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Multiplying Binomials

Multiplying Binomials. ALGEBRA 1 LESSON 9-3. Simplify (2 y – 3)( y + 2). ( 2 y – 3 )( y + 2) = ( 2 y – 3 )( y ) + ( 2 y – 3 )(2) Distribute 2 y – 3. = 2 y 2 – 3 y + 4 y – 6 Now distribute y and 2. = 2 y 2 + y – 6 Simplify. 9-3. First. Outer. Inner. Last.

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Multiplying Binomials

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  1. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (2y – 3)(y + 2). (2y – 3)(y + 2) = (2y – 3)(y) + (2y – 3)(2) Distribute 2y – 3. = 2y2 – 3y + 4y – 6Now distribute y and 2. = 2y2 + y – 6 Simplify. 9-3

  2. First Outer Inner Last (4x + 2)(3x – 6) = (4x)(3x) (4x)(–6) (2)(3x) + (2)(–6) + + = 12x2 24x 6x 12 – + – = 12x2 18x 12 – – Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (4x + 2)(3x – 6). The product is 12x2 – 18x – 12. 9-3

  3. = 6x2 – x2 – 3x + 4x – 3x – 2 Group like terms. = 5x2 – 2x – 2 Simplify. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Find the area of the shaded region. Simplify. area of outer rectangle = (3x + 2)(2x – 1) area of hole = x(x + 3) area of shaded region = area of outer rectangle – area of hole = (3x + 2)(2x – 1) –x(x + 3) Substitute. = 6x2 – 3x + 4x – 2 –x2 – 3xUse FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). 9-3

  4. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify each product using any method. 1. (x + 3)(x – 6) 2. (2b – 4)(3b – 5) 3. (3x – 4)(3x2 + x + 2) 4. Find the area of the shaded region. x2 – 3x – 18 6b2 – 22b + 20 9x3 – 9x2 + 2x – 8 2x2 + 3x – 1 9-3

  5. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (For help, go to Lessons 8–4 and 9-3.) Simplify. 1. (7x)22. (3v)23. (–4c)24. (5g3)2 Multiply to 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) 9-4

  6. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Solutions 1. (7x)2 = 72 • x2 = 49x22. (3v)2 = 32 • v2 = 9v2 3. (–4c)2 = (–4)2 • c2 = 16c24. (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 9-4

  7. 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 9-4

  8. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find (y + 11)2. (y + 11)2 = y2 + 2y(11) + 72Square the binomial. = y2 + 22y + 121 Simplify. b. Find (3w – 6)2. (3w – 6)2 = (3w)2 –2(3w)(6) + 62Square the binomial. = 9w2 – 36w + 36 Simplify. 9-4

  9. The Punnett square below models the possible combinations of color genes that parents who carry both genes can pass on to their offspring. Since WW is of the outcomes, the probability that a guinea pig has white fur is . 1 4 1 4 B W B W BB BW BW WW Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Among guinea pigs, the black fur gene (B) is dominant and the white fur gene (W) is recessive. This means that a guinea pig with at least one dominant gene (BB or BW) will have black fur. A guinea pig with two recessive genes (WW) will have white fur. 9-4

  10. You can model the probabilities found in the Punnett square with the expression ( B + W)2. Show that this product gives the same result as the Punnett square. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (B + W)2 = (B)2 – 2(B)(W) + (W)2Square the binomial. 1 4 1 4 1 4 1 2 1 4 The expressions B2 and W 2 indicate the probability that offspring will have either two dominant genes or two recessive genes is . The expression BW indicates that there is chance that the offspring will inherit both genes. These are the same probabilities shown in the Punnett square. = B2 + BW + W 2Simplify. 1 4 1 2 1 2 Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (continued) 9-4

  11. = 802 + 2(80 • 1) + 12Square the binomial. = 6400 + 160 + 1 = 6561 Simplify. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find 812 using mental math. 812 = (80 + 1)2 b. Find 592 using mental math. 592 = (60 – 1)2 = 602 – 2(60 • 1) + 12Square the binomial. = 3600 – 120 + 1 = 3481 Simplify. 9-4

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

  13. 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 – 32Find the difference of squares. = 1600 – 9 = 1591 Simplify. 9-4

  14. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find each square. 1. (y + 9)22. (2h – 7)2 3. 4124. 292 5. Find (p3 – 7)(p3 + 7). 6. Find 32 • 28. y2 + 18y + 81 4h2 – 28h + 49 1681 841 p6 – 49 896 9-4

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