The Factor Theorem is widely used to factor a polynomial and locate its roots. The polynomial remainder theorem is a specific instance of this. It is one of the ways of factoring polynomials.

In this article, we will look at how to prove the Factor Theorem. We will also see how to prove it using the remainder theorem. Finally, we will look at examples with answers.

## Proof of the Factor Theorem

We will use this section to show the factor theorem proof, which allows us to factor the polynomial.

Consider a polynomial* f*(x) which is divided by (*x-c*). This would mean that *f*(*c*)=0, and we can add *f*(c) without changing the polynomial:

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

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

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

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

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

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

This means that (*x-c*) is a factor of the polynomial *f*(*x*).

## Proof of the Factor Theorem by the remainder theorem

Consider a polynomial* f*(x), which is divided by (*x-c*). This means that *f*(c)=0. Therefore, we have:

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

If (*x-c*) is a factor of *f*(x), then the remainder must be zero given that (*x*–*c*) exactly divides *f*(x).

Therefore, we must have *f*(*c*)=0.

The following statements are equivalent for any polynomial *f*(x)

- The remainder is zero when the polynomial
*f*(x) is exactly divided by (*x – c*) - (
*x-c*) is a factor of*f*(*x*) *c*is one solution to*f*(*x*)*c*is a zero of the function*f*(*x*), or*f*(c) =0

## How to use the factor theorem?

The following are the steps that we can use to apply 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)*.

Rather than using the polynomial long division method to identify the factors, the factor theorem and synthetic division approach are better options. By using this theorem, we can eliminate known zeros from polynomials while keeping all unknown zeros unaffected. This allows us to find the lower degree polynomials easily.

The factor theorem can also be defined in another way. We usually get a reminder when a polynomial is divided by a binomial. When a polynomial is split by one of its binomial factors, the quotient obtained is known as a depressed polynomial. The factor theorem is demonstrated as follows if the remainder is zero:

The polynomial, say *f*(*x*) has a factor (*x*–*c*) if *f*(*c*)= 0, where *f*(x) is a polynomial of degree *n*, where *n* is greater than or equal to 1 for any real number *c*.

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## Factor theorem – Examples with answers

We can use the steps detailed above to apply the Factor theorem and solve the following problems.

### EXAMPLE 1

Examine whether *x*+2 is a factor of $latex F(x) = {x}^3 + 3 {x}^3 + 5x + 6$.

**Solution:**

To begin with, we will test if *x*+2 is a factor of the polynomial, where *x* = -2.

We can substitute the values of *x* in the function in order to test it.

$latex F(x) = {x}^3 + 3 {x}^3 + 5x + 6$

$$F(-2) = {(-2)}^3 + 3 {(-2)}^3 + 5(-2) + 6$$

$latex F(-2) = (-8 )+ 12 – 10 + 6$

$latex F(-2) = 0$

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 $latex {x}^3 + 3 {x}^3 + 5x + 6$ or *F*(x).

### EXAMPLE 2

Determine if x-3 is a factor of $latex: f(x) = {x}^3 – 3{x}^2 -5x +5$.

**Solution:**

We will test if *x*-3 is a factor of the polynomial, where *x* = 3.

We will substitute the values of *x* in the function in order to test it.

$latex f(x) = {x}^3 – 3{x}^2 -5x +5$

$latex f(3) = {(3)}^3 – 3{(3)}^2 -5(3) +5$

$latex f(3) = 27 – 3{(9)} -15 +5$

$latex f(3) = -10 $

According to Factor theorem, if *F*(c) = 0, then (*x* – *c*) is a factor of *f*(x).

Then, we have *f*(3)= -10.

Therefore, (x -3) is not a factor of $latex {x}^3 – 3{x}^2 -5x +5$ or *f*(x).

### EXAMPLE 3

Determine if 2*x* -1 is a factor of $latex f(x)=4{x}^2 + 2 x -2$.

**Solution:**

We will test if 2*x*-1 is a factor of polynomial, where $latex x = \frac{1}{2}$.

Substituting values of *x* in the function, we have:

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

$latex f(\frac{1}{2})=4{(\frac{1}{2})}^2 + 2 (\frac{1}{2}) -2 $

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

$latex f(\frac{1}{2}) = 0 $

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

Therefore, we have, $latex f(\frac{1}{2})= 10$.

This means that (2x – 1) is not a factor of $latex 4{x}^2 + 2 x -2 $ or *f*(x).

### EXAMPLE 4

Determine if 3*x*-3 is a factor of $latex f(x)=2{x}^2 + 2x – 4$

**Solution:**

We will test if 3*x*-3 is a factor of the polynomial using *x*=1.

Substituting the value of *x* in the function, we have.

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

$latex f(1)=2{(1)}^2 + 2 (1) -4 $

$latex f(1)= 2 + 2 -4 $

$latex: f(1) = 0 $

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

Then, we have, *f*(1)= 0.

Therefore, (3x – 3) is a factor of $latex 2{x}^2 + 2x – 4$ or *f*(x).

### EXAMPLE 5

Is *x*+4 a factor of $latex p(x)= \frac{{x}^2}{3}+4x -\frac{7}{3}$?

**Solution:**

We will test if *x*+4 is a factor of polynomial using *x* =-4:

We have to substitute the value of *x* in the polynomial to test it. Therefore, we have:

$latex p(x)= \frac{{x}^2}{3} + 4x -\frac{1}{2}$

$latex p(-4)= \frac{{(-4)}^2}{3} + 4(-4) -\frac{7}{3}$

$latex p(-4)= \frac{16}{3} – 16 -\frac{7}{3} $

$latex: p(-4) = 3-16 $

$latex: p(-4) = -13 $

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

Therefore, we have *p*(-4)= -13.

This means that (x + 4) is not a factor of $latex \frac{{x}^2}{3} + 4x -\frac{7}{3}$ or *p*(x).

## See also

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