1. Unit 2: Factoring Part I MM 218 McGrath

2. Vocubulary A factor is a number, variable, or algebraic expression multiplying another number, variable or algebraic expression. Example: 4*3 (both the “4” and the “3” are factors) Example: 2x*3y 2 (both the “2x” and the “3y 2 ” are factors)

3. Vocabulary A greatest common factor (GCF) is the largest factor common to 2 or more terms Example: Find the GCF of 12 and 18 Factors of 12: 1, 2, 3, 4, 6, 12 Factors of 18: 1, 2, 3, 6, 9, 18 Common Factors:1, 2, 3, 6 Greatest Common Factor: 6

4. GCF Cont’d Example: Find the GCF of 6x 3 and 15x 2 Factors of 6 : 1, 2, 3, 6 Factors of x 3 : x*x*x Factors of 15: 1, 3, 5, 15 Factors of x 2 : x*x Common Factors:1, 3, 6, x*x Greatest Common Factor: 6x 2

5. Factoring out a GCF This is distribution in reverse! Distribute: 2(x + 3) = 2(x) + 2(3) = 2x + 6 Factoring: 2x + 6 = 2(x) + 2(3) = 2(x + 3)

6. Factoring out the GCF Factor: 3x – 12 Factor 12y 2 – 5y

7. Factoring out the GCF Factor: 4x 2 y 3 + 10x 5 y - 6x 3 y 2

8. Practice with GCFs 1.2a + 2b + 2c 2.3xy – 4xy 2 + 5x 2 y 3 3.7x 2 + 21x + 14

9. A few more… x + 3 = x( ) + 3( ) = x(x – 2) + 3(x – 2) =

10. Factor by Grouping This is the same as doing the GCF, but now we do it THREE times! Factor x(z + w) + 2(z + w) Example: Factor ad + 3a – d 2 – 3d Solution: ad + 3a – d 2 – 3d = ad – d 2 + 3a – 3d = = d(a – d) + 3(a – d) = = (a – d)(d + 3)

11. Practice with Factoring by Grouping 1.4x 3 + 2x 2 – 6x – 3 2.3xy + 21x - 2y – 14

12. Factoring Trinomials Factoring by Grouping STEPS: 1. Multiply the first and last term. 2.Identify the coefficient of the x term. 3.Find two numbers that MULTIPLY to the product of the first and last term (Step 1) and ADD to the coefficient of the x term (Step 2). 4.Rewrite the original problem, replacing the middle term with the numbers from Step 3. 5.Factor by grouping.

13. ExampleSteps Factor: 10a 2 – a – 2 We need to find two numbers that: 1) multiply to -20; and 2) add to -1 The only two numbers that work are -5 and +4. =10a 2 – 5a + 4a – 2 Rewrite the equation by substituting the new values in for the middle term. =5a(2a – 1) + 2(2a – 1) =(2a – 1)(5a + 2) Factor by grouping.

14. Factor: x 2 – 12x + 32

15. Factor: 3x 2 – 6x + 3

16. Factor: 4x 2 – 14x – 30