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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 (2x 2 − 5 x)(− 7 x + 4) is − 14 x 3. 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)(x 2 + 3x − 7) is a degree 4 polynomial because its leading term is 5x 4. The degrees of 5, (x − 2), (x + 3), and (x 2 + 3x − 7) are 0, 1, 1, and 2, respectively. Notice that 0 + 1 + 1 + 2 = 4. The degrees of (2x 3 − 7), (x 5 − 3 x + 5), (x − 1), and (5x 7 + 6x − 9) are 3, 5, 1, and 7, respectively. The degree of their product,
(2x 3 − 7)(x 5 − 3 x + 5)(x − 1)(5x 7 + 6x − 9),
equals 16 since its leading term is 10x 16. 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) is factored into a product of polynomials, then the degree of p(x) equals the sum of the degrees of its factors.
Examples.
- The degree of (4x 3 + 27x − 3)(3x 6 − 27 x 3 + 15) equals 3 + 6 = 9.
- The degree of −7(x + 4)(x − 1)(x − 3)(x − 3)(x 2 + 1) equals 0 + 1 + 1 + 1 + 1 + 2 = 6.
Degree of a polynomial bounds the number of roots
Suppose p(x) is a polynomial that has n roots, and that p(x) is not the constant polynomial p(x) = 0. Let’s name the roots of p(x) as α 1 , α 2 , ..., α (^) n.
Any root of p(x) gives a linear factor of p(x), so p(x) = (x − α 1 )(x − α 2 ) · · · (x − α (^) n )q(x)
for some polynomial q(x). Because the degree of a product is the sum of the degrees, the degree of p(x) is at least n.
The degree of p(x) (if p(x) = 0) is greater than or equal to the number of roots that p(x) has.
Examples.
- 5 x 4 − 3 x 3 + 2x − 17 has at most 4 roots.
- 4 x 723 − 15 x 52 + 37x 14 − 7 has at most 723 roots.
- Aside from the constant polynomial p(x) = 0, if a function has a 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) and cos(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) and cos(x), then you can ignore this paragraph.)
Fundamental Theorem of Algebra
A monic polynomial is a polynomial whose leading coefficient equals 1. So x 4 − 2 x 3 + 5x − 7 is monic, and x − 2 is monic, but 3x 2 − 4 is not monic.
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:
do not have roots, and of monic linear polynomials.
Completely factored
A polynomial is completely 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)(x 2 + 2)(x 2 + 5) and 2(x 2 + 3)(x 2 − x + 4) are completely factored.
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) = − 2 x 5 + 10x 4 + 2x 3 − 38 x 2 + 4x − 48. Written in this form, its difficult to see what the roots of p(x) are. But after being completely factored, p(x) = −2(x + 2)(x − 3)(x − 4)(x 2 + 1). The roots of 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)(x 2 + 1) is completely factored because x 2 + 1 has no roots.)
Completely factored linears
To completely factor a linear polynomial, just factor out its leading coeffi- cient:
ax + b = a
x +
b a
For example, to completely factor 2x + 6, write it as the product 2(x + 3).
Completely factored quadratics
What a completely factored quadratic polynomial looks like will depend on how many roots it has.
0 Roots. If the quadratic polynomial ax 2 + bx + c has 0 roots, then it can be completely factored by factoring out the leading coefficient:
ax 2 + bx + c = a
x 2 +
b a
x +
c a
(The graphs of ax 2 +bx+c and x 2 + ba x+ (^) ac differ by a vertical stretch or shrink that depends on a. A vertical stretch or shrink of a graph won’t change the number of x-intercepts, so x 2 + (^) ab x + ca won’t have any roots since ax 2 + bx + c doesn’t have any roots. Thus, x 2 + (^) ab x + (^) ac is completely factored.)
Example. The discriminant of 4x 2 − 2 x+2 equals (−2) 2 −4(4)(2) = 4−32 = −28, a negative number. Therefore, 4x 2 − 2 x + 2 has no roots, and it is completely factored as 4(x 2 − 12 x + 12 ).
2 Roots. If the quadratic polynomial ax 2 + bx + c has 2 roots, we can name them α 1 and α 2. Roots give linear factors, so we know that (x − α 1 ) and (x − α 2 ) are factors of ax 2 + bx + c. That means that there is some polynomial q(x) such that
ax 2 + bx + c = q(x)(x − α 1 )(x − α 2 )
The degree of ax 2 + 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 − α 1 ) plus the degree of (x − α 2 ) equals 2. In other words, the degree of q(x) plus 1 plus 1 equals 2. 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) is just a constant number. Because the leading term of ax 2 + bx + c – namely ax 2 – is the product of the leading terms of q(x), (x − α 1 ), and (x − α 2 ) – namely the number q(x), x, and x – it must be that q(x) = a. Therefore,
ax 2 + bx + c = a(x − α 1 )(x − α 2 )
Example. The discriminant of 2x 2 + 4x − 2 equals 4 2 − 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:
Summary. The following chart summarizes the discussion above.
roots of ax 2 + bx + c completely factored form of ax 2 + bx + c
no roots a(x 2 + ba x + ca )
2 roots: α 1 and α 2 a(x − α 1 )(x − α 2 )
1 root: α 1 a(x − α 1 )(x − α 1 )
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.
