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Factoring by Grouping

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. 6 y 2 + 12 y – 4 2. 9 r 3 + 15 r 2 + 21 r 3. 30 h 3 – 25 h 2 – 40 h 4. 16 m 3 – 12 m 2 – 36 m Find each product.

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Factoring by Grouping

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  1. 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. 6y2 + 12y – 4 2. 9r3 + 15r2 + 21r 3. 30h3 – 25h2 – 40h4. 16m3 – 12m2 – 36m Find each product. 5. (v + 3)(v2 + 5) 6. (2q2 – 4)(q – 5) 7. (2t – 5)(3t + 4) 8. (4x – 1)(x2 + 2x + 3) Check Skills You’ll Need 9-8

  2. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Solutions 1. 6y2 + 12y – 4 2. 9r3 + 15r2 + 21r 6y2 = 2 • 3 • y • y; 9r3 = 3 • 3 • r • r • r; 12y = 2 • 2 • 3 • y; 4 = 2 • 2; 15r2 = 3 • 5 • r • r; 21r = 3 • 7 • r; GCF = 2 GCF = 3r 3. 30h3 – 25h2 – 40h4. 16m3 – 12m2 – 36m 30h3 = 2 • 3 • 5 • h • h • h; 16m3 = 2 • 2 • 2 • 2 • m • m • m; 25h2 = 5 • 5 • h • h; 12m2 = 2 • 2 • 3 • m • m; 40h = 2 • 2 • 2 • 5 • h; 36m = 2 • 2 • 3 • 3 • m; GCF = 5h GCF = 2 • 2 • m = 4m 5. (v + 3)(v2 + 5) = (v)(v2) + (v)(5) + (3)(v2) + (3)(5) = v3 + 5v + 3v2 + 15 = v3 + 3v2 + 5v + 15 9-8

  3. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Solutions (continued) 6. (2q2 – 4)(q – 5) = (2q2)(q) + (2q2)(–5) + (–4)(q) + (–4)(–5) = 2q3 – 10q2 – 4q + 20 7. (2t – 5)(3t + 4) = (2t)(3t) + (2t)(4) + (–5)(3t) + (–5)(4) = 6t2 + 8t – 15t – 20 = 6t2 – 7t – 20 8. (4x – 1)(x2 + 2x + 3) = (4x)(x2) + (4x)(2x) + (4x)(3) + (–1)(x2) + (–1)(2x) + (–1)(3) = 4x3 + 8x2 + 12x – x2 – 2x – 3 = 4x3 + (8 – 1)x2 + (12 – 2)x – 3 = 4x3 + 7x2 + 10x – 3 9-8

  4. Check: 6x3 + 3x2 – 4x – 2 (2x + 1)(3x2 – 2) = 6x3 – 4x + 3x2 – 2 Use FOIL. = 6x3 + 3x2 – 4x – 2 Write in standard form. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 6x3 + 3x2 – 4x – 2. 6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1) Factor the GCF from each group of two terms. = (2x + 1)(3x2 – 2) Factor out (2x + 1). Quick Check 9-8

  5. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor 2t3 + 3t2 + 4t + 6. 2t3 + 3t2 + 4t + 6 t2(2t + 3) + 2(2t + 3) Factor by grouping. (2t + 3)(t2 + 2) Factor again. Quick Check 9-8

  6. Factoring by Grouping ALGEBRA 1 LESSON 9-8 Factor each expression. 1. 10p3 – 25p2 + 4p – 10 2. 36x4 – 48x3 + 9x2 – 12x 3. 16a3 – 24a2 + 12a – 18 (5p2 + 2)(2p – 5) 3x(4x2 + 1)(3x – 4) 2(4a2 + 3)(2a – 3) 9-8

  7. 2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property. 2x = –3 Solve for x. 3 2 x = – or x = 4 3 2 Check: Substitute – for x. Substitute 4 for x. (2x + 3)(x – 4) = 0 (2x + 3)(x – 4) = 0 3 2 3 2 [2(– ) + 3](– – 4) 0 [2(4) + 3](4 – 4) 0 1 2 (11)(0) = 0 (0)(– 5 ) = 0 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 Quick Check 10-4

  8. (x + 7)(x – 6) = 0 Factor using x2+ 7x – 6x – 42 x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property. Solve for x. x = –7 or x = 6 Factoring to Solve Quadratic Equations ALGEBRA 1 Lesson 10-4 Solve x2 + 7x – 6x– 42 = 0 by factoring. x (x+ 7) – 6 (x+7) = 0 Factor by grouping. Quick Check 10-4

  9. 3x2 + 7x – 9x = 21 3x2 + 7x – 9x – 21= 0 Subtract 21 from each side. (3x + 7)(x – 3) = 0 Factor 3x2+ 7x – 9x– 21. 3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property 3x = –7 Solve for x. 7 3 x = – or x = 3 Factoring to Solve Quadratic Equations ALGEBRA 1 Lesson 10-4 Solve 3x2 + 7x – 9x = 21 by factoring. Quick Check 10-4

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