# 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.** x ^{3} + x^{2} = x^{2} (x + 1)**

The** roots** are: x = 0, and x = −1

2.** 2x ^{4} + 4x^{2} = 2x^{2} (x^{2} + 2)**

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

3.** x ^{2} − 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.

**a ^{2} − b^{2} = (a + b) · (a − b) **

#### Examples

1 x^{2} − 4 = (x + 2) · (x − 2)

The roots are x = −2 and x = 2.

2 x^{4} − 16 = (x^{2} + 4) · (x^{2} − 4) = (x + 2) · (x − 2) · (x^{2} + 4)

The roots are x = −2 and x = 2.

### Perfect Square Trinomial

A perfect square trinomial is equal to a squared binomial.

**a ^{2} ± 2 a b + b^{2} = (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) = ax^{2} + bx + c,** make it equal to zero and solve the quadratic equation**. If the solutions to the equation are x_{1}_{} and x_{2}_{}, the decomposed polynomial will be:

**ax ^{2} + bx + c = a · (x − x_{1}) · (x − x_{2})**

#### Examples

**The roots are x = 3 and x = 2. **

**The roots are x = 3 and x = −2. **

x^{4} − 10x^{2} + 9

x^{2} = t

x^{4} − 10x^{2} + 9 = 0

t^{2} − 10t + 9 = 0

**x ^{4} − 10x^{2} + 9 = (x + 1) · (x − 1) · (x + 3) · (x − 3)**

x^{4} − 2x^{2} − 3

x^{2} = t

t^{2} − 2t − 3 = 0

x^{4} − 2x^{2} + 3 = (x^{2} + 1) · (x + )

## Factoring a Polynomial whose Degree Exceedes Two

Use the remainder theorem and Ruffini's rule.

#### Example

P(x) = 2x^{4} + x^{3} − 8x^{2} − 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 · 1^{4} + 1^{3} − 8 · 1^{2} − 1 + 6 = 2 + 1− 8 − 1 + 6 = 0

**3 Divided by Ruffini's Rule:**

**4**** For the exact division, D = d · c**.

(x − 1) · (2x^{3} + 3x^{2} − 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 · 1^{3} + 3 · 1^{2} − 5 ** · ** 1 − 6≠ 0

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

(x −1) · (x +1) · (2x^{2} +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 · 2^{2} + 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) = 2x^{4} + x^{3} − 8x^{2} − 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