x n ) cient of the equals k 1 n. That means that each of the roots of g(x) whichare the i arefactorsofthedegree0coe cient of g(x). Now it s not true that every polynomial has integer roots, but many of the polynomials you will come across do, so the two paragraphs above o er a powerful hint as to what the roots of a polynomial might be. 16
7 When searching for roots of a polynomial whose coe cients are all integers, check the factors of the degree 0 coe cient. Example. 3and 7arebothrootsof(x 3)(x +7). Notice that (x 3)(x +7)=x +8x 4, and that 3 and 7areboth factors of 4. Example. Suppose p(x) =3x 4 +3x 3 3x +3x 6. Th a degree 4 polynomial, so it will have at most 4 roots. There n t a really easy way to find the roots of a degree 4 polynomial, so to find the roots of p(x), we have to start by guessing. The degree 0 coe cient of p(x) 6, so a good place to check for roots in the factors of 6. The factors of 6are1, 1,,, 3, 3, 6, and 6, so we have eight quick candidates for what the roots of p(x) mightbe. Aquickcheckshows that of these eight candidates, exactly two are roots of p(x) namely 1 and. That to say, p(1) = 0 and p( ) = 0. * * * * * * * * * * * * * Factoring cubics It follows from the Fundamental Theorem of Algebra that a cubic polynomial either the product of a constant and three linear polynomials, or else it the product of a constant, one linear polynomial, and one quadratic polynomial that has no roots. In either case, any cubic polynomial guaranteed to have a linear factor, and thus guaranteed to have a root. You re going to have to guess what that root by looking at the factors of the degree 0 coe cient. (There a cubicformula thatlikethequadraticformulawilltellyoutherootsof acubic,buttheformuladi cult to remember, and you d need to know about complex numbers to be able to use it.) Once you ve found a root, factor out the linear factor that the root gives you. You will now be able to write the cubic as a product of a monic linear 163
8 factors of 3 are 1, 1, 3, and 3. Check these factors to see if any of them are roots. After checking, you ll see that 1 a root. That means that x 1 a factor of x3 3x + 4x 3. Therefore, we can divide x3 3x + 4x 3 by x 1 polynomial to get another and polynomial a quadratic polynomial. Completely factor the quadratic and then polynomial you will and have a quadratic completely x3 3x+4x 3 polynomial. factored the Completely cubic. factor the quadratic and then you will have completely factored the =x cubic. x+3 Problem. Completely factor x 3 3x +4x 3. Problem. Thus, Completely factor x3 3~ + 4x 3. Solution. Start x3 3x+4x 3=(x 1)(x x+3) by guessing a root. The degree 0 coe cient 3, and the factors Solution. of 3are1, Start by guessing 1, 3, anda root. 3. Check The these degreefactors 0 coefficient to see if any 3, of and them the are factors roots. of 3 are 1, 1, 3, and 3. Check these to see if any of them are After roots. checking, you ll see that 1 a root. That means that x 1afactor of After x 3 checking, 3x +4x you ll 3. Therefore, see that 1we acan root. divide That xmeans 3 3x that +4xx 13byx a factor 1 to of get x3 another 3x + polynomial 4x 3. Therefore, divide x3 3x + 4x 3 by x 1 to get another polynomial x 3 3x +4x 3 =x x +3 The dcriminant ofx3 3x+4x 3 x x x equals (_1) 4()(3) = 1 4 = =x x+3 3, Thus, a negative number. Therefore, x x + 3 has no roots, so to completely Thus, factor x x 3 3x +4x 3 = (x 1)(x x +3 we just have to factor out the leading x +3) coefficient as follows: x x13=(x x3 3x+4x 3=(x 1)(x x+3) ~x+~). x33x1 ~~+z3 (xl) /\ a (zx_x+3 ) /N