Factoring by Grouping ALGEBRA 1 LESSON 9-8 (For help, go to Lessons 9-2 and 9-3.) Find the GCF of the terms of each polynomial. 1.6y 2 + 12y – 42.9r 3.

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Factoring by Grouping ALGEBRA 1 LESSON 9-8 (For help, go to Lessons 9-2 and 9-3.) Find the GCF of the terms of each polynomial. 1.6y y – 42.9r r r 3.30h 3 – 25h 2 – 40h4.16m 3 – 12m 2 – 36m Find each product. 5.(v + 3)(v 2 + 5)6.(2q 2 – 4)(q – 5) 7.(2t – 5)(3t + 4)8.(4x – 1)(x 2 + 2x + 3) 9-8 Check Skills You’ll Need

Factoring by Grouping ALGEBRA 1 LESSON y y – 42.9r r r 6y 2 = 2 3 y y; 9r 3 = 3 3 r r r; 12y = y; 4 = 2 2;15r 2 = 3 5 r r; 21r = 3 7 r; GCF = 2GCF = 3r 3.30h 3 – 25h 2 – 40h4.16m 3 – 12m 2 – 36m 30h 3 = h h h;16m 3 = m m m; 25h 2 = 5 5 h h;12m 2 = m m; 40h = h;36m = m; GCF = 5hGCF = 2 2 m = 4m 5.(v + 3)(v 2 + 5) = (v)(v 2 ) + (v)(5) + (3)(v 2 ) + (3)(5) = v 3 + 5v + 3v = v 3 + 3v 2 + 5v + 15 Solutions 9-8

Factoring by Grouping ALGEBRA 1 LESSON 9-8 Solutions (continued) 7.(2t – 5)(3t + 4) = (2t)(3t) + (2t)(4) + (–5)(3t) + (–5)(4) = 6t 2 + 8t – 15t – 20 = 6t 2 – 7t – 20 6.(2q 2 – 4)(q – 5) = (2q 2 )(q) + (2q 2 )(–5) + (–4)(q) + (–4)(–5) = 2q 3 – 10q 2 – 4q (4x – 1)(x 2 + 2x + 3) = (4x)(x 2 ) + (4x)(2x) + (4x)(3) + (–1)(x 2 ) + (–1)(2x) + (–1)(3) = 4x 3 + 8x x – x 2 – 2x – 3 = 4x 3 + (8 – 1)x 2 + (12 – 2)x – 3 = 4x 3 + 7x x – 3 9-8

Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 6x 3 + 3x 2 – 4x – 2. 6x 3 + 3x 2 – 4x – 2 = 3x 2 (2x + 1) – 2(2x + 1)Factor the GCF from each group of two terms. = (2x + 1)(3x 2 – 2)Factor out (2x + 1). = 6x 3 – 4x + 3x 2 – 2Use FOIL. Check: 6x 3 + 3x 2 – 4x – 2 (2x + 1)(3x 2 – 2) = 6x 3 + 3x 2 – 4x – 2 Write in standard form. Quick Check 9-8

Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 2t 3 + 3t 2 + 4t t 3 + 3t 2 + 4t + 6 t 2 (2t + 3) + 2(2t + 3)Factor by grouping. (2t + 3)(t 2 + 2)Factor again. 9-8 Quick Check

Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor each expression. 1.10p 3 – 25p 2 + 4p – x 4 – 48x 3 + 9x 2 – 12x 3.16a 3 – 24a a – 18 (5p 2 + 2)(2p – 5) 3x(4x 2 + 1)(3x – 4) 2(4a 2 + 3)(2a – 3) 9-8

Factoring to Solve Quadratic Equations ALGEBRA 1 Lesson 10-4 Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property. (2x + 3)(x – 4) = 0 2x + 3 = 0 or x – 4 = 0Use the Zero-Product Property. 2x = –3 Solve for x. x = – 3232 orx = 4 Check: Substitute – for x Substitute 4 for x. (2x + 3)(x – 4) = 0 [2(– ) + 3](– – 4) [2(4) + 3](4 – 4) 0 (0)(– 5 ) = (11)(0) = 0 Quick Check 10-4

Factoring to Solve Quadratic Equations ALGEBRA 1 Lesson 10-4 Solve x 2 + 7x – 6x – 42 = 0 by factoring. (x + 7)(x – 6) = 0Factor using x 2 + 7x – 6x – 42 x + 7 = 0orx – 6 = 0Use the Zero-Product Property. x = –7orx = 6 Solve for x Quick Check x (x+ 7) – 6 (x +7) = 0 Factor by grouping.

Factoring to Solve Quadratic Equations ALGEBRA 1 Lesson 10-4 Solve 3x 2 + 7x – 9x = 21 by factoring. 3x 2 + 7x – 9x = 21Subtract 21 from each side. (3x + 7)(x – 3) = 0Factor 3x 2 + 7x – 9x – 21. 3x + 7 = 0orx – 3 = 0Use the Zero-Product Property 3x = –7Solve for x. x = – or x = Quick Check 3x 2 + 7x – 9x – 21= 0