Multiplying Binomials ALGEBRA 1 LESSON 9-3 (For help, go to Lesson 9-2.) Find each product. 1. 4r(r – 1) 2. 6h(h2 + 8h – 3) 3. y2(2y3 – 7) Simplify. Write each answer in standard form. 4. (x3 + 3x2 + x) + (5x2 + x + 1) 5. (3t3 – 6t + 8) + (5t3 + 7t – 2) 6. w(w + 1) + 4w(w – 7) 7. 6b(b – 2) – b(8b + 3) 8. m(4m2 – 6) + 3m2(m + 9) 9. 3d2(d3 – 6) – d3(2d2 + 4) 9-3

Multiplying Binomials ALGEBRA 1 LESSON 9-3 Solutions 1. 4r(r – 1) = 4r(r) – 4r(1) = 4r 2 – 4r 2. 6h(h2 + 8h – 3) = 6h(h2) + 6h(8h) – 6h(3) = 6h3 + 48h2 – 18h 3. y2(2y3 – 7) = y2(2y3) – 7y2 = 2y5 – 7y2 4. x3 + 3x2 + x 5. 3t3 – 6t + 8 + 5x2 + x + 1 + 5t3 + 7t – 2 x3 + 8x2 + 2x + 1 8t3 + t + 6 6. w(w + 1) + 4w(w – 7) 7. 6b(b – 2) – b(8b + 3) = w(w) + w(1) + 4w(w) – 4w(7) = 6b(b) – 6b(2) – b(8b) – b(3) = w2 + w + 4w2 – 28w = 6b2 – 12b – 8b2 – 3b = (1 + 4)w2 + (1 – 28)w = (6 – 8)b2 + (–12 – 3)b = 5w2 – 27w = –2b2 – 15b 9-3

Multiplying Binomials ALGEBRA 1 LESSON 9-3 Solutions (continued) 8. m(4m2 – 6) + 3m2(m + 9) = m(4m2) – m(6) + 3m2(m) + 3m2(9) = 4m3 – 6m + 3m3 + 27m2 = (4 + 3)m3 + 27m2 – 6m = 7m3 + 27m2 – 6m 9. 3d2(d3 – 6) – d3(2d2 + 4) = 3d2(d3) – 3d2(6) – d3(2d2) – d3(4) = 3d5 – 18d2 – 2d5 – 4d3 = (3 – 2)d5 – 4d3 – 18d2 = d5 – 4d3 – 18d2 9-3

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 – 6 Now distribute y and 2. = 2y2 + y – 6 Simplify. 9-3

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

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. = 6x2 – 3x + 4x – 2 –x2 – 3x Use FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3). = 6x2 – x2 – 3x + 4x – 3x – 2 Group like terms. = 5x2 – 2x – 2 Simplify. 9-3

Multiplying Binomials ALGEBRA 1 LESSON 9-3 Simplify the product (3x2 – 2x + 3)(2x + 7). Method 1: Multiply using the vertical method. 3x2  –   2x  +  3 2x  +  7 21x2  –  14x  +  21   Multiply by 7. 6x3  –  4x2  +   6x Multiply by 2x. 6x3  + 17x2  –   8x  +  21 Add like terms. 9-3

Multiplying Binomials ALGEBRA 1 LESSON 9-3 (continued) Method 2: Multiply using the horizontal method. = (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3) (2x + 7)(3x2 – 2x + 3) = 6x3 – 4x2 + 6x + 21x2 – 14x + 21 = 6x3 + 17x2 – 8x + 21 The product is 6x3 + 17x2 – 8x + 21. 9-3

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

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 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

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 9-4

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

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

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 . 1 4 BB BW BW WW B W B W 9-4

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. 1 2 ( B + W)2 = ( B)2 – 2( B)( W) + ( W)2 Square the binomial. 1 2 = B2 + BW + W 2 Simplify. 1 4 2 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. 1 4 2 9-4

Multiplying Special Cases ALGEBRA 1 LESSON 9-4 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

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. 9-4

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. 9-4

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 9-4