# Factoring

## Common Factoring

If we write a polynomial as a product of other polynomials, this is called factoring. Notice in the polynomial below that both 45 and 18 are multiples of 9 and both x^{3} and x are multiples of x. Therefore, the polynomial can be divided by both 9 and x (ie. both 9 and x are factors of the polynomial). In this case, 9x is the greatest common factor.

45x^{3} + 18x

= 9x(5x^{2} + 2)

When we want to solve a polynomial equation, it may be possible to use factoring to find the solution. After rearranging the equation and factoring the polynomial, set each factor equal to zero to find all the solutions.

**Example**: Solve 63x^{2 }= 7x for x.

*Solution:*

63x^{2}= 7x

63x^{2} - 7x = 0

7x(9x - 1) = 0

If two factors multiplied together give 0, then at least one of them must be 0.

7x = 0 or 9x - 1 = 0

x = 0 9x = 1

x = 1/9

## Factoring Trinomials

Just as two binomials can be multiplied together to form a trinomial, trinomials can sometimes be factored resulting in two binomials. In order to factor a trinomial, the second term often has to be broken into two parts. To accomplish this, we do the following steps:

- Write the trinomial in order of decreasing powers of x.
- Look for 2 numbers that have the following properties:
- Their sum gives the coefficient of the 2nd term in the trinomial
- Their product gives the coefficient of the 3rd term in the trinomial

**Example: **Solve x^{2} + x - 6 = 0

*Solution*: First, we factor the left-hand side of this equation.

x^{2} + x - 6 = 0

x^{2} + 3x - 2x - 6 = 0

(x^{2} + 3x) + (-2x - 6) = 0

x(x + 3) + (-2)(x + 3) = 0

(x + 3)(x - 2) = 0

Here, 3 and -2 add together to equal 1, which is the coefficient of the second term; and the product of 3 and -2 is -6, which the coefficient of the third term. Therefore, by this method, the factors are (x + 3)(x - 2). You'll note that you can skip the intermediate steps once you're comfortable with this method.

Now we can solve by setting each factor in turn equal to zero to obtain the possible solutions.

x + 3 = 0 or x - 2 = 0

x = -3 x = 2

**Example: **Solve x^{2} = 2x + 15 for x.

x^{2} = 2x + 15

x^{2} - 2x - 15 = 0

x^{2} -5x + 3x - 15 = 0

x(x - 5) + 3(x - 5) = 0

(x - 5)(x + 3) = 0

x - 5 = 0 or x + 3 = 0

x = 5 x = -3

In the above examples, you could also have used the "quadratic formula" to solve; however, this only applies to quadratic equations.

**Example:**

**Example:**