1. Objective The student will be able to: factor quadratic trinomials. Trial and Error Method SOL: A.2c Designed by Skip Tyler, Varina High School

2. Factoring Chart This chart will help you to determine which method of factoring to use. TypeNumber of Terms 1. GCF 2 or more 2. Diff. Of Squares 2 3. Trinomials 3

3. Review: (y + 2)(y + 4) Multiply using FOIL or using the Box Method. Box Method: y + 4 y y 2 +4y + 2 +2y +8 Combine like terms. FOIL: y 2 + 4y + 2y + 8 y 2 + 6y + 8

4. 1) Factor. y 2 + 6y + 8 Put the first and last terms into the box as shown. What are the factors of y 2 ? y and y y2y2 + 8

5. 1) Factor. y 2 + 6y + 8 Place the factors outside the box as shown. y2y2 + 8 y y What are the factors of + 8? +1 and +8, -1 and -8 +2 and +4, -2 and -4

6. The second box works. Write the numbers on the outside of box for your solution. 1) Factor. y 2 + 6y + 8 Which box has a sum of + 6y? y2y2 + 8 y y y2y2 y y+ 1+ 2 y + 8 y + 4 + y + 8y+ 4y + 2y

7. 1) Factor. y 2 + 6y + 8 (y + 2)(y + 4) Here are some hints to help you choose your factors. 1) When the last term is positive, the factors will have the same sign as the middle term. 2) When the last term is negative, the factors will have different signs.

8. x2x2 - 63 2) Factor. x 2 - 2x - 63 Put the first and last terms into the box as shown. What are the factors of x 2 ? x and x

9. 2) Factor. x 2 - 2x - 63 Place the factors outside the box as shown. x2x2 - 63 x x What are the factors of - 63? Remember the signs will be different!

10. 2) Factor. x 2 - 2x - 63 Use trial and error to find the correct combination! Do any of these combinations work? The second one has the wrong sign! x2x2 - 63 x x + 21 - 3 +21x -3x x2x2 - 63 x x- 7 + 9 -7x +9x

11. 2) Factor. x 2 - 2x - 63 Change the signs of the factors! Write your solution. (x + 7)(x - 9) x2x2 - 63 x x+ 7 - 9 +7x -9x

12. 5x 2 + 14 3) Factor. 5 x 2 - 17x + 14 Put the first and last terms into the box as shown. What are the factors of 5x 2 ? 5x and x

13. 3) Factor. 5 x 2 - 17x + 14 5x 2 + 14 x 5x What are the factors of + 14? Since the last term is positive, the signs of the factors are the same! Since the middle term is negative, the factors must be negative!

14. 3) Factor. 5 x 2 - 17x + 14 When the coefficient is not 1, you must try both combinations! Do any of these combinations work? The second one! Write your answer. 5x 2 + 14 x 5x - 7 - 2 - 2x -35x 5x 2 + 14 x 5x - 2 - 7 - 7x -10x

15. 3) Factor. 5 x 2 - 17x + 14 (5x - 7)(x - 2) It is not the easiest of things to do, but the more problems you do, the easier it gets! Trust me! 4) Factor 2 x 2 + 9x + 10 (x + 2)(2x + 5)

16. 5) Factor. 6y 2 - 13y - 5 (2y - 5)(3y + 1) 6) 12x 2 + 11x - 5 (4x + 5)(3x - 1) 7) 5x - 6 + x 2 x 2 + 5x - 6 (x - 1)(x + 6)