# Factor Theorem

How to use the factor theorem to factor a polynomial: theorem, formula, 3 examples, and their solutions.

## Theorem

### Theorem

If f(a) = 0,
then f(x) has the factor (x - a).

This is the factor theorem.

Then if f(a) = 0, f(b) = 0, ...,
then f(x) has the factors (x - a), (x - b), ... .

So, by finding the zeros,
you can factor f(x).

## Formula

### Formula

Recall that
f(a) is the remainder of the synthetic division
of f(x)/(x - a).

Synthetic Substitution

So, if the remainder is 0, f(a) = 0,
then (x - a) is the factor of f(x).

By repeating this,
you can find the factors of f(x)
and factor f(x).

## Example 1

### Solution

Do the synthetic division.

f(x) = x3 + 3x2 - 16x + 12

Write the coefficients of the terms
in descending order:
1 3 -16 12.

Draw an L shape form like this.

Pick a number
that seems to make the remainder, f(a), 0.

1 seems to be good.
So write 1
on the left side of the form.

Tip: If the sum of the coefficients is 0,
then 1 is the right number.

↓: 1 = 1

↗: 1⋅1 = 1

↓: 3 + 1 = 4

↗: 4⋅1 = 4

↓: -16 + 4 = -12

↗: -12⋅1 = -12

↓: 12 - 12 = 0

The remainder is 0.
So f(1) = 0.

So 1 is the right number.

If the remainder is not 0,
then you picked the wrong number.
Then pick another number.

This is a process of trial and error.
So it's natural to choose the wrong numbers.

1 4 -12 means
x2 + 4x - 12.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

2 seems to be good.
So write 2
on the left side of the form.

↓: 1 = 1

↗: 1⋅2 = 2

↓: 4 + 2 = 6

↗: 6⋅2 = 12

↓: 12 - 12 = 0

The remainder is 0.
So f(2) = 0.

So 2 is the right number.

1 6 means
x + 6.

(x + 6) is a binomial factor.

So stop finding the factors.

And write the factors
from this.

For the number 1,
the remainder, f(1), is 0.

So write the factor (x - 1).

For the number 2,
the remainder, f(2), is 0.

So write the factor (x - 2).

1 6 means
x + 6.

So write (x + 6).

This is the quotient.

So (given) = (x - 1)(x - 2)(x + 6).

So
(x - 1)(x - 2)(x + 6)

## Example 2

### Solution

Do the synthetic division.

f(x) = x4 - 2x3 - 4x2 - 2x + 3

Write the coefficients of the terms
in descending order:
1 -2 -4 -2 3.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

1 seems to be good.
So write 1
on the left side of the form.

↓: 1 = 1

↗: 1⋅1 = 1

↓: -2 + 1 = -1

↗: -1⋅1 = -1

↓: -4 - 1 = -5

↗: -5⋅1 = -5

↓: 2 - 5 = -3

↗: -3⋅1 = -3

↓: 3 - 3 = 0

The remainder is 0.
So f(1) = 0.

So 1 is the right number.

1 -1 -5 3 means
x3 - x2 - 5x + 3.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

-1 seems to be good.
So write -1
on the left side of the form.

↓: 1 = 1

↗: 1⋅(-1) = -1

↓: -1 - 1 = -2

↗: -2⋅(-1) = 2

↓: -5 + 2 = -3

↗: -3⋅(-1) = 3

↓: -3 + 3 = 0

The remainder is 0.
So f(-1) = 0.

So -1 is the right number.

1 -2 -3 means
x2 - 2x - 3.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

-1 seems to be good.
So write -1 again
on the left side of the form.

↓: 1 = 1

↗: 1⋅(-1) = -1

↓: -2 - 1 = -3

↗: -3⋅(-1) = 3

↓: -3 + 3 = 0

The remainder is 0.
So f(-1) = 0.

So -1 is the right number.

1 -3 means
x - 3.

(x - 3) is a binomial factor.

So stop finding the factors.

And write the factors
from this.

For the number 1,
the remainder, f(1), is 0.

So write the factor (x - 1).

For two numbers -1,
the remainders are both 0.

So write the factor (x + 1)2.

1 -3 means
x - 3.

So write (x - 3).

This is the quotient.

So (given) = (x - 1)(x + 1)2(x - 3).

So
(x - 1)(x + 1)2(x - 3)

## Example 3

### Solution

Do the synthetic division.

f(x) = x4 + x3 - 5x2 + x + 6

Write the coefficients of the terms
in descending order:
1 1 -5 1 -6.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

2 seems to be good.
So write 2
on the left side of the form.

↓: 1 = 1

↗: 1⋅2 = 2

↓: 1 + 2 = 3

↗: 3⋅2 = 6

↓: -5 + 6 = 1

↗: 1⋅2 = 2

↓: 1 + 2 = 3

↗: 3⋅2 = 6

↓: -6 + 6 = 0

The remainder is 0.
So f(2) = 0.

So 2 is the right number.

1 3 1 3 means
x3 + 3x2 + x + 3.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number
that seems to make the remainder 0.

-3 seems to be good.
So write -3
on the left side of the form.

↓: 1 = 1

↗: 1⋅(-3) = -3

↓: 3 - 3 = 0

↗: 0⋅(-3) = 0

↓: 1 + 0 = 1

↗: 1⋅(-3) = -3

↓: 3 - 3 = 0

The remainder is 0.
So f(-3) = 0.

So -3 is the right number.

1 0 1 means
x2 + 1.

x2 + 1 is always (+).
There's no root for x2 + 1 = 0
So this is not factorable.

So stop the synthetic division.

And write the factors
from this.

For the number 2,
the remainder, f(2), is 0.

So write the factor (x - 2).

For the number -3,
the remainder, f(-3), is 0.

So write the factor (x + 3).

1 0 1 means
x2 + 1.

So write (x2 + 1).

This is the quotient.

So (given) = (x - 2)(x + 3)(x2 + 1).

So
(x - 2)(x + 3)(x2 + 1)