Factor Theorem

Now that you've learned about the remainder theorem and remainders when performing polynomial division, we can examine a special case of the remainder theorem: the factor theorem.

The factor theorem states that if $\left(x-a\right)$ is a factor of the polynomial you're dividing, there will be no remainders when performing polynomial division; furthermore, f(a) will equal zero.

The factor theorem is more readily understood by examining regular numbers first. In elementary school, you learned that when dividing a number by another number, the second number is a factor of the first if there are no remainders.

In other words, both 4 and 5 are factors of 20. $20÷5=4$ , and $20÷4=5$ . There are no remainders.

On the other hand, when you divide 20 by 3, you get 6 with a remainder of $\frac{2}{3}$ . Therefore, 3 is not a factor of 20.

Similarly, if $\left(x-a\right)$ is a factor of a polynomial ${ax}^2+bx+c$ , there will be no remainder when dividing the polynomial by $\left(x-a\right)$ . In other words, the remainder will be 0.

Using the Factor Theorem With Trial and Error

We can often use trial and error to find factors of polynomial functions, armed with our knowledge of the factor theorem.

Recall that if $\left(x-a\right)$ is a factor of a polynomial, $f\left(a\right)$ will equal 0. In other words, when replacing every x in the polynomial with an $\left(x=a\right)$ , the resulting equation will equal 0.

Therefore, we can use trial and error to try different numbers until we find one (or several) that when inserted into the polynomial, equal zero.

For example, we might try:

$f\left(1\right)$

$f\left(-1\right)$

$f\left(2\right)$

$f\left(-2\right)$

If, when inserting (2) into the polynomial in every instance we see x, we find that the equation equals 0 – in other words, $f\left(2\right)=2$ we can infer that (x-2) is a factor of the polynomial and there will be no remainders when dividing the polynomial by $\left(x-2\right)$ .

Similarly, if you insert -2 into the polynomial instead of x, and it equals 0, we can deduce that $\left(x+2\right)$ is a factor of the polynomial.

That might seem a bit confusing at first, but think of it this way: if $f\left(-2\right)=0$ , then $(x-\left(-2\right))$ is a factor of the polynomial. $(x-\left(-2\right))$ is the same as $(x+2)$ .

Another way to think about it is as follows. Using our example, the factor theorem states that $(x+2)$ is a factor of the polynomial in question. Therefore, $x+2=0$ ; subtract 2 from both sides and we get $x=-2$ , which is what we're doing when replacing every x in the polynomial with (-2), which we write as $f\left(-2\right)$ .

Using Prime Factorization to Guess Factors

There's a little shortcut we can often use to accurately guess the correct factors of a polynomial. Look at the constant at the end of the polynomial and find the prime factors of that constant.

For example, consider the cubic $x^3-14x^2+63x-90$ . This polynomial, factorized, is $\left(x-3\right)\left(x-6\right)\left(x-5\right)$ . We can confirm this using the factor theorem:

$f\left(3\right)=3^3-14{\left(3\right)}^2+63\left(3\right)-90=0$

$f\left(6\right)=6^3-14{\left(6\right)}^2+63\left(6\right)-90=0$

$f\left(5\right)=5^3-14{\left(5\right)}^2+63\left(5\right)-90=0$

Note, however, that both 2, 3, and 5 are prime factors of 90. And, since both 3 and 2 go into 90 with no remainders, 6 goes into 90 as well. While 2 is a prime factor of 90, we can test $f\left(2\right)$ and find that it does not equal 0: $2^3-14\left(2^2\right)+63\left(2\right)-90=-12$ ; hence, $(x-2)$ is not a factor of $x^3-14x^2+63x-90$ .

This makes sense when you think about it. Where is the -90 coming from at the end of the polynomial? When you multiply out $\left(x-3\right)\left(x-6\right)\left(x-5\right)$ , the product of -3, -5, and -6 is -90 (since there is a triple negative, as we are dealing with a cubic, the sign doesn't change).

In this particular case, random guessing starting from $1,-1,2,-2$ , and so on is still possible. However, it would have been a bit lengthier; this shortcut helps you save time.

Here's another example: $x^3-18x^2+95x-126$ , the roots of which are 2, 7, and 9. Fully factorized, we get $\left(x-2\right)\left(x-7\right)\left(x-9\right)$ . Note that the prime factors of 126 are 2, 3, 3, and 7 – in other words, 2, $3^2$ (which is 9), and 7.

Using the Factor Theorem on Cubics and Quartics

When dealing with higher-degree polynomials, such as cubics and quartics, you can find one factor $\left(x-a\right)$ first and then perform polynomial long division.

If you factorize a cubic, you will be left with the factor timed by a quadratic: $\left(x-a\right)\left({ax}^2+bx+c\right)$ . To factorize fully, you can then factorize the quadratic independently.

Factor Theorem Practice Questions

1. Using the random guessing method, find the factors of $x^2-3x+2$ . Remember that if $f\left(a\right)=0$ , a is a factor.

let's check values starting from 1:

$f\left(1\right)={\left(1\right)}^2-3\times \left(1\right)+2=1-3+2=0$ . So, $x-1$ is a factor.

$f\left(2\right)={\left(2\right)}^2-3\times \left(2\right)+2=4-6+2=0$ . So, $x-2$ is a factor.

These are the only two factors needed for this quadratic, so we can stop here.

The factored form of the quadratic $x^2-3x+2$ is $\left(x-1\right)\left(x-2\right)$ .

2. By examining the factors of the constant, find the factors of $x^3-12x^2+47x-60$ .

The constant term in the cubic polynomial $x^3-12x^2+47x-60$ is -60, and the factors of 60 are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30$ , and $\pm 60$ . The leading coefficient is 1 (from the term $x^3$ ), and the factors of 1 are $\pm 1$ .

Therefore, any rational root of the cubic equation must be in the set of numbers $\left\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30,\pm 60\right\}$ .

We would then substitute these values into the equation to see if they make the equation equal to zero and thus are roots.

If we check these values, we'll find that 4, 3, and 5 are roots of the polynomial.

So, $\left(x-4\right),\left(x-3\right)$ , and $\left(x-5\right)$ are factors of the polynomial $x^3-12x^2+47x-60$ .

Therefore, $x^3-12x^2+47x-60=\left(x-4\right)\left(x-3\right)\left(x-5\right)$ .

Topics related to the Factor Theorem

Polynomial Function

Irreducible Polynomials

Factoring

Flashcards covering the Factor Theorem

Algebra II Flashcards

College Algebra Flashcards

Practice tests covering the Factor Theorem

Algebra II Diagnostic Tests

College Algebra Diagnostic Tests

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