Lesson 8-5 Warm-Up

1. Lesson 8-5 Warm-Up

2. “Factoring Trinomials of the Type x2 + bx +c” (8-5) • What is a “trinomial”? • How do you factor a trinomial? • Trinomial: a polynomial that consists of three unlike terms • Examples: x2 + 7x + 12 x2 + bx + c • To factor a trinomial of the form x2 + bx + c, you must find two numbers that have a sum of b and a product of c • Example: Factor x2 + 7x + 12 • Notice that the coefficient of the middle term, b or 7, is the sum of 3 and 4. Also, the constant, c or 12, is the product of 3 and 4. Therefore, you can now create two binomials whose product is x 2 + 7x + 12. • x 2 + 7x + 12. = (x +3)(x + 4) • Check: Does (x +3)(x + 4) = x 2 + 7x + 12? • (x +3)(x + 4) = x 2 + 4x + 3x + 12 FOIL • = x 2 + 7x + 12  Combine like terms. • S

3. “Factoring Trinomials of the Type x2 + bx +c” (8-5) • How do you find two numbers that have a sum of b and a product of c? • Method 1: Create a Table: Title one column “Factors of (Constant)” or “Factors of “c” and the other column “Sum of the Factors”. Then, fill in the table with the number pairs that are factors of the constant. • Example: Factor x2 + 7x + 12 • To factor this polynomial, we’ll need to find factors pairs of 12 (two numbers whose product is 12) whose sum is 7. To do this create a table. • S

4. “Factoring Trinomials of the Type x2 + bx +c” (8-5) • Method 2: Use an Area Model in Reverse: Arrange the Algebra Tiles that model the trinomial into a rectangle. The sides of the rectangle (length and width) are the factors of the trinomial. Tip: Think about how to end with the number of desired “1” tiles. • Example: Factor x2 + 7x + 12 • S n n n n x+ 4 x2 x x x x 3n+ 1 3n+ 1 x+ 3 x+ 3 x 1 1 1 1 x 1 1 1 1 x 1 1 1 1 x+ 4 n n 2n+ 7

5. “Factoring Trinomials of the Type x2 + bx +c” (8-5) • Example: Factor d2 – 17d + 42 • To factor this polynomial, we’ll need to find factors pairs of 42 (two numbers whose product is 42) whose sum is -17. To do this create a table. • So, d2 - 17x + 42 = (d - 3)(d - 14) • Check: Does (d -3)(d - 14) = d 2 - 17x + 42? • (d -3)(d - 14) = d 2 – 3d – 14d + 42 FOIL • = d2 – 17d + 12  Combine like terms.

6. Factors of 15 Sum of Factors 1 and 15 16 3 and 5 8 Check:x2 + 8x + 15 (x + 3)(x + 5) = x2 + 8x + 15 Factoring Trinomials of the Type x2 + bx + c LESSON 8-5 Additional Examples Factor x2 + 8x + 15. Find the factors of 15. Identify the pair that has a sum of 8. x2 + 8x + 15 = (x + 3)(x + 5). = x2 + 5x + 3x + 15

7. Factors of 20 Sum of Factors –1 and –20 –21 –2 and –10 –12 –4 and –5 –9 Factoring Trinomials of the Type x2 + bx + c LESSON 8-5 Additional Examples Factor c2 – 9c + 20. Since the middle term is negative, find negative factors of 20 (a negative times a negative equals a positive). Identify the pair that has a sum of –9. c2 – 9c + 20 = (c – 5)(c – 4)

8. Factors of –48 Sum of Factors 1 and –48 –47 48 and –1 47 2 and –24 –22 24 and –2 22 3 and –16 –13 16 and –3 13 Factors of –24 Sum of Factors 1 and –24 –23 24 and –1 23 2 and –12 –10 12 and –2 10 3 and –8 –5 Factoring Trinomials of the Type x2 + bx + c LESSON 8-5 Additional Examples a.Factor x2 + 13x – 48. b.Factor n2 – 5n – 24. Identify the pair of factors of –48 that has a sum of 13. Identify the pair of factors of –24 that has a sum of –5. x2 + 13x – 48 = (x + 16)(x – 3) n2 – 5n – 24 = (n + 3)(n – 8)

9. Find the factors of –60. Identify the pair that has a sum of 17. Factors of –60 Sum of Factors 1 and –60 –59 60 and –1 59 2 and –30 –28 30 and –2 28 3 and –20 –17 20 and –3 17 Factoring Trinomials of the Type x2 + bx + c LESSON 8-5 Additional Examples Factor d + 17dg – 60g . 2 2 d2 + 17dg – 60g2= (d – 3g)(d + 20g)

10. Factoring Trinomials of the Type x2 + bx + c LESSON 8-5 Lesson Quiz Factor each expression. 1.c2 + 6c + 9 2.x2 – 11x + 18 3.g2 – 2g – 24 4.y2 + y – 110 5.m2 – 2mn + n2 (c + 3)(c + 3) (x – 2)(x – 9) (g – 6)(g + 4) (y + 11)(y – 10) (m – n)(m – n)