1. Sect. 5.4 Factoring Trinomials  Factoring x 2 + bx + c Using a Trial Factor Table  Factoring ax 2 + bx + c Using a Grouping Factor Table 5.41

2. Remember FOIL?  (x + )(x + )  (x – )(x – )  (x + )(x – ) Examine the trinomial: You can predict the Binomial operation signs 5.42

3. Factoring Trinomials with Leading Coefficient=1 x 2 + bx + c b and c are #s 1. Write the trinomial in descending powers of one variable. 2. First - remove any common factor (there may not be one) 3. (not in book) Write down the binomial factor pair with middle operators + or – 4. Create a “Trial Factor Table”: 4.1 List the possible factor pairs of the 3 rd term’s coefficient. 4.2 List their corresponding sums. 5. Use the factor pair where the sum of the factors is the coefficient of the middle term. Therefore: (x – 3 )(x – 4 ) factor pairs of 12 factor sum = -7 -1 x -12 -1 + -12 = -13 no -2 x -6 -2 + -6 = -8 no -3 x -4 -3 + -4 = -7 yes! 5.43

4. Using a Factor Table for x 2 + bx + c - Organized Trial & Error  Let’s use x 2 + 13x + 36 as an example  Factors must both be sums: (x + ?)(x + ?)  Pairs=c=36 Sum=b=13  1, 3637  2, 1820  3, 1215  4, 913 ok quit!  x 2 + 13x + 36 = (x + 4)(x + 9) 5.44

5. Factoring Practice: 5.45

6. Factoring Practice: 5.46

7. Factoring Practice: 5.47

8. “Can’t Factor”? When There are 2 Variables x 2 – x – 7 x 2 + 3x – 42 x 2 – 2xy – 48y 2 5.48

9. 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 5.49

10. Factoring ax 2 + bx + c (a not 1 or 0) Factoring by Guessing 1. Write the trinomial in descending powers of one variable. 2. Factor out any greatest common factor (including 1, if that is necessary to make the coefficient of the first term positive). 3. Determine the signs of the factors: (assume a is positive) If c is + then both factors would need to have the sign of b If c is – then the factors must have different signs 4. Try various combinations of the factors of the first terms and the last terms until either: You find a pair of binomial factors that work, or Try all possible combinations but none work. (unfactorable) 5. Check the factorization by multiplication. Let’s guess: 3p 2 – 4p – 4 = (3p )(p ) 5.410

11. Use Grouping to Factor  16 + 24x + 5x 2  5x 2 + 24x + 16 (first rearrange) ac = 5(16) = 2· 2· 2· 2· 5 b = 24 1 · 80 1 + 80 = 81 No 2 · 40 2 + 40 = 42 No 4 · 20 4 + 20 = 24 YES  5x 2 + 4x + 20x + 16  x(5x + 4) + 4(5x + 4)  (5x + 4)(x + 4) ta-da! 5.411

12. The ac Grouping Method: ax 2 + bx + c Split bx into 2 Terms: Use a Grouping Factor Table  Let’s use 3x 2 – 10x – 8 as an example  ac = 3(-8) = -24 One factor is positive, the other negative and larger.  Pairs=ac=3(-8) = - 2 · 2 · 2 · 3 Sum=b=-10  1, -24-23  2, -12-10 quit!  3, -8 -5  4, -6 -2  3x 2 – 10x – 8 =  3x 2 + 2x – 12x – 8 = split the middle  x(3x + 2) – 4(3x + 2) = do grouping  (3x + 2)(x – 4) 5.412

13. ac Factoring Practice 1: 3x 2 + 10x – 8 5.413

14. ac Factoring Practice 2: 6x 6 – 19x 5 + 10x 4 5.414

15. ac Factoring Practice 3: 6x 4 – 116x 3 – 80x 2 5.415

16. What Next?  Section 5.5 – Factoring Perfect Squares & Differences of Squares5.5 5.416