Look for the greatest common factor (GCF). In the following example, 10x^2 + 15x -- 45, the GCF is five. The GCF is always pulled out before the rest of the factoring beings, 5(x + x)(x + x), regardless of the number of terms within the polynomials.
Examine the polynomial to see if it contains only two terms, also called a binomial.
The formula a^2 -- b^2 is an example of a two-termed polynomial and is described as a difference of squares. The formula for difference of squares expressions is a^2 -- b^2 = (a -- b)(a + b). Notice that you have one of each signs within the factored binomials.
Examine the polynomial to see if it is a trinomial, that is, it contains only three terms, e.g. a^2 + 2ab + b^2 and a^2 -- 2ab + b^2. This is a special case example. The first term and last term are perfect squares and the inside term is twice the product of the outside terms. In this case, both signs within the factored binomials will be either positive or negative.
Examine the polynomial to see if it contains four terms, e.g. 2xy -- 8x -- 3y + 12. To factor four terms, you must divide the polynomial expression down the middle and factor one side at a time. After pulling out the GCF, the binomials within the parentheses must match.
Apply the trial and error method to any trinomials that do not follow the special case example. For example, examine the expression x^2 -- 7xy + 12y^2. The outside terms are squared but the inside term is not the product of the outside terms, meaning 7 does not equal (12 x 1). Instead, the middle term must equal the sum of the products of the first and last terms when factored out. (x -- 4y)(x -- 3y) -4xy + (-3xy) = -7xy.
Examine the expression 15x^2 -- 10. Pull out the GCF five and then divide the polynomial by the GCF to create the parenthetical binomials. Five divides into 15x^2 three times and into 10 twice. 5(3x^2 -- 2). This is as far as you can go with this example.
Examine the expression h^2 -- 16. Both h^2 and 16 are square roots, but notice the subtraction sign. This is an example of a difference of squares.
Factor the binomial out. This is the process of finding the square root of each square and placing it into parentheses. The square root of h^2 is h and the square root of -16 is four. (h + 4)(h -- 4).
Distribute to double check your work. h x h = h^2, h x -4 = -4h, h x 4 = 4h and 4 x -4 = -16. Combine like terms, 4h -- 4h = 0, thereby canceling each other out. You are left with the original binomial, which confirms the factoring process.
Examine the expression m^2 -- 6my + 9y^2. This is an example of a perfect square, both the first and last terms are squared.
Factor the trinomial out. (m -- 3y)(m -- 3y).
Distribute to check your work. m^2 -- 3my -- 3my + 9y^2. After combining like terms, -3my -- 3my = -6my, the trinomial is the same as the original. The simplified answer to this expression is (m -- 3y)^2.
Examine the polynomial 2xy + 12 -- 3y -- 8x. Normally, the polynomial would be divided down the center and you would factor one side at a time. However, in this example, the terms need to be reordered so that the variables are listed from greatest to least, which makes factoring by grouping easier. 2xy -- 8x -- 3y + 12.
Divide the expression down the middle to address one pair of polynomials at a time, e.g. 2xy -- 8x and then -3y -- 12.
Pull out the GCF of 2xy -- 8x and then factor. 2x(y -- 4). Pull out the GCF of -3y -- 12 and then factor. -3(y -- 4). Notice that the parenthesis match. This is the key to making factoring by grouping work.
Write the parenthetical binomials in the answer, (y -- 4). Place the terms outside of the parenthesis into the answer, (y -- 4)(2x -- 3)
Examine the trinomial 7a^2 + 17a -- 12.
Look for the GCF, there is none. Is the first term a square? Yes, but the last term is not, so this is not an example of a perfect square but an example of the trial by error method.
Write the easy part first. The square root of 7a^2 is 7a x a, so fill that into the factored set of parenthesis, (7a + ...)(a - ...). Because the middle term is positive and the last term is negative, use one of each sign.
List all of the factors of the last term, 12. 1 x 12, 2 x 6 and 3 x 4. Select the group of factors that will equal the middle term after factoring. For example, if you use two and six, the expression is (7a + 6)(a -- 2). However, after distributing, you see that you are left with the middle terms 14a + 6a, which equals 20a. Instead, use four and three, (7a -- 4)(a + 3)
Distribute to check your work. 7a x a = 7a^2, 7a x 3 = 21a, -4 times a = -4a and -4 x 3 = -12. Combine like terms, 21a -- 4a = 17a.