# Factor Theorem

How to use the factor theorem to factor polynomials: theorem, proof, examples and their solutions.

## Theorem

If *f*(*a*) = 0,

then (*x* - *a*) is the factor of *f*(*x*).

## Proof

*f*(*x*) = (*x* - *a*)⋅(quotient) + *f*(*a*)

Remainder theorem

If *f*(*a*) = 0,

then *f*(*x*) = (*x* - *a*)⋅(quotient).

So if *f*(*a*) = 0,

then (*x* - *a*) is the factor of *f*(*x*).

## How to Use

So if *f*(*a*) = 0 and *f*(*b*) = 0,

then (*x* - *a*) and (*x* - *b*) are the factors of *f*(*x*).

This means you can factor a polynomial

by finding the zeros of the function.

## Example 1

To factor the given polynomial,

find the zeros of the function.

The zeros are the numbers

that make the remainder of the synthetic division 0.

Use the synthetic division.

(divisor's zero: 1)

The remainder is 0.

Then *f*(1) = 0.

Synthetic substitution

So (*x* - 1) is the factor of *f*(*x*).

To factor further,

use the synthetic division.

(divisor's zero: 2)

The remainder is 0.

Then *f*(2) = 0.

So (*x* - 2) is the factor of *f*(*x*).

1, 6 means (*x* + 6). (brown)

(*x* + 6) is the factor that can't be factored further.

The previously found zeros are 1 and 2.

So *f*(*x*) = (*x* - 1)(*x* - 2)(*x* + 6).

## Example 2

Use the synthetic division.

(divisor's zero: 1)

The remainder is 0.

Then *f*(1) = 0.

Synthetic substitution

So (*x* - 1) is the factor of *f*(*x*).

Use the synthetic division.

(divisor's zero: -1)

The remainder is 0.

Then *f*(-1) = 0.

So (*x* + 1) is the factor of *f*(*x*).

Use the synthetic division.

(divisor's zero: -1)

The remainder is 0.

Then *f*(-1) = 0.

So (*x* + 1) is another factor of *f*(*x*).

1, -3 means (*x* - 3). (brown)

(*x* - 3) is the factor that can't be factored further.

The previously found zeros are 1, -1, and -1.

So *f*(*x*) = (*x* - 1)(*x* + 1)^{2}(*x* - 3).

## Example 3

Use the synthetic division.

(divisor's zero: 2)

The remainder is 0.

Then *f*(2) = 0.

Synthetic substitution

So (*x* - 2) is the factor of *f*(*x*).

Use the synthetic division.

(divisor's zero: -3)

The remainder is 0.

Then *f*(-3) = 0.

So (*x* + 3) is the factor of *f*(*x*).

1, 0, 1 means (*x*^{2} + 1). (brown)

(*x*^{2} + 1) is the factor that can't be factored further.

The previously found zeros are 2 and -3.

So *f*(*x*) = (*x* - 2)(*x* + 3)(*x*^{2} + 1).