Factoring Polynomials Any natural number that is greater than 1 can be factored into a product of prime numbers. For example 20 = (2)(2)(5) and 30 = (2)(3)(5). In this chapter we’ll learn an analogous way to factor polynomials. Degree of a product is the sum of degrees of the factors Let’s take a look at some products of polynomials that we saw before in the chapter on “Basics of Polynomials”: The leading term of (2x2 5x)( 7x +4)is 14x3. This is an example of a degree 2 and a degree 1 polynomial− − whose product− equals 3. Notice that 2+1=3 The product 5(x 2)(x +3)(x2 +3x 7) is a degree 4 polynomial because its leading term is 5−x4. The degrees of− 5, (x 2), (x +3),and(x2 +3x 7) are 0, 1, 1, and 2, respectively. Notice that 0+1+1+2− = 4. − The degrees of (2x3 7), (x5 3x +5),(x 1), and (5x7 +6x 9) are 3, 5, 1, and 7, respectively.− The degree− of their product,− − (2x3 7)(x5 3x +5)(x 1)(5x7 +6x 9), − − − − equals 16 since its leading term is 10x16. Once again, we have that the sum of the degrees of the factors equals the degree of the product: 3+5+1+7 = 16. These three examples suggest a general pattern that always holds for fac- tored polynomials (as long as the factored polynomial does not equal 0): If a polynomial p(x)isfactoredintoaproductofpolynomials, then the degree of p(x)equalsthesumofthedegreesofitsfactors. Examples. The degree of (4x3 +27x 3)(3x6 27x3 +15)equals3+6=9. • − − The degree of 7(x +4)(x 1)(x 3)(x 3)(x2 + 1) equals • 0+1+1+1+1+2=6.− − − − Degree of a polynomial bounds the number of roots Suppose p(x)isapolynomialthathasn roots, and that p(x) is not the constant polynomial p(x)=0.Let’snametherootsofp(x)asα1, α2,...,αn. 122 Any root of p(x)givesalinearfactorofp(x), so p(x)=(x α )(x α ) (x α )q(x) − 1 − 2 ··· − n for some polynomial q(x). Because the degree of a product is the sum of the degrees, the degree of p(x)isatleastn. The degree of p(x)(ifp(x) =0)isgreaterthanor equal to the number of roots that p(x)has. Examples. 5x4 3x3 +2x 17 has at most 4 roots. • − − 4x723 15x52 +37x14 7hasatmost723roots. • − − Aside from the constant polynomial p(x)=0,ifafunctionhasa • graph that has infinitely many x-intercepts, then the function cannot be a polynomial. If it were a polynomial, its number of roots (or alternatively, its number of x-intercepts) would be bounded by the degree of the polynomial, and thus there would only be finitely many x-intercepts. To illustrate, if you are familiar with the graphs of the functions sin(x)andcos(x), then you’ll recall that they each have infinitely many x-intercepts. Thus, they cannot be polynomials. (If you are unfamiliar with sin(x)andcos(x), then you can ignore this paragraph.) ************* Fundamental Theorem of Algebra A monic polynomial is a polynomial whose leading coefficient equals 1. So x4 2x3 +5x 7ismonic,andx 2ismonic,but3x2 4isnotmonic. − − − − Carl Friedrich Gauss was the boy who discovered a really quick way to see that 1 + 2 + 3 + +100=5050. In 1799, a grown-up··· Gauss proved the following theorem: 123 Any polynomial is the product of a real number, and a collection of monic quadratic polynomials that do not have roots, and of monic linear polynomials. This result is called the Fundamental Theorem of Algebra. It is one of the most important results in all of mathematics, though from the form it’s written in above, it’s probably difficult to immediately understand its importance. The explanation for why this theorem is true is somewhat difficult, and it is beyond the scope of this course. We’ll have to accept it on faith. Examples. 4x2 12x +8canbefactoredintoaproductofanumber,4,and • two− monic linear polynomials, x 1andx 2. That is, 4x2 12x +8=4(x 1)(x 2).− − − − − x5 +2x4 7x3 +14x2 10x +20canbefactoredintoaproduct • −of a number,− 1, a monic− linear polynomial, x 2, and two monic quadratic polynomials− that don’t have roots, x2 −+2 and x2 +5. That is x5 +2x4 7x3 +14x2 10x +20= (x 2)(x2 +2)(x2 +5). (We− can check− the discriminants− of x2 +2and− −x2 +5toseethatthe two quadratics don’t have roots.) 