Factoring Special Cases

1. Factoring Special Cases ALGEBRA 1 LESSON 9-7 (For help, go to Lessons 8–4 and 9-4.) Simplify each expression. 1. (3x)22. (5y)23. (15h2)24. (2ab2)2 Simplify each product. 5. (c – 6)(c + 6) 6. (p – 11)(p – 11) 7. (4d + 7)(4d + 7) 5-6

2. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Solutions 1. (3x)2 = 32 • x2 = 9x22. (5y)2 = 52 • y2 = 25y2 3. (15h2)2 = 152 • (h2)2 = 225h44. (2ab2)2 = 22 • a2 • (b2)2 = 4a2b4 5. (c – 6)(c + 6) is the difference of squares.(c – 6)(c + 6) = c2 – 62 = c2 – 36 6. (p – 11)(p – 11) is the square of a binomial.(p – 11)2 = p2 – 2p(11) + 112 = p2 – 22p + 121 7. (4d + 7)(4d + 7) is the square of a binomial.(4d + 7)2 = (4d)2 + 2(4d)(7) + 72 = 16d2 + 56d + 49 5-6

3. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor m2 – 6m + 9. m2 – 6m + 9 = m • m – 6m + 3 • 3Rewrite first and last terms. = m • m – 2(m • 3) + 3 • 3 Does the middle term equal 2ab? 6m = 2(m • 3) = (m – 3)2Write the factors as the square of a binomial. 5-6

4. = (4h)2 + 2(4h)(5) + 52Does the middle term equal 2ab? 40h = 2(4h)(5) Factoring Special Cases ALGEBRA 1 LESSON 9-7 The area of a square is (16h2 + 40h + 25) in.2. Find the length of a side. 16h2 + 40h + 25 = (4h)2 + 40h + 52Write 16h2 as (4h)2 and 25 as 52. = (4h + 5)2Write the factors as the square of a binomial. The side of the square has a length of (4h + 5) in. 5-6

5. Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor each expression. 1.y2 – 18y + 81 2. 9a2 – 24a + 16 3.p2 – 169 4. 36x2 – 225 5. 5m2 – 45 6. 2c2 + 20c + 50 (y – 9)2 (3a – 4)2 (p + 13)(p – 13) (6x + 15)(6x – 15) 5(m + 3)(m – 3) 2(c + 5)2 5-6