The dcriminant of x x +3equals( 1) 4()(3) = 1 4 = 3, anegativenumber. The dcriminant of Therefore,x x x + 3 equals x +3(_1) has no 4()(3) roots, so = to 1 completely 4 = 3, factor a negative x number. x +3 we just Therefore, have tox factor x out + 3the has leading no roots, coe so cient to as completely follows: x factor The x x final +3= answer x 1 x + 3. x +3 we just have to factor out leading coefficient as follows: x x13=(x ~x+~). (x_1)(x_ ~+~) 137 x33x1 ~~+z3 (xl) /\ a (zx_x+3 ) /N The final answer The final answer (x 1) x 1 x + 3 (x_1)(x_ ~+~)
9 3x3 3x 15xI6 = (x+)(3x 9x+3) Problem. Completely factor 3x 3 3x 15x +6. Problem. Completely factor 3w3 3w 15w + 6. Solution. The factors of 6 are {1, 1,,, 3, 3, 6, 6}. Check to see that Solution. aroot.thendividebyx The factors of 6 are {1, +tofindthat 1,,,3, 3,6, 6}. Check to see that The dcriminant of 3w 9w +3 equals 45, and thus 3w 9w +3 has two a root. Then divide by x + to find that roots and can be factored 3x 3 further. 3x 15x +6 The leading coefficient 3w3 of = 3x 3x 9x +3 x 3w 15w + 9w , and we can use the quadratic = 3w 9x+3 formula so to check that the roots x+of 3w 9w + 3 are ~ and~ From what we so learned above 3x about 3 3x factoring 15x +6 quadratics, = (x +)(3x we know 9x that +3) 3w 9w + 3 3(w ~~ ~)(x ~ 3x3 3x 15xI6 V s) = (x+)(3x 9x+3) 3x33x15i /\ (x+~ (3x_cbc. / The dcriminant of 3x 39x +3equals45, c~ / 3+S\,~i apositivenumber, / andthus 3x The dcriminant 9x +3hastworootsandcanbefactoredfurther. of 3w 9w +3 equals 45, and thus 3w 9w +3 has two roots The and leading can be coe factored cient further. of 3x 9x +33,andwecanusethequadratic formula To The summarize, leading coefficient of 3w to check that the roots of 3x 9x+3 are 3+p 5 and 3 p 9w + 3 3, and we can use the 5 quadratic. From what formula to check that the roots of 3w 9w + 3 are ~ and~ we learned 3w3 above about factoring quadratics, we know that 3x From what 9x +3= 3+ 3(x p 5 3 )(x p 3w 15w + 6 = (w + )(3w we learned above about 9w + 3) 5 factoring quadratics, we know that 3w 9w + 3 ). 3(w ~~ ~)(x ~ V s) I 3+~/~~i 3 Vg =(wi)3~çw )çw 3x33x15i / 3+v g ~,i 3 ~./~ =3(w+)~w )k~x (x+~ /\ (3x_cbc. / / 3+S\ / 3 c~,~i To summarize, To summarize, 3x 3 3x 15x +6=(x +)(3x 9x +3) 3w3 3+ p 3w 15w + 6 = (w + )(3w 9w + 3) 5 p 3 5 =(x +)3I x 3+~/~~i x 3 Vg =(wi)3~çw 3+ p )çw 5 p 3 5 =3(x +) / x 3+v g ~,i x 3 ~./~ =3(w+)~w )k~x 165
10 Factoring quartics Degree 4 polynomials are tricky. As with cubic polynomials, you should begin by checking whether the factors of the degree 0 coe cient are roots. If one of them a root, then you can use the same basic steps that we used with cubic polynomials to completely factor the polynomial. The problem with degree 4 polynomials that there s no reason that a degree 4 polynomial has to have any roots take (x +1)(x +1) for example. Because a degree 4 polynomial might not have any roots, it might not have any linear factors, and it s very hard to guess which quadratic polynomials it might have as factors. * * * * * * * * * * * * * 166
11 Exerces Completely factor the polynomials given in #18 1.) 10x +0.) x +5 3.) x 1x 18 4.) 10x +3 5.) 3x 10x +5 6.) 3x 4x +5 7.) x +6x 3 8.) 5x +3x 9.) Find a root of x 3 5x +10x ) Find a root of 15x 3 +35x +30x ) Find a root of x 3 x x 3. Completely factor the polynomials in # ) x 3 x + x ) 5x 3 9x +8x 0 14.) x 3 +17x 3 15.) 4x 3 0x +5x 3 16.) x 4 5x ) How can the Fundamental Theorem of Algebra be used to show that any polynomial of odd degree has at least one root? 167
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