Factoring Special Cases ALGEBRA 1 LESSON 7 7

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

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

Factoring Special Cases ALGEBRA 1 LESSON 7 -7 Factor a 2 – 16 = a 2 – 42 = (a + 4)(a – 4) Rewrite 16 as 42. Factor. Check: Use FOIL to multiply. (a + 4)(a – 4) a 2 – 4 a + 4 a – 16 a 2 – 16 7 -7

Factoring Special Cases ALGEBRA 1 LESSON 7 -7 Factor 9 b 2 – 25. 9 b 2 – 225 = (3 b)2 – 52 = (3 b + 5)(3 b – 5) Rewrite 9 b 2 as (3 b)2 and 25 as 52. Factor. 7 -7

Factoring Special Cases ALGEBRA 1 LESSON 7 -7 Factor 5 x 2 – 80 = 5(x 2 – 16) Factor out the GCF of 5. = 5(x + 4)(x – 4) Factor (x 2 – 16). Check: Use FOIL to multiply the binomials. Then multiply by the GCF. 5(x + 4)(x – 4) 5(x 2 – 16) 5 x 2 – 80 7 -7

Factoring by Grouping ALGEBRA 1 LESSON 7 -7 Factor 6 x 3 + 3 x 2 – 4 x – 2 = 3 x 2(2 x + 1) – 2(2 x + 1) = (2 x + 1)(3 x 2 – 2) Check: 6 x 3 + 3 x 2 – 4 x – 2 Factor the GCF from each group of two terms. Factor out (2 x + 1)(3 x 2 – 2) = 6 x 3 – 4 x + 3 x 2 – 2 Use FOIL. = 6 x 3 + 3 x 2 – 4 x – 2 Write in standard form. 7 -7

Factoring by Grouping ALGEBRA 1 LESSON 7 -7 Factor 8 t 4 + 12 t 3 + 16 t + 24 = 4(2 t 4 + 3 t 3 + 4 t + 6) Factor out the GCF, 4. = 4[t 3(2 t + 3) + 2(2 t + 3)] Factor by grouping. = 4(2 t + 3)(t 3 + 2) Factor again. 7 -7

Factoring by Grouping ALGEBRA 1 LESSON 7 -7 Factor 24 h 2 + 10 h – 6. Step 1: 24 h 2 + 10 h – 6 = 2(12 h 2 + 5 h – 3) Factor out the GCF, 2. Step 2: 12 • – 3 = – 36 Find the product ac. Step 3: Factors – 2(18) = – 36 – 3(12) = – 36 – 4(9) = – 36 Sum – 2 + 18 = 16 – 3 + 12 = 9 – 4 + 9 = 5 Find two factors of ac that have a sum b. Use mental math to determine a good place to start. Step 4: 12 h 2 – 4 h + 9 h – 3 Rewrite the trinomial. Step 5: 4 h(3 h – 1) + 3(3 h – 1) Factor by grouping. (4 h + 3)(3 h – 1) Factor again. 24 h 2 + 10 h – 6 = 2(4 h + 3)(3 h – 1) 7 -7 Include the GCF in your final answer.

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

Factoring by Grouping ALGEBRA 1 LESSON 7 -7 Factor each expression. 1. 10 p 3 – 25 p 2 + 4 p – 10 (5 p 2 + 2)(2 p – 5) 2. 36 x 4 – 48 x 3 + 9 x 2 – 12 x 3 x(4 x 2 + 1)(3 x – 4) 3. 16 a 3 – 24 a 2 + 12 a – 18 2(4 a 2 + 3)(2 a – 3) 7 -7