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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.
Fundamental Theorem of Algebra
A monic polynomial is a polynomial whose leading coe cient equals 1. So x 4 2 x 3 + 5x 7 is monic, and x 2 is monic, but 3x 2 4 is not monic.
The following result tells us how to factor polynomials. It essentially tells us what the “prime polynomials” are:
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 di cult to immediately understand its importance. The explanation for why this theorem is true is somewhat di cult, and it is beyond the scope of this course. We’ll have to accept it on faith.
Examples.
- 4 x 2 12 x + 8 can be factored into a product of a number, 4, and two monic linear polynomials, x 1 and x 2. That is, 4 x 2 12 x + 8 = 4(x 1)(x 2). 157
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.
Fundamental Theorem of Algebra
A monic polynomial is a polynomial whose leading coe cient 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:
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 di cult to immediately understand its importance. The explanation for why this theorem is true is somewhat di cult, and it is beyond the scope of this course. We’ll have to accept it on faith.
Examples.
- 4 x^2 12 x + 8 can be factored into a product of a number, 4, and two monic linear polynomials, x 1 and x 2. That is, 4 x^2 12 x + 8 = 4(x 1)(x 2). 130
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).
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In this chapter we’ll learn an analogous way to factor polynomials.
Fundamental Theorem of Algebra
A monic polynomial is a polynomial whose leading coefficient equals 1. So
— 2x3 + 5x — 7 is monic, and x — 2 is monic, but 3x2 — 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:
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 +8 can be factored into a product of a number, 4, and
two monic linear polynomials, x — 1 and x — 2. That is,
4~2_ 12x+8=4(x—1)(x—2).
- x 5 + 2x 4 7 x 3 + 14x 2 10 x + 20 can be factored into a product of a number, 1, a monic linear polynomial, x 2, and two monic quadratic polynomials that don’t have roots, x 2 + 2 and x 2 + 5. That is x 5 + 2x 4 7 x 3 + 14x 2 10 x + 20 = (x 2)(x 2 + 2)(x 2 + 5). (We can check the discriminants of x 2 + 2 and x 2 + 5 to see that these two quadratics don’t have roots.)
- 2 x 4 2 x 3 + 14x 2 6 x + 24 = 2(x 2 + 3)(x 2 x + 4). Again, x 2 + 3 and x 2 x + 4 do not have roots.
Notice that in each of the above examples, the real number that appears in the product of polynomials – 4 in the first example, 1 in the second, and 2 in the third – is the same as the leading coe cient for the original polynomial. This always happens, so the Fundamental Theorem of Algebra can be more precisely stated as follows:
If p(x) = a (^) n x n^ + a (^) n 1 x n ^1 + · · · + a 0 , then p(x) is the product of the real number a (^) n , and a collection of monic quadratic polynomials that 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 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 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 o↵ 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 di cult 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
and (x ↵ 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 ↵ 1 )(x ↵ 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 ↵ 1 ) plus the degree of (x ↵ 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 ↵ 1 ), and (x ↵ 2 ) – namely the number q(x), x, and x – it must be that q(x) = a. Therefore,
ax 2 + bx + c = a(x ↵ 1 )(x ↵ 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:
p 32 =
p (4)(4)(2) = 4
p
Now we use the quadratic formula to find that the roots are 4 + 4
p 2 2(2)
p 2 4
p 2
and 4 4
p 2 2(2)
p 2 4
p 2
Therefore, 2x 2 + 4x 2 is completely factored as
2
x ( 1 +
p
x ( 1
p
= 2(x + 1
p 2)(x + 1 +
p
1 Root. If ax 2 + bx + c has exactly 1 root (let’s call it ↵ 1 ) then (x ↵ 1 ) is a factor of ax 2 + bx + c. Hence,
ax 2 + bx + c = g(x)(x ↵ 1 )
for some polynomial g(x).
Because the degree of a product is the sum of the degrees of the factors, g(x) must be a degree 1 polynomial, and it can be completely factored into something of the form (x ) where , 2 R. Therefore,
ax 2 + bx + c = (x )(x ↵ 1 )
Notice that is a root of (x )(x ↵ 1 ), so is a root of ax 2 + bx + c since they are the same polynomial. But we know that ax 2 + bx + c has only one root, namely ↵ 1 , so must equal ↵ 1. That means that
ax 2 + bx + c = (x ↵ 1 )(x ↵ 1 )
The leading term of ax 2 +bx+c is ax 2. The leading term of (x ↵ 1 )(x ↵ 1 ) is x 2. Since ax 2 + bx + c equals (x ↵ 1 )(x ↵ 1 ), they must have the same leading term. Therefore, ax 2 = x 2. Hence, a = . Replace with a in the equation above, and we are left with
ax 2 + bx + c = a(x ↵ 1 )(x ↵ 1 )
Example. The discriminant of 3x 2 6 x+3 equals ( 6) 2 4(3)(3) = 36 36 = 0, so there is exactly one root. We find the root using the quadratic formula:
( 6) +
p 0 2(3)
Therefore, 3x 2 6 x + 3 is completely factored as 3(x 1)(x 1).
