Multiplying Special Cases ALGEBRA 1 LESSON 7 3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 (For help, go to Topic 6. ) Simplify. 1. (7 x)2 2. (3 v)2 3. (– 4 c)2 4. (5 g 3)2 Use FOIL to find each product. 5. (j + 5)(j + 7) 6. (2 b – 6)(3 b – 8) 7. (4 y + 1)(5 y – 2) 8. (x + 3)(x – 4) 9. (8 c 2 + 2)(c 2 – 10) 10. (6 y 2 – 3)(9 y 2 + 1) 7 -3

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

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

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 a. Find (y + 11)2 = y 2 + 2 y(11) + 72 = y 2 + 22 y + 121 Square the binomial. Simplify. b. Find (3 w – 6)2 = (3 w)2 – 2(3 w)(6) + 62 = 9 w 2 – 36 w + 36 Square the binomial. Simplify. 7 -3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 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. 1 Since WW is of the outcomes, the probability that a guinea pig has 4 1 white fur is. 4 B B W BB BW WW 7 -3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 (continued) You can model the probabilities found in the Punnett square with the 1 1 expression ( B + W)2. Show that this product gives the same result 2 2 as the Punnett square. (12 B + 12 W)2 = ( 12 B)2 – 2(12 B)( 12 W) + ( 12 W)2 1 1 1 = 4 B 2 + 2 BW + 4 W 2 1 1 Square the binomial. Simplify. The expressions 4 B 2 and 4 W 2 indicate the probability that offspring will 1 have either two dominant genes or two recessive genes is 4. The expression 1 BW indicates that there is 1 chance that the offspring will 2 2 inherit both genes. These are the same probabilities shown in the Punnett square. 7 -3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 a. Find 812 using mental math. 812 = (80 + 1)2 = 802 + 2(80 • 1) + 12 Square the binomial. = 6400 + 160 + 1 = 6561 Simplify. b. Find 592 using mental math. 592 = (60 – 1)2 = 602 – 2(60 • 1) + 12 Square the binomial. = 3600 – 120 + 1 = 3481 Simplify. 7 -3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 Find (p 4 – 8)(p 4 + 8) = (p 4)2 – (8)2 = p 8 – 64 Find the difference of squares. Simplify. 7 -3

Multiplying Special Cases ALGEBRA 1 LESSON 7 -3 Find 43 • 37 = (40 + 3)(40 – 3) Express each factor using 40 and 3. = 402 – 32 Find the difference of squares. = 1600 – 9 = 1591 Simplify. 7 -3

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