# Factor Polynomials

To factor a polynomials is to write that polynomial as the product of irreducible (cannot be factored) polynomials.

Example

\[ x^2 + 6 x + 8 = (x + 2)(x+4) \]

Examples with detailed solutions on factoring polynomials are presented along with questions and their solutions.

A factor polynomial calcultor is included and could be used to check answers.

## Factoring Polynomials Formulas

1: Factor out a common factor using the distributive property in reverse. \[ \color{red}{a} x + \color{red}{a} y = \color{red}{a} (x + y) \]

2: Sevearl grouped common factor. \[ \color{red}{a} x + \color{red}{a} y + \color{blue}{b} x + \color{blue}{b} y = \color{red}{a} (x + y) + \color{blue}{b}(x + y) = (\color{red}{a} + \color{blue}{b} ) (x + y) \]

3: Difference of two squares (1). \[ x^2 - y^2 = (x + y)(x - y) \]

4: Difference of two squares (2). \[ (x + y)^2 - z^2 = (x + y + z)(x + y - z) \]

5: Sum of two cubes. \[ x^3 + y^3 = (x + y)(x^2 - x y + y^2) \]

6: Difference of two cubes. \[ x^3 - y^3 = (x - y)(x^2 + x y + y^2) \]

7: Difference of fourth powers. \[ x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x + y)(x - y)(x^2 + y^2) \]

8: Perfet square \[ x^2 + 2xy + y^2 = (x + y)^2 \]

9: Perfet square \[ x^2 - 2xy + y^2 = (x - y)^2 \]

10: Perfect cube \[ x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3 \]

11: Perfect cube \[ x^3 - 3x^2y + 3xy^2 - y^3 = (x - y)^3 \]

## Examples in Factoring Polynomials with Solutions

### Example 1 - Factoring Out Common Factor

Factor the binomial \( \quad 12 x - 8 \)

__Solution__

The coefficients \( 12 \) and \( 8 \) have the greatest common factor equal to \( 4 \) and the given expression may be written as.

\( 12 x - 8 = \color{red}4 (3) x - \color{red}{4} (2) \)

Factor \( 4 \) out using formula 1 above.

\( 12 x - 8 = 4 (3 x - 2 ) \)

Check the answer by expanding the term on the right \( 4 (3 x - 2 ) \) to obtain the given binomial.

### Example 2 - Factoring the Difference of two Squares

Factor the binomial \( \quad 9 - 4x^2 \)

__Solution__

Rewrite the given expression as the difference of two squares then apply formula 1 given above.

\( 9 - 4x^2 = 3^2 - (2x)^2 = (3 - 2x)(3 + 2x) \)

As a practice, expand \( (3 - 2x)(3 + 2x) \) to obtain the given binomial.

### Example 3 - Factoring Trinomials

Factor the trinomial \( \quad 9x^2 + 3x - 2 \)

__Solution__

To factor the above trinomial, we need to write it in the form.

\( 9x^2 + 3x - 2 = (a x + m)(b x + n) \)

Expand the product on the right above

\( 9x^2 + 3x - 2 = abx^2 + x(m b + n a) + m n \)

and find the coefficients \( a, m, b \) and \( n \)

For the polynomial on the left to be equal to the polynomial on the right we need to have equal corresponding coefficients, hence

\( a b = 9 \)

\( m b + n a = 3 \)

\( m n = -2 \)

Trial values for \(a \) and \(b\) in the equation \( a b = 9 \) are:

\( a = 1 \) and \( b = 9 \) or \( a = 3 \) and \( b = 3 \)

Trial values for \( m \) and \( n \) in the equation \( m n = -2 \) are:

\( m = 1 , n = -2 \), \( m = 2 , n = -1\) , \( m = -1 , n = 2 \) and \( m = -2 , n = 1 \).

We now select values of the coefficients \( a, m, b \) and \( n \) such that \( m b + n a = 3 \)

\( a = 3 , b = 3, m = 2 \) and \( n = -1 \)

and write the polynomial in factored form

\( 9x^2 + 3x - 2 = (ax + m)(bx + n) = (3x + 2)(3x - 1) \)

As a practice, multiply the factored form \( (3x + 2)(3x - 1) \) and simplify to obtain the given trinomial.