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: ↵ 1 and ↵ 2 a(x ↵ 1 )(x ↵ 2 )
1 root: ↵ 1 a(x ↵ 1 )(x ↵ 1 )
When searching for roots of a polynomial whose coe cients are all integers, check the factors of the degree 0 coe cient.
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 coe cient 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.
Factoring cubics
It follows from the Fundamental Theorem of Algebra that a cubic poly- nomial is either the product of a constant and three linear polynomials, or else it is the product of a constant, one linear polynomial, and one quadratic polynomial that has no roots. In either case, any cubic polynomial is guaranteed to have a linear factor, and thus is guaranteed to have a root. You’re going to have to guess what that root is by looking at the factors of the degree 0 coe cient. (There is a “cubic formula” that like the quadratic formula will tell you the roots of a cubic, but the formula is di 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
polynomial and a quadratic polynomial. Completely factor the quadratic and then you will have completely factored the cubic.
Problem. Completely factor 2x 3 3 x 2 + 4x 3.
Solution. Start by guessing a root. The degree 0 coe cient is 3, and the 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 is a root. That means that x 1 is a factor of 2x 3 3 x 2 + 4x 3. Therefore, we can divide 2x 3 3 x 2 + 4x 3 by x 1 to get another polynomial
2 x 3 3 x 2 + 4x 3 x 1
= 2x 2 x + 3
Thus,
2 x 3 3 x 2 + 4x 3 = (x 1)(2x 2 x + 3)
The discriminant of 2x 2 x + 3 equals ( 1) 2 4(2)(3) = 1 24 = 23, a negative number. Therefore, 2x 2 x + 3 has no roots, so to completely factor 2x 2 x + 3 we just have to factor out the leading coe cient as follows: 2 x 2 x + 3 = 2
x 2 12 x + (^32)
The final answer is 2(x 1)
x 2
x +
164
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 is a root. That means that x —1 is a factor
of 2x3 — 3x2 + 4x — 3. Therefore, we can divide 2x3 — 3x2 + 4x — 3 by x — 1
to get another polynomial
2x3—3x2+4x—3 2
=2x —x+
Thus,
2x3—3x2+4x—3=(x—1)(2x2—x+3)
The discriminant of 2x2 — x + 3 equals (_1)2 — 4(2)(3) = 1 — 24 = —23,
a negative number. Therefore, 2x2 — x + 3 has no roots, so to completely
factor 2x2 — x +3 we just have to factor out the leading coefficient as follows:
2x2—x-1-3=2(x2— ~x+~).
2x3-3x1 ~~+z- /
(x-l) (zx2_x+3’) /N a
The final answer is
2(x_1)(x2_ ~+~) 137
polynomial and a quadratic polynomial. Completely factor the quadratic and
then you will have completely factored the cubic.
Problem. Completely factor 2x3 — 3~2 + 4x — 3.
Solution. Start by guessing a root. The degree 0 coefficient is —3, and the
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 is a root. That means that x —1 is a factor
of 2x3 — 3x2 + 4x — 3. Therefore, we can divide 2x3 — 3x2 + 4x — 3 by x — 1
to get another polynomial
2x3—3x2+4x—3 2
=2x —x+
Thus,
2x3—3x2+4x—3=(x—1)(2x2—x+3)
The discriminant of 2x2 — x + 3 equals (_1)2 — 4(2)(3) = 1 — 24 = —23,
a negative number. Therefore, 2x2 — x + 3 has no roots, so to completely
factor 2x2 — x +3 we just have to factor out the leading coefficient as follows:
2x2—x-1-3=2(x2— ~x+~).
2x3-3x1 ~~+z- /
(x-l) (zx2_x+3’) /N a
The final answer is
2(x_1)(x2_ ~+~)
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 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
Completely factor the polynomials given in #1-
1.) 10x + 20
2.) 2 x + 5
3.) 2 x 2 12 x 18
4.) 10x 2 + 3
5.) 3x 2 10 x + 5
6.) 3x 2 4 x + 5
7.) 2 x 2 + 6x 3
8.) 5x 2 + 3x 2
9.) Find a root of x 3 5 x 2 + 10x 8.
10.) Find a root of 15x 3 + 35x 2 + 30x + 10.
11.) Find a root of x 3 2 x 2 2 x 3.
Completely factor the polynomials in #12-16.
12.) x 3 x 2 + x + 1
13.) 5x 3 9 x 2 + 8x 20
14.) 2 x 3 + 17x 3
15.) 4x 3 20 x 2 + 25x 3
16.) x 4 5 x 2 + 4
17.) How can the Fundamental Theorem of Algebra be used to show that any polynomial of odd degree has at least one root?
