# Factor Polynomials

Tutorial with detailed solutions on factoring polynomials.

## Factoring Polynomials Formulas

1:    One common factor.   a x + a y = a (x + y)
2:    Sevearl grouped common factor.   a x + a y + b x + b y = a(x + y) + b(x + y) = (a + 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

Factor the binomial 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, multiply (3 - 2x)(3 + 2x) to obtain the given expression.

### Example 2

Factor the trinomial 9x 2 + 3x - 2
Solution
To factor the above trinomial, we need to write it in the form.
9x
2 + 3x - 2 = (ax + m)(bx + n)
Expand the product on the right above
9x
2 + 3x - 2 = abx 2 + x(mb + na) + mn
For the polynomial on the left to be equal to the polynomial on the right we need to have equal corresponding coefficients, hence
ab = 9
mb + na = 3
mn = -2
Trial values for a and b are: a = 1 and b = 9 or a = 3 and b = 3
Trial values for m and n are: m = 1 and n = -2, m = 2 and n = -1, m = -1 and n = 2 and m = -2 and n = 1.
Trying various values for a, b, m and n among the list above, we arrive at:
9x
2 + 3x - 2 = (3x + 2)(3x - 1)
As a practice, multiply (3x + 2)(3x - 1) and simplify to obtain the given trinomial.

### Example 3

Factor the polynomial 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 1 above
= (x + 2)(x + 4)(x - 4)
Check the above answer by expanding the obtained result.

## Exercises

Factor the polynomials.
1:       (x + 1)
2 - 4
2:       x
2 + 5x + 6
3:       x
3 - 1
4:       x
3 - x 2 - 25x + 25

Solutions to above exercises
1:       (x + 1)
2 - 4 = (x - 1)(x + 3)
2:       x
2 + 5x + 6 = (x + 2)(x + 3)
3:       x
3 - 1 = (x - 1)(x 2 + x + 1)
4:       x
3 - x 2 - 25x + 25 = (x - 1)(x + 5)(x - 5)

## 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