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. = 2y 2 – 3y + 4y – 6Now distribute y and 2. = 2y 2 + y – 6Simplify. 9-3

2. Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify (4x + 2)(3x – 6). The product is 12x 2 – 18x – 12. Last (2)(–6)+ Outer (4x)(–6) + Inner (2)(3x) + 24x6x6x12 –+– = 12x 2 18x – 12 – First = (4x)(3x) (4x + 2)(3x – 6) 9-3

3. 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. = 6x 2 – 3x + 4x – 2 –x 2 – 3xUse FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). = 6x 2 – x 2 – 3x + 4x – 3x – 2Group like terms. = 5x 2 – 2x – 2Simplify. 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)(3x 2 + x + 2) 4.Find the area of the shaded region. x 2 – 3x – 186b 2 – 22b + 20 9x 3 – 9x 2 + 2x – 8 2x 2 + 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) 2 2.(3v) 2 3.(–4c) 2 4.(5g 3 ) 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.(8c 2 + 2)(c 2 – 10)10.(6y 2 – 3)(9y 2 + 1) 9-4

6. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 1.(7x) 2 = 7 2 x 2 = 49x 2 2.(3v) 2 = 3 2 v 2 = 9v 2 3.(–4c) 2 = (–4) 2 c 2 = 16c 2 4.(5g 3 ) 2 = 5 2 (g 3 ) 2 = 25g 6 5.(j + 5)(j + 7) = (j)(j) + (j)(7) + (5)(j) + (5)(7) = j 2 + 7j + 5j + 35 = j 2 + 12j + 35 6.(2b – 6)(3b – 8) = (2b)(3b) + (2b)(–8) + (–6)(3b) + (–6)(–8) = 6b 2 – 16b – 18b + 48 = 6b 2 – 34b + 48 Solutions 9-4

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

8. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find (y + 11) 2. (y + 11) 2 = y 2 + 2y(11) + 7 2 Square the binomial. = y 2 + 22y + 121Simplify. b. Find (3w – 6) 2. (3w – 6) 2 = (3w) 2 –2(3w)(6) + 6 2 Square the binomial. = 9w 2 – 36w + 36Simplify. 9-4

9. 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. 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. 1414 1414 BBBW BWWW BWBW BWBW 9-4

10. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 (continued) 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. 1212 1212 ( B + W ) 2 = ( B ) 2 – 2 ( B )( W ) + ( W ) 2 Square the binomial. 1212 1212 1212 1212 1212 1212 = B 2 + BW + W 2 Simplify. 1414 1212 1414 The expressions B 2 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. 1414 1414 1414 1212 1212 9-4

11. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 a. Find 81 2 using mental math. 81 2 = (80 + 1) 2 = 80 2 + 2(80 1) + 1 2 Square the binomial. = 6400 + 160 + 1 = 6561Simplify. b. Find 59 2 using mental math. 59 2 = (60 – 1) 2 = 60 2 – 2(60 1) + 1 2 Square the binomial. = 3600 – 120 + 1 = 3481Simplify. 9-4

12. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find (p 4 – 8)(p 4 + 8). (p 4 – 8)(p 4 + 8) = (p 4 ) 2 – (8) 2 Find the difference of squares. = p 8 – 64Simplify. 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. = 40 2 – 3 2 Find the difference of squares. = 1600 – 9 = 1591Simplify. 9-4

14. Multiplying Special Cases ALGEBRA 1 LESSON 9-4 Find each square. 1.(y + 9) 2 2.(2h – 7) 2 3.41 2 4.29 2 5.Find (p 3 – 7)(p 3 + 7).6.Find 32 28. y 2 + 18y + 814h 2 – 28h + 49 1681841 p 6 – 49896 9-4