ALGEBRA 1 LESSON 9 3 Multiplying Binomials Simplify

ALGEBRA 1 LESSON 9 -3 Multiplying Binomials Simplify (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

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

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

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

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

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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. 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases (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. 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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. 9 -4

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

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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. 9 -4

ALGEBRA 1 LESSON 9 -4 Multiplying Special Cases 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 9 -4