1. First look for common factors:

• e.g. 10x2+ 6x – 8 (factor out common 2) = 2(5x2+ 3x – 4)
• e.g. 9x4y – 3x3y2 + 5x2y2 (factor out common x2y) = x2y(9x2 – 3xy +5y)

2. Once you have removed all common factors decide which of the following type of expression you have:

• A fully factored expression which means you are done!
• A trinomial of type x2 + Bx + C (see #3 below).
• A trinomial of type Ax2 + Bx + C (see #4 below).
• A difference of squares (see #5 below)

3. There are three varieties of trinomials of the type x2 + Bx + C. For all types the middle term is the sum of the two number factors and the last term is the product of the two number factors.

1. e.g. x2 + 8x + 15 = (x + 3)(x + 5) since 3 + 5 = 8 and 3 x 5 = 15
2. e.g. x2 – 6x + 8 = (x – 4)(x – 2) since -4 + -2 = -6 and -4 x -2 = 8
3. e.g. x2 – 3x – 28 = (x – 7)(x + 4) since -7 = 4 = -3 and -7 x 4 = -28

4. There are a number of different methods to factor these type of trinomials (Ax2 + Bx + C) including trial and error. I will show you one method only:

1. e.g. 6x2 + 19x + 15
• step 1: decide which two numbers add to the middle term (19) and multiply to the product of the two outside terms (90).
• step 2: rewrite the trinomial with those two terms in the middle: 6x2 + 9x + 10x + 15
• step 3: factor out the common factors from the first 2 terms and then from the second 2 terms: 3x(2x + 3) + 5(2x + 3)
• step 4: pull out the common factor and you are done: (2x + 3)(3x + 5)
2. e.g. 3x2 – 4x – 15
• step 1: decide which two numbers add to the middle term (-4) and multiply to the product of the two outside terms (-45).
• step 2: rewrite the trinomial with those two terms in the middle: 3x2 – 9x + 5x – 15
• step 3: factor out the common factors from the first 2 terms and then from the second 2 terms: 3x(x – 3) + 5(x- 3)
• step 4: pull out the common factor and you are done: (x – 3)(3x + 5)

5. Difference of squares have only two terms, each of which must be a perfect square. To factor, you simply take the square root of the first term and add the square root of the second term. You then repeat, taking the square root of the first term and subtracting the square root of the second term.

1. e.g. 4x2 – 25 = (2x + 5)(2x – 5)
2. e.g. 9x2 – 64y2 = (3x + 8y)(3x – 8y)
3. e.g. (x – 2)2 – (x + 3)2 =((x – 2) + (x + 3))((x – 2) – (x + 3)) = (2x + 1)(-5) or -5(2x + 1)