In mathematics, factoring, also referred to as factorization, involves breaking down a number or mathematical objects (if possible) into a product of several factors.


1. Factor 24:

24 = 2 × 2 × 2 × 3

It is also possible to factor other mathematical objects, such as polynomials.

2. Factor x2 - 16:

x2 - 16 = (x - 4)(x + 4)

The above is an example of an expression that is relatively easy to factor. The format of the expression, a2 - b2, is referred to as a difference of squares.

When you see an expression of this format, you can factor it to (a - b)(a + b) as shown above.

Solving algebraic equations using factoring

In algebra, one method for solving equations is to factor them when possible. This is because factoring gives us an equation in the form of a product of expressions that we can set equal to 0. If the product of two (or more) expressions is equal to 0, as is the case when we factor polynomials, at least one of the expressions must equal 0. This allows us to separate each expression and set it equal to 0 to solve for the value(s) of x.


Solve x2 - 16 = 0

Since we factored this equation in example (2.) above, we can rewrite the above equation as:

(x - 4)(x + 4) = 0

We then know that at least one of the terms, x - 4 or x + 4 must equal 0 for the statement to be true, and can solve for x in either case:

x - 4 = 0

x = 4

x + 4 = 0

x = -4

Therefore, the solutions to the above equation are x = -4 and x = 4.

Using the FOIL method to factor

To factor more complicated polynomials, different methods need to be used including the FOIL method and completing the square. FOIL stands for "First Outer Inner Last," which refers to a method for multiplying binomials. In the context of factoring, the FOIL method is used to help visualize the binomials that make up a polynomial. Essentially, factoring is the opposite of expanding a binomial, and can be thought of as performing the FOIL method, backwards. To factor using the FOIL method, use the following steps, and refer to the example below.

  1. Set up a product of binomials. Write 2 empty parentheses that will be filled with 2 binomials that are equivalent to the original equation.
  2. Write values for the first term in each binomial such that the product of the values is equal to the first term of the expression being factored.
  3. Find a product of two values that is equal to the third term in the expression being factored, that when added, equals the coefficient of the second term in the equation. Write each value as the second term in each binomial with the appropriate sign.


Factor x2 + 3x - 28:

1. (  )(  )
2. (x  )(x  )
3. (x + 7)(x - 4)

We can check that x2 + 3x - 28 = (x + 7)(x - 4) by expanding using FOIL.

This works out to (a + b)(c + d) = ac + ad + bc + bd, or in the case of the example above:

(x)(x) + (x)(-4) + (7)(x) + (7)(-4)

x2 - 4x + 7x - 28

x2 + 3x - 28

This is the same equation we started with, confirming that we factored it correctly. If we were to solve for this equation using the factored form:

x + 7 = 0

x = -7

x - 4 = 0

x = 4

The solutions to the equation above are x = -7 and x = 4.