- 12 = 3 · 4, so 4 is a factor of 12.
- −30 = − 2 · 15, so 15 is a factor of −30.
- 1, −1, n and −n are all factors of an integer n. That’s because n = n · 1 and n = (−n)(−1).
Important special case. If α 1 , α 2 ,... α (^) n ∈ Z, then each of these numbers are factors of the product α 1 α 2 · · · α (^) n. For example, 2, 10, and 7 are each factors of 2 · 10 · 7 = 140.
Check factors of degree 0 coefficient when searching for
roots
If k, α 1 , and α 2 are all integers, then the polynomial q(x) = k(x − α 1 )(x − α 2 ) = kx^2 − k(α 1 + α 2 )x + kα 1 α (^2)
has α 1 and α 2 as roots, and each of these roots are factors of the degree 0 coefficient of q(x). (The degree 0 coefficient is kα 1 α 2 .)
More generally, if k, α 1 , α 2 ,... , α (^) n ∈ Z, then the degree 0 coefficient of the polynomial
g(x) = k(x − α 1 )(x − α 2 ) · · · (x − α (^) n )
equals kα 1 α 2 · · · α (^) n. That means that each of the roots of g(x) – which are the α (^) i – are factors of the degree 0 coefficient 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 offer a powerful hint as to what the roots of a polynomial might be.
When searching for roots of a polynomial whose coefficients are all integers, check the factors of the degree 0 coefficient.
Example. 3 and −7 are both roots of 2(x − 3)(x + 7). Notice that 2(x − 3)(x + 7) = 2x 2 + 8x − 42, and that 3 and −7 are both factors of −42.
Example. Suppose p(x) = 3x 4 + 3x 3 − 3 x 2 + 3x − 6. This is a degree 4 polynomial, so it will have at most 4 roots. There isn’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 coefficient of p(x) is −6, so a good place to check for roots is in the factors of −6. The factors of −6 are 1, −1, 2, −2, 3, −3, 6, and −6, so we have eight quick candidates for what the roots of p(x) might be. A quick check shows that of these eight candidates, exactly two are roots of p(x) – namely 1 and −2. That is to say, p(1) = 0 and p(−2) = 0.
Problem. Completely factor 3x 3 − 3 x 2 − 15 x + 6.
Solution. The factors of 6 are { 1 , − 1 , 2 , − 2 , 3 , − 3 , 6 , − 6 }. Check to see that −2 is a root. Then divide by x + 2 to find that
3 x 3 − 3 x 2 − 15 x + 6 x + 2
= 3x 2 − 9 x + 3
so 3 x 3 − 3 x 2 − 15 x + 6 = (x + 2)(3x 2 − 9 x + 3)
The discriminant of 3x 2 − 9 x + 3 equals 45, and thus 3x 2 − 9 x + 3 has two roots and can be factored further. The leading coefficient of 3x 2 − 9 x + 3 is 3, and we can use the quadratic
formula to check that the roots of 3x 2 − 9 x + 3 are 3+
√ 5 2 and^
3 −√ 5
- From what we learned above about factoring quadratics, we know that 3x 2 − 9 x + 3 =
3(x − 3+
√ 5 2 )(x^ −^
3 −√ 5 2 ). To summarize, 3 x 3 − 3 x 2 − 15 x + 6 = (x + 2)(3x 2 − 9 x + 3)
= (x + 2)
x −
x −
= 3(x + 2)
x −
x −
Completely factoring quartics
Degree 4 polynomials are tricky. As with cubic polynomials, you should begin by checking whether the factors of the degree 0 coefficient are roots. If one of them is 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 is that there’s no reason that a degree 4 polynomial has to have any roots – take (x 2 + 1)(x 2 + 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.
Exercises
For #1-5, determine the degree of the given polynomial.
1.) (x + 3)(x − 2)
2.) (3x + 5)(4x 2 + 2x − 3)
3.) −17(3x 2 + 20x − 4)
4.) 4(x − 1)(x − 1)(x − 1)(x − 2)(x 2 + 7)(x 2 + 3x − 4)
5.) 5(x − 3)(x 2 + 1)
6.) (True/False) 7x 5 + 13x 4 − 3 x 3 − 7 x 2 + 2x − 1 has 8 roots.
Completely factor the polynomials given in #7-
7.) 10x + 20
8.) − 2 x + 5
9.) − 2 x 2 − 12 x − 18
10.) 10x 2 + 3
11.) 3x 2 − 10 x + 5
12.) 3x 2 − 4 x + 5
13.) − 2 x 2 + 6x − 3
14.) 5x 2 + 3x − 2
15.) Find a root of x 3 − 5 x 2 + 10x − 8.
16.) Find a root of 15x 3 + 35x 2 + 30x + 10.
17.) Find a root of x 3 − 2 x 2 − 2 x − 3.
