Factoring by Grouping

When you have 4 terms Remember– To factor out a GCF, we write the GCF on the outside and divide each term by the GCF. Ex) x5y – x3y + 3y = y(x5 – x3 + 3) You try: 3x2y - 9xy + 18xy2 3xy(x – 3 + 6y)

So, if instead of a variable, our GCF is a binomial, we do the same thing: 3(x – 2) + 5y(x – 2) Our GCF is: (x – 2) When we pull that out we get… (x – 2) (3 + 5y)

You try: x(a – 4) – 3( a – 4) Did you get: (a - 4)(x - 3)??? m(p + 3) + q(p + 3) Did you get: (p + 3)(m + q)???

What if the GCF is not obvious? What if I have a problems like: x2 + 3x + 2xy + 6y That is when we have to group terms together. Always try to group the 1st 2 and the last 2. x(x + 3) + 2y(x + 3) (x + 3)(x + 2y) Take out the GCF of each group What is in parentheses will be the same (most of the time)

Let’s try: y2 – 5wy + 4y - 20w y (y – 5w) + 4(y – 5w) (y – 5w)(y + 4) 2) ab + 7b – 3a – 21 b(a + 7) – 3(a + 7) (a + 7)(b – 3) If the middle sign is a negative, you must remove a negative from the second set. Remember, when you divide a number by a negative, the sign changes

What if what is in my parentheses is not the same??? x2 – 5x3 + 15x – 3 x2(1 – 5x) + 3(5x – 1) x2(1 – 5x) - 3(1 – 5x) (1 – 5x)(x2 – 3) You try: ax - bx + by – ay x(a - b) + y (b - a) x(a - b) – y (a – b) (a – b) (x – y) Is -1(5x – 1) the same thing as (1 -5x)???? Yes!!!! So, change that +3 to a -3 and then the insides of the ( ) will be the same thing!!

Try these: 1. 5xy2 – 20x + 3y3 – 12y 2x3 + 10x – x2y - 5y 12n3 + 15n2 + 4n + 5 Answers: 1. (5x + 3y)(y2 – 4) (5x + 3y)(y + 2)(y – 2) 2. (2x – y)(x2 + 5) 3. (3n2 + 1)(4n + 5)

Homework 1. 2x3 + 4x2 + x + 2 2. 2x3 + 6x2 + 3x + 9