The factor theorem is commonly used to factor a polynomial and for finding its roots. The polynomial remainder theorem is a specific instance of this. It is one of the ways to factor polynomials.

In this article, we will learn about the factor theorem in detail. We will look at how to prove the factor theorem. Then, we will use the theorem to solve some practice problems.

## What is the factor theorem?

The factor theorem is an algebraic theorem that connects a polynomial’s factors and zeros. The polynomial remainder theorem is a specific version of this theorem.

The factor theorem tells us that a polynomial has a factor (*x-c*) if and only if *f*(c)=0. Then, c is a solution of the polynomial.

The factor theorem can also be used to remove known zeros from a polynomial while leaving all unknown zeros intact, resulting in a polynomial of lower degree with easier to find zeros.

## Proof of the factor theorem:

To prove the factor theorem, we can consider a polynomial* f*(*x*), which is divided by (*x-c*). This would mean that *f*(c)=0.

We write in the following way:

$latex f(x) = (x-c)q(x) + f(c)$

where, *f*(x) is the target polynomial and *q*(x) is the quotient polynomial.

Given that *f*(c) = 0, we have:

$latex f(x)= (x-c)q(x)+ f(c) $

$latex f(x) = (x-c) q (x)+0$

$latex f(x) = (x-c) q (x)$

Therefore, we have proved that (*x-c*) is a factor of the polynomial *f*(x).

## How to use the factor theorem:

The following are the steps that we can follow to use the factor theorem and identify the factors of a polynomial:

**Step 1**: If *f*(-c)=0, then (*x+ c*) is a factor of the polynomial *f(x)*.

** Step 2:** If

*p*(

*d/c*)= 0, then (

*cx-d*) is a factor of the polynomial

*f(x)*.

** Step 3:** If

*p*(

*-d/c*)= 0, then (

*cx+d*) is a factor of the polynomial

*f(x)*.

** Step 4:** If

*p*(

*c*)=0 and

*p*(

*d*) =0, then (

*x – c*) and (

*x -d*) are factors of the polynomial

*p(x)*.

The factor theorem can be an easier method for finding the factors of a polynomial as opposed to the long division method. We can use this theorem to remove known zeros while leaving all unknown zeros intact in order to find the lower degree polynomial.

If the remainder is zero, the factor theorem is shown as follows:

If *f*(c)=0, the polynomial* f*(x) has a factor (*x-c*), where *f*(x) is a polynomial of degree *n*.

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## Factor theorem – Examples with answers

Using the formula detailed above, we can solve various factor theorem examples. Each of the following examples has its respective detailed solution. Try to solve the problems yourself before looking at the solution so that you can practice and fully master this topic.

**EXAMPLE 1**

Determine if (*x*+1) is a factor of the polynomial $latex f(x) = {x}^2 + x – 2$.

##### Solution

We will test if *x*+1 is a factor of the polynomial, where *x*= -1. This should satisfy the condition *f*(*c*)=0, Substituting *x* into the polynomial, we have:

$latex f(x) = {x}^2 – x -2$

$latex f(-1) = {(-1)}^2 – (-1) -2$

$latex f(-1) = 1 +1 – 2 $

$latex f(-1) = 0$

We have *f*(-1) = 0. According to the factor theorem, if *f*(c) = 0, then (*x*–*c*) is a factor of *f*(x).

Therefore, (x+2) is a factor of the polynomial given.

**EXAMPLE 2**

Determine if (2*x-*1) is a factor of the polynomial $latex f(x) = 2{x}^2 – x – 1$.

##### Solution

We will test if 2*x*-1 is a factor of the polynomial, where $latex x = \frac{1}{2}$. This should satisfy the condition where *f*(c) = 0. Substituting *x* in the polynomial, we have:

$latex f(\frac{1}{2}) = 2{x}^2 – x – 1$

$latex f(\frac{1}{2}) = 2{(\frac{1}{2})}^2 – (\frac{1}{2}) – 1$

$latex f(\frac{1}{2}) = 2(\frac{1}{4}) – (\frac{1}{2}) – 1$

$latex f(\frac{1}{2}) = \frac{1}{2} – (\frac{1}{2}) – 1$

We got $latex f(\frac{1}{2}) = -1$. According to the factor theorem, if *f*(c) = 0, then (*x*–*c*) is a factor of *f*(x).

