1. Sect. 5.3 Common Factors & Factoring by Grouping  Definitions Factor Common Factor of 2 or more terms  Factoring a Monomial into two factors  Identifying Common Monomial Factors  Factoring Out Common Factors  Arranging 4 Term Polynomials into 2 Groups  Factoring Out Common Binomials 5.31

2. What’s a Factor? 84 is a product that can be expressed by many different factorizations: 84 = 2(42) or 84 = 7(12) or 84 = 4(7)(3) or 84 = 2(2)(3)(7) Only one example, 84 = 2(2)(3)(7), shows 84 as the product of prime integers. Factoring is the reverse of multiplication. product = (factor)(factor)(factor) … (factor) 5.32

3. Factoring Monomials  12x 3 also can be expressed in many ways: 12x 3 = 12(x 3 ) 12x 3 = 4x 2 (3x) 12x 3 = 2x(6x 2 )  Usually, we only look for two factors  Your turn – factor these monomials into two factors:  4a = 2(2a) or 4(a)  x 3 = x(x 2 ) or x 2 (x)  14y 2 = 14(y 2 ) or 14y(y) or 7(2y 2 ) or 7y(2y) or y(14y) or …  43x 5 = 43(x 5 ) or 43x(x 4 ) or x 3 (43x 2 ) or 43x 2 (x 3 ) or … 5.33

4. Common Factors  Sometimes multi-termed polynomials can be factored  Looking for common factors in 2 or more terms … is the first step in factoring polynomials  Remember a(b + c) = ab + ac (distributive law)  Consider that a is a common factor of ab + ac so we can factor ab + ac into a(b + c)  For x 2 + 3x the only common factor is x, so x 2 + 3x = x (? + ?) = x(x + 3)  Another example: 4y 2 + 6y – 10  The common factor is 2 4y 2 + 6y – 10 = 2(? + ? – ?) = 2(2y 2 + 3y – 5)  Check by multiplying: 2(2y 2 ) + 2(3y) – 2(5) = 4y 2 + 6y – 10 5.34

5. Find the Greatest Common Factor  7a – 21 = 7(? – ?) = 7(a – 3)  19x 3 + 3x = x(? + ?) = x(19x 2 + 3)  18y 3 – 12y 2 + 6y = 6y(? – ? + ?) = 6y(3y 2 – 2y + 1) 5.35

6. Introduction to Factoring by Grouping: Factoring Out Binomials  x 2 (x + 7) + 3(x + 7) = (x + 7)(? + ?) = (x + 7)(x 2 + 3)  y 3 (a + b) – 2(a + b) = (a + b)(? – ?) = (a + b)(y 3 – 2)  You try: 2x 2 (x – 1) + 6x(x – 1) + 17(x – 1) = (x – 1)(? + ? – ?) (x – 1)(2x 2 + 6x + 17) 5.36

7. Factoring by Grouping  For polynomials with 4 terms: 1. Arrange the terms in the polynomial into 2 groups such that each group has a common monomial factor 2. Factor out the common monomials from each group (the binomial factors produced will be either identical or opposites) 3. Factor out the common binomial factor  Example: 2c – 2d + cd – d 2  2(c – d) + d(c – d)  (c – d)(2 + d) 5.37

8. Factor by Grouping  8t 3 + 2t 2 – 12t – 3  2t 2 (4t + 1) – 3(4t + 1)  (4t + 1)(2t 2 – 3)  4x 3 – 6x 2 – 6x + 9  2x 2 (2x – 3) – 3(2x – 3)  (2x – 3)(2x 2 – 3)  y 4 – 2y 3 – 12y – 3  y 3 (y – 2) – 3(4y + 1)  Oops – not factorable via grouping 5.38

9. What Next?  Next time: Section 5.4 – Factoring Trinomials5.4 5.39