2x4 2x3 +14x2 6x +24=2(x2 +3)(x2 x + 4). Again, x2 +3 • and−x2 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 2 in the third – is the same as the leading coefficient− for the original polynomial. This always happens, so the Fundamental Theorem of Algebra can be more precisely stated as follows: n n 1 If p(x)=anx + an 1x − + + a0, − ··· then p(x)istheproductoftherealnumberan, and a collection of monic quadratic polynomials that 124 do not have roots, and of monic linear polynomials. Completely factored Apolynomialiscompletely factored if it is written as a product of a real number (which will be the same number as the leading coefficient 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 2) and (x 2)(x2 +2)(x2 +5) and 2(x2 +3)(x2 x +4)arecompletelyfactored.− − − − − One reason it’s nice to completely factor a polynomial is because if you do, then it’s easy to read off what the roots of the polynomial are. Example. Suppose p(x)= 2x5 +10x4 +2x3 38x2 +4x 48. Written in this form, its difficult to see− what the roots of−p(x)are.Butafterbeing− completely factored, p(x)= 2(x +2)(x 3)(x 4)(x2 +1). Therootsof this polynomial can be read− from the monic− linear− factors. They are 2, 3, and 4. − (Notice that p(x)= 2(x +2)(x 3)(x 4)(x2 +1)iscompletelyfactored because x2 +1hasnoroots.)− − − Completely factored linears To completely factor a linear polynomial, just factor out its leading coeffi- cient: b ax + b = a x + a For example, to completely factor 2x +6,writeitastheproduct2(x +3). Completely factored quadratics What a completely factored quadratic polynomial looks like will depend on how many roots it has. 125 0Roots.If the quadratic polynomial ax2 + bx + c has 0 roots, then it can be completely factored by factoring out the leading coefficient: b c ax2 + bx + c = a x2 + x + a a 2 2 b c (The graphs of ax +bx+c and x + ax+ a differ by a vertical stretch or shrink that depends on a. A vertical stretch or shrink of a graph won’t change the 2 b c 2 number of x-intercepts, so x + ax + a won’t have any roots since ax + bx + c 2 b c doesn’t have any roots. Thus, x + ax + a is completely factored.) Example. The discriminant of 4x2 2x+2 equals ( 2)2 4(4)(2) = 4 32 = 28, a negative number. Therefore,− 4x2 2x +2− hasnoroots,anditis− − completely− factored as 4(x2 1x + 1). − − 2 2 2Roots. If the quadratic polynomial ax2 + bx + c has 2 roots, we can name them α1 and α2. Roots give linear factors, so we know that (x α1) 2 − and (x α2)arefactorsofax + bx + c. That means that there is some polynomial− q(x)suchthat ax2 + bx + c = q(x)(x α )(x α ) − 1 − 2 The degree of ax2 + bx + c equals 2. 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 α1)plusthedegreeof(x α2)equals2.Inotherwords, the degree of q(x−)plus1plus1equals2. − Zero is the only number that you can add to 1 + 1 to get 2, so q(x) must have degree 0, which means that q(x)isjustaconstantnumber. Because the leading term of ax2 + bx + c –namelyax2 –istheproductof the leading terms of q(x), (x α1), and (x α2)–namelythenumberq(x), x,andx –itmustbethatq(−x)=a. Therefore,− ax2 + bx + c = a(x α )(x α ) − 1 − 2 Example. The discriminant of 2x2 +4x 2equals42 4(2)( 2) = 16+16 = 32, 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 discriminant: √32 = (4)(4)(2) = 4√2. 126 Now we use the quadratic formula to find that the roots are 4+4√2 4+4√2 − = − = 1+√2 2(2) 4 − and 4 4√2 4 4√2 − − = − − = 1 √2 2(2) 4 − − Therefore, 2x2 +4x 2iscompletelyfactoredas − 2 x ( 1+√2) x ( 1 √2) =2(x +1 √2)(x +1+√2) − − − − − − 2 1Root.If ax + bx + c has exactly 1 root (let’s call it α1)then(x α1)is afactorofax2 + bx + c. Hence, − ax2 + bx + c = g(x)(x α ) − 1 for some polynomial g(x). Because the degree of a product is the sum of the degrees of the factors, g(x)mustbeadegree1polynomial,anditcanbecompletelyfactoredinto something of the form λ(x β)whereλ, β R. Therefore, − ∈ ax2 + bx + c = λ(x β)(x α ) − − 1 2 Notice that β is a root of λ(x β)(x α1), so β is a root of ax + bx + c since they are the same polynomial.− But− we know that ax2 + bx + c has only one root, namely α1, so β must equal α1.