Therefore, (2x–1) is **not** a factor of $latex f(x) = 2{x}^2 – x – 1$.

**EXAMPLE 3**

Is (*x*+3) a factor of the polynomial $latex f(x) = {x}^2 +x – 6$?

##### Solution

We can use *x*=-3 to test if *x*+3 is a factor of the polynomial. This satisfies the condition where *f*(c) = 0. Therefore, we have:

$latex f(x) = {x}^2 +x – 6$

$latex f(-3) = {-3}^2 + (-3) – 6$

$latex f(-3) = (9) -3 – 6$

$latex f(-3) =0$

We see that $latex f(-3) = 0$. According to the factor theorem, if *f*(c) = 0, then (*x*–*c*) is a factor of *f*(x).

Therefore, (*x*+3) is a factor of $latex f(x) = {x}^2 +x – 6$.

**EXAMPLE 4**

Determine which of the following polynomial functions have the polynomial factor (*x*+5):

$latex f(x) = {x}^3-x -6$

$latex g(x) = {x}^2-3x +4$

$latex h(x) = {x}^2-25$.

##### Solution

We are going to test each of the polynomials:

$latex f(x) = {x}^3-x -6$

$latex g(x) = {x}^2-3x +4$

$latex h(x) = {x}^2-25$

Assume that *x*+5 is a polynomial factor, then, we use *x* =5.

First,

$latex f(x) = {x}^3-x -6$

$latex f(-5) = {(-5)}^3-(-5) -6$

$latex f(-5) = -75 + 5 -6 $

$latex f(-5) = -76 $

Second,

$latex g(x) = {x}^2-3x +4$

$latex g(-5) = {(-5)}^2-4(-5) +4$

$latex g(-5) = 25+20 +4$

$latex g(-5) = 49$

Third,

$latex h(x) = {x}^2-25$

$latex h(-5) = {(-5)}^2-25$

$latex h(-5) = 25 – 25$

$latex h(-5) = 0 $

We see that only *h*(-5) = 0 satisfies factor theorem. Therefore, only *h*(*x*) is a polynomial function that has the factor *x+5*.

**EXAMPLE 5**

Find the exact solution of the function $latex f(x) = {x}^2 + 5x – 36$.

##### Solution

To find the solution of the function, we assume that (*x-c*) is a polynomial factor, where *x = c.*

To satisfy the factor theorem, we need to have *F*(c) = 0. Therefore,

$latex f(x) = {x}^2+ 5x -36$.

$latex {x}^2+ 5x -36 = 0$.

By factoring, we have:

$latex (x+9)(x-4) = 0$.

This means that *x*+3 and *x*-2 are the polynomial factors of the function.

Then, *x*+9=0, which means that *x*=-9 and *x*-4=0, which means that *x*=4.

Therefore, the solutions of the function *f*(*x*) are -9 and 4.

**EXAMPLE 6**

Find the factors of this polynomial, $latex F(x)= {x}^2 -25$.

##### Solution

According to the factor theorem, if we have that (*x-c*) is a factor of *f*(*x*), then we must have *f*(c)=0.

By factoring, we have,

$latex F(x)= {x}^2 -25$

$latex {x}^2 -25 = o$

then,

$latex (x+5)(x-5) = 0$.

Therefore, *x*+5 and *x*-5 are polynomial factors.

## Factor theorem – Practice problems

Solve the following factor theorem problems and test your knowledge on this topic. Use the factor theorem detailed above to solve the problems. If you have problems with these exercises, you can study the examples solved above.

## See also

Interested in learning more about the factor theorem? Take a look at these pages:

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