1. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 5 Polynomials and Factoring

2. 5-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Factoring Trinomials of the Type ax 2 + bx + c Factoring with FOIL The Grouping Method 5.3

3. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. To Factor ax 2 + bx + c Using FOIL

4. 5-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 3x 2  14x  5 Solution 1. First, check for a common factor. There is none other than 1 or  1. 2. Find the First terms whose product is 3x 2. The only possibilities are 3x and x: (3x + )(x + ) 3. Find the Last terms whose product is  5. Possibilities are (  5)(1), (5)(  1) Important!: Since the First terms are not identical, we must also consider the above factors in reverse order: (1)(  5), and (  1)(5).

5. 5-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 4. Knowing that the First and Last products will check, inspect the Outer and Inner products resulting from steps (2) and (3) Look for the combination in which the sum of the products is the middle term. (3x  5)(x + 1) = 3x 2 + 3x  5x  5 = 3x 2  2x  5 (3x  1)(x + 5) = 3x 2 + 15x  x  5 = 3x 2 + 14x  5 (3x + 5)(x  1) = 3x 2  3x + 5x  5 = 3x 2 + 2x  5 (3x + 1)(x  5) = 3x 2  15x + x  5 = 3x 2  14x  5 Wrong middle term Correct middle term!

6. 5-6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 18x 3  141x 2  24x Solution 1. First, we factor out the largest common factor, 3x: 3x(6x 2  47x  8) 2. Factor 6x 2  47x  8. Since 6x 2 can be factored as 3x  2x or 6x  x, we have two possibilities (3x + )(2x + ) or (6x + )(x + ) 3. There are several pairs of factors of  8. List each way:  8, 11,  8  2, 4 4,  2 8,  1  1, 8 2,  4  4, 2

7. 5-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example TrialProduct (6x + 4)(x  2) 6x 2  12x + 4x  8 = 6x 2  8x  8 (6x  4)(x + 2) 6x 2 + 12x  4x  8 = 6x 2 + 8x  8 (6x + 1)(x  8) 6x 2  48x + x  8 = 6x 2  47x  8 We do not need to consider (3x + )(2x + ). The complete factorization is 3x(6x + 1)(x  8). Wrong middle term Correct middle term

8. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Tips for Factoring ax 2 + bx + c To factor ax 2 + bx + c (a > 0): 1.Make sure that any common factor has been factored out. 2.Once the largest common factor has been factored out of the original trinomial, no binomial factor can contain a common factor (other than 1 or –1). 3.If c is positive, then the signs in both binomial factors must match the sign of b.

9. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. 4.Reversing the signs in the binomials reverses the sign of the middle term of their product. 5.Organize your work so that you can keep track of which possibilities you have checked. 6.Remember to include the largest common factor—if there is one—in the final factorization. 7.Always check by multiplying.

10. 5-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 14x + 5  3x 2 Solution It is an important problem-solving strategy to find a way to make problems look like problems we already know how to solve. Rewrite the equation in descending order. 14x + 5  3x 2 =  3x 2 + 14x + 5 Factor out the  1:  3x 2 + 14x + 5 =  1(3x 2  14x  5) =  1(3x + 1)(x  5) The factorization of 14x + 5  3x 2 is  1(3x + 1)(x  5).

11. 5-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 6x 2  xy  12y 2 Solution No common factors exist, we examine the first term, 6x 2. There are two possibilities: (2x + )(3x + ) or (6x + )(x + ). The last term  12y 2, has the following pairs of factors: 12y,  y6y,  2y4y,  3y and  12y, y  6y, 2y  4y, 3y as well as each of the pairing reversed. Some trials like (2x  6y)(3x + 2y) and (6x + 4y) (x  3y), cannot be correct because (2x  6y) and (6x + 4y) contain a common factor, 2.

12. 5-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example TrialProduct (2x + 3y)(3x  4y)6x 2  8xy + 9xy  12y 2 = 6x 2 + xy  12y 2 Our trial is incorrect, but only because of the sign of the middle term. To correctly factor, we simply change the signs in the binomials. (2x  3y)(3x + 4y)6x 2 + 8xy  9xy  12y 2 = 6x 2  xy  12y 2 The correct factorization is (2x  3y)(3x + 4y).

13. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. To Factor ax 2 + bx + c, Using the Grouping Method 1.Factor out the largest common factor, if one exists. 2.Multiply the leading coefficient a and the constant c. 3.Find a pair of factors of ac whose sum is b. 4.Rewrite the middle term, bx, as a sum or difference using the factors found in step (3). 5.Factor by grouping. 6.Include any common factor from step (1) and check by multiplying.

14. 5-14 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor 4x 2  5x  6 Solution 1. First, we note that there is no common factor (other than 1 or  1). 2. We multiply the leading coefficient, 4 and the constant,  6: (4)(  6) =  24. 3. We next look for the factorization of  24 in which the sum of the factors is the coefficient of the middle term,  5.

15. 5-15 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Pairs of Factors of  24 Sums of Factors 1,  24  23  1, 24 23 2,  12  10  2, 12 10 3,  8 55  3, 8 5 4,  6 22  4, 6 2 We would normally stop listing pairs of factors once we have found the one we are after.

16. 5-16 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 4. Next, we express the middle term as a sum or difference using the factors found in step (3):  5x =  8x + 3x. 5. We now factor by grouping as follows: 4x 2  5x  6 = 4x 2  8x + 3x  6 = 4x(x  2) + 3(x  2) = (x  2)(4x + 3) 6. Check: (x  2)(4x + 3) = 4x 2 + 3x  8x  6 = 4x 2  5x  6 The factorization of 4x 2  5x  6 is (x  2)(4x + 3).

17. 5-17 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor 8x 3 + 10x 2  12x Solution 1. We factor out the largest common factor, 2x: 8x 3 + 10x 2  12x = 2x(4x 2 + 5x  6) 2. To factor 4x 2 + 5x  6 by grouping, we multiply the leading coefficient, 4 and the constant term (  6): 4(  6) =  24. 3. We next look for pairs of factors of  24 whose sum is 5. Pairs of Factors of  24 Sums of Factors 3,  8 55  3, 8 5

18. 5-18 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 4. We then rewrite the 5x in 4x 2 + 5x  6 using 5x =  3x + 8x 5. Next, we factor by grouping: 4x 2 + 5x  6 = 4x 2  3x + 8x  6 = x(4x  3) + 2(4x  3) = (x + 2)(4x  3) 6. The factorization of the original trinomial 8x 3 + 10x 2  12x is 2x(x + 2)(4x  3).