### Example 4 - Factor by Grouping

Factor the polynomial \( \quad x^3 + 2x^2 - 16x - 32 \)

__Solution__

Group terms that have common factors.

\( x^3 + 2x^2 - 16x - 32 = (x^3 + 2x^2) - (16x + 32) \)

Factor the grouped terms

\( = x^2(x + 2) - 16(x + 2) \)

Factor \( x + 2 \) out

\( = (x + 2)(x^2 - 16) \)

The term \( x^2 - 16 \) is the difference of two squares and can be factored using formula 3(1) above

\( = (x + 2)(x + 4)(x - 4) \)

Check the above answer by expanding the obtained result.

## Questions

Part A

Factor the polynomials.

1: \( (x + 1)^2 - 4 \)

2: \( x^2 + 5x + 6 \)

3: \( x^3 - 1 \)

4: \( x^3 - x^2 - 25x + 25 \)

5: \( 27 + 8 x^3 \)

Part B

Prove that \( x^3 - 3 x^2 y + 3 x y^2 - y^3 = (x - y)^3 \).

## Solutions to Above Questions

Part A

1: \( (x + 1)^2 - 4 = (x - 1)(x + 3) \) , using difference of two squares , formula 4.

2: \( x^2 + 5x + 6 = (x + 2)(x + 3) \)

using method of trinomials above starting with: \( \quad x^2 + 5x + 6 = (x + a) (x + b) \)

expand the term \( (x + a) (x + b) \): \( \quad x^2 + 5x + 6 = x^2 + x(a+b) + ab \)

which gives: \( a +b = 5 \) and \( a b = 6 \)

which gives: \( a +b = 5 \) and \( a b = 6 \)

trial values give: \( a = 2 \) and \( b = 3 \)

3: \( x^3 - 1 = (x - 1)(x^2 + x + 1) \) , using difference of two cubes, formula 6 above

4: \( x^3 - x^2 - 25x + 25 = (x - 1)(x + 5)(x - 5) \)

grouping: \( x^3 - x^2 - 25x + 25 = (x^3 - x^2) - (25 x - 25) \)

using common factors \( x^2 \) and \( 25\): \( \quad = x^2 (x - 1) - 25( x - 1) \)

using common factor \( x - 1 \): \( \quad = (x - 1) (x^2 - 25)\)

using difference of squares: \( \quad = (x - 1) (x+5)(x-5)\)

5: Rewrite given expression as: \( \quad 27 + 8 x^3 = 3^3 + (2x)^3 \)

which is a sum of two cubes.

use formula 5 above to write: \( \quad 3^3 + (2x)^3 = (3+2x)(3^2-3(2x)+(2x)^2) \)

expand the term \( (3^2-3(2x)+(2x)^2) \) and factor as follows: \( \quad 27 + 8 x^3 = (3+2x)(4x^2-6x+9) \)

Note that the polynomial \( 4x^2-6x+9 \) cannot be factored over the real numbers.

Part B

Group as follows \( \quad x^3 - 3 x^2 y + 3 x y^2 - y^3 = (x^3 - y^3) - (3 x^2 y - 3 x y^2 ) \)

Factor common terms in the expression \( (3 x^2 y - 3 x y^2 ) \) and rewrite the above as: \( \quad = (x^3 - y^3) - 3 x y (x - y) \)

Use formula 6 to rewrite \( (x^3 - y^3) \) in factored form and rewrite the above as: \( \quad = (x - y)(x^2 + x y + y^2) - 3 x y (x - y) \)

Factor out the common term \( x - y \) and rewrite the above as: \( \quad = (x - y)(x^2 + x y + y^2 - 3 xy) \)

Group like terms in the expression \( x^2 + x y + y^2 - 3 xy \) and rewrite as: \( \quad = (x - y)(x^2 - 2 x y + y^2) \)

Use formula 9 to rewrite the expression \( x^2 - 2 x y + y^2 \) in factored form and rewrite as: \( \quad = (x - y)(x - y)^2 \)

The above may be written as: \( \quad = (x - y)^3 \)

Hence \( \quad x^3 - 3 x^2 y + 3 x y^2 - y^3 = (x - y)^3 \)

## More references and links to polynomial functions

Multiplicity of Zeros and Graphs Polynomials.Find Zeros of Polynomial Functions - Problems

Polynomial Functions, Zeros, Factors and Intercepts