# Factoring Polynomials

## Methods for Factoring a Polynomial

### Remove the Common Factor

It consists by applying the distributive property.

a · b + a · c + a · d = a (b + c + d)

#### Examples

1.  x3 + x2 = x2 (x + 1)

The roots are: x = 0, and x = −1

2.  2x4 + 4x2 = 2x2 (x2 + 2)

It only has a root x = 0; since the polynomial, x2 + 2, does not have any value that annuls it.

3.  x2 − ax − bx + ab = x (x − a) − b (x − a) = (x − a) · (x − b)

The roots are x = a and x = b.

## Remarkable Identities

### Difference of Squares

A difference of squares is equal to the sum of the difference.

a2 − b2 = (a + b) · (a − b)

#### Examples

1  x2 − 4 = (x + 2) · (x − 2)

The roots are x = −2 and x = 2.

2  x4 − 16 = (x2 + 4) · (x2 − 4) = (x + 2) · (x − 2) · (x2 + 4)

The roots are x = −2 and x = 2.

### Perfect Square Trinomial

A perfect square trinomial is equal to a squared binomial.

a2 ± 2 a b + b2 = (a ± b)2

#### Examples

The root is x = −3, which is said to be a double root.

The root is x = 2.

### Second-Degree Trinomial

To factor the second-degree trinomial P(x) = ax2 + bx + c, make it equal to zero and solve the quadratic equation. If the solutions to the equation are x1 and x2, the decomposed polynomial will be:

ax2 + bx + c = a · (x − x1) · (x − x2)

#### Examples

The roots are x = 3 and x = 2.

The roots are x = 3 and x = −2.

x4 − 10x2 + 9

x2 = t

x4 − 10x2 + 9 = 0

t2 − 10t + 9 = 0

x4 − 10x2 + 9 = (x + 1) · (x − 1) · (x + 3) · (x − 3)

x4 − 2x2 − 3

x2 = t

t2 − 2t − 3 = 0

x4 − 2x2 + 3 = (x2 + 1) · (x + )

## Factoring a Polynomial whose Degree Exceedes Two

Use the remainder theorem and Ruffini's rule.

#### Example

P(x) = 2x4 + x3 − 8x2 − x + 6

1 Take the divisors of the independent term: ±1, ±2, ±3.

2 By applying the remainder theorem, it will be known that the value of the division is exact.

P(1) = 2 · 14 + 13 − 8 · 12 − 1 + 6 = 2 + 1− 8 − 1 + 6 = 0

3 Divided by Ruffini's Rule:

4 For the exact division, D = d · c.

(x − 1) · (2x3 + 3x2 − 5x − 6 )

A root is x = 1.

Continue doing the same operations to the second factor.

Try again for 1 because the first factor could be squared.

P(1) = 2 · 13 + 3 · 12 − 5 · 1 − 6≠ 0

P(−1) = 2 · (− 1)3 + 3 · (− 1)2 − 5 · (− 1) − 6 = −2 + 3 + 5 − 6 = 0

(x −1) · (x +1) · (2x2 +x −6)

Another root is x = −1.

The third factor can be found by applying the quadratic equation or the previous method, but the disadvantage is that it will only give whole roots.

Rule out 1 and keep trying for 1.

P(−1) = 2 · (−1)2 + (−1) − 6 ≠ 0

P(2) = 2 · 22 + 2 − 6 ≠ 0

P(−2) = 2 · (−2)2 + (−2) − 6 = 2 · 4 − 2 − 6 = 0

(x − 1) · (x + 1) · (x + 2) · (2x − 3 )

Extract the common factor 2 in the last binomial.

2x − 3 = 2 (x − 3/2)

The factored polynomial is:

P(x) = 2x4 + x3 − 8x2 − x + 6 = 2 (x −1) · (x +1) · (x +2) · (x − 3/2)

The roots are: x = 1, x = − 1, x = −2 and x = 3/2