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Multiplying Binomials. ALGEBRA 1 LESSON 9-3. (For help, go to Lesson 9-2.). Find each product. 1. 4 r ( r – 1) 2. 6 h ( h 2 + 8 h – 3) 3. y 2 (2 y 3 – 7) Simplify. Write each answer in standard form. 4. ( x 3 + 3 x 2 + x ) + (5 x 2 + x + 1)

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multiplying binomials
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 binomials2
Multiplying Binomials

ALGEBRA 1 LESSON 9-3

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 + x5. 3t3 – 6t + 8

+ 5x2 + x + 1 + 5t3 + 7t – 2

x3 + 8x2 + 2x + 1 8t3 + t + 6

Solutions

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 binomials3
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 binomials4
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

multiplying binomials5

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

multiplying binomials6

= 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

multiplying binomials7

Method 1: Multiply using the vertical method.

3x2  –   2x  +  3

2x  +  7

6x3  + 17x2  –   8x  +  21 Add like terms.

Multiplying Binomials

ALGEBRA 1 LESSON 9-3

Simplify the product (3x2 – 2x + 3)(2x + 7).

21x2  –  14x  +  21Multiply by 7.

6x3  –  4x2  +   6xMultiply by 2x.

9-3

multiplying binomials8

(2x + 7)(3x2 – 2x + 3)

= (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3)

= 6x3 – 4x2 + 6x + 21x2 – 14x + 21

= 6x3 + 17x2 – 8x + 21

Multiplying Binomials

ALGEBRA 1 LESSON 9-3

(continued)

Method 2: Multiply using the horizontal method.

The product is 6x3 + 17x2 – 8x + 21.

9-3

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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
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

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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

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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

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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

multiplying special cases14

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

multiplying special cases15

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

multiplying special cases16

= 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

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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

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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

multiplying special cases19
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