Multiplying Special Cases

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Presentation transcript:

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (For help, go to Lessons 8–4 and 9-3.) Simplify. 1. (7x)2 2. (3v)2 3. (–4c)2 4. (5g3)2 Find each product. 5. (j + 5)(j + 7) 6. (2b – 6)(3b – 8) 7. (4y + 1)(5y – 2) 8. (x + 3)(x – 4) 9. (8c2 + 2)(c2 – 10) 10. (6y2 – 3)(9y2 + 1) 5-5

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Solutions 1. (7x)2 = 72 • x2 = 49x2 2. (3v)2 = 32 • v2 = 9v2 3. (–4c)2 = (–4)2 • c2 = 16c2 4. (5g3)2 = 52 • (g3)2 = 25g6 5. (j + 5)(j + 7) = (j)(j) + (j)(7) + (5)(j) + (5)(7) = j2 + 7j + 5j + 35 = j2 + 12j + 35 6. (2b – 6)(3b – 8) = (2b)(3b) + (2b)(–8) + (–6)(3b) + (–6)(–8) = 6b2 – 16b – 18b + 48 = 6b2 – 34b + 48 5-5

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Solutions (continued) 7. (4y + 1)(5y – 2)) = (4y)(5y) + (4y)(–2) + (1)(5y) + (1)(–2) = 20y2 – 8y + 5y – 2 = 20y2 – 3y – 2 8. (x + 3)(x – 4) = (x)(x) + (x)(-4) + (3)(x) + (3)(–4) = x2 – 4x + 3x – 12 = x2 – x – 12 9. (8c2 + 2)(c2 – 10) = (8c2)(c2) + (8c2)(–10) + (2)(c2) + (2)(–10) = 8c4 – 80c2 + 2c2 – 20 = 8c4 – 78c2 – 20 10. (6y2 – 3)(9y2 + 1) = (6y2)(9y2) + (6y2)(1) + (–3)(9y2) + (–3)(1) = 54y4 + 6y2 – 27y2 – 3 = 54y4 – 21y2 – 3 5-5

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find (p4 – 8)(p4 + 8). (p4 – 8)(p4 + 8) = (p4)2 – (8)2 Find the difference of squares. = p8 – 64 Simplify. 5-5

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find 43 • 37. 43 • 37 = (40 + 3)(40 – 3) Express each factor using 40 and 3. = 402 – 32 Find the difference of squares. = 1600 – 9 = 1591 Simplify. 5-5

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find each square. 1. (y + 9)2 2. (2h – 7)2 3. 412 4. 292 5. Find (p3 – 7)(p3 + 7). 6. Find 32 • 28. y2 + 18y + 81 4h2 – 28h + 49 1681 841 p6 – 49 896 5-5