# 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