ALGEBRA 1 Lesson 8-5 Warm-Up

ALGEBRA 1 “Factoring Trinomials of the Type x 2 + bx +c” (8-5) What is a “trinomial”? How do you factor a trinomial? Trinomial: a polynomial that consists of three unlike terms Examples: x 2 + 7x + 12 x 2 + bx + c To factor a trinomial of the form x 2 + bx + c, you must find two numbers that have a sum of b and a product of c Example: Factor x 2 + 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 x 2 + 7x = (x +3)(x + 4) Check: Does (x +3)(x + 4) = x 2 + 7x + 12? (x +3)(x + 4) = x 2 + 4x + 3x + 12FOIL = x 2 + 7x + 12  Combine like terms. S

ALGEBRA 1 “Factoring Trinomials of the Type x 2 + 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 x 2 + 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

ALGEBRA 1 “Factoring Trinomials of the Type x 2 + 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 x 2 + 7x + 12 S x2x2 nnn x nn 2n + 7 3n + 1 x + 4 3n + 1 xxx n x x x 1 x + 3 x

ALGEBRA 1 “Factoring Trinomials of the Type x 2 + bx +c” (8-5) Example: Factor d 2 – 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, d x + 42 = (d - 3)(d - 14) Check: Does (d -3)(d - 14) = d x + 42? (d -3)(d - 14) = d 2 – 3d – 14d + 42 FOIL = d 2 – 17d + 12  Combine like terms.

ALGEBRA 1 Factor x 2 + 8x Find the factors of 15. Identify the pair that has a sum of 8. Factors of 15Sum of Factors 1 and and 5 8 x 2 + 8x + 15 = (x + 3)(x + 5). = x 2 + 5x + 3x + 15 Check: x 2 + 8x + 15 (x + 3)(x + 5) = x 2 + 8x + 15 Factoring Trinomials of the Type x 2 + bx + c LESSON 8-5 Additional Examples

ALGEBRA 1 Factor c 2 – 9c 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. c 2 – 9c + 20 = (c – 5)(c – 4) Factors of 20Sum of Factors –1 and –20–21 –2 and –10–12 –4 and –5–9 Factoring Trinomials of the Type x 2 + bx + c LESSON 8-5 Additional Examples

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

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

ALGEBRA 1 Factor each expression. 1.c 2 + 6c + 92.x 2 – 11x g 2 – 2g – 24 4.y 2 + y – 1105.m 2 – 2mn + n 2 (c + 3)(c + 3)(x – 2)(x – 9)(g – 6)(g + 4) (y + 11)(y – 10)(m – n)(m – n) Factoring Trinomials of the Type x 2 + bx + c LESSON 8-5 Lesson Quiz