multiplying binomials algebra 1 lesson 9-3 (for help, go to lesson 9-2.) find each product. 1.4r(r...

Multiplying BinomialsALGEBRA 1 LESSON 9-3ALGEBRA 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


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

Solutions

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


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


Simplify (4x + 2)(3x – 6).

The product is 12x2 – 18x – 12.

Last

(2)(–6)+

Outer

(4x)(–6)+

Inner

(2)(3x)+

24x 6x 12– + –= 12x2

= 12x2 18x– 12–

First

= (4x)(3x)(4x + 2)(3x – 6)

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


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

21x2 – 14x + 21 Multiply by 7.

6x3 – 4x2 + 6x Multiply by 2x.

Method 1: Multiply using the vertical method.

3x2 – 2x + 3

2x + 7

6x3 + 17x2 – 8x + 21 Add like terms.

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(continued)

Method 2: Multiply using the horizontal method.

= 6x3 – 4x2 + 6x + 21x2 – 14x + 21

= 6x3 + 17x2 – 8x + 21

The product is 6x3 + 17x2 – 8x + 21.

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

(2x + 7)(3x2 – 2x + 3)

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 CasesALGEBRA 1 LESSON 9-4ALGEBRA 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


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

Solutions

9-4

Multiplying Special Cases

7. (4y + 1)(5y – 2)) = (4y)(5y) + (4y)(–2) + (1)(5y) + (1)(–2)

= 20y2 – 8y + 5y – 2

= 20y2 – 3y – 2

ALGEBRA 1 LESSON 9-4ALGEBRA 1 LESSON 9-4

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

Solutions (continued)

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


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 .

141

4

BB BW

BW WW

B

W

B W

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.

12

12

( B + W)2 = ( B)2 – 2( B)( W) + ( W)2 Square the binomial.12

12

12

12

12

12

= B2 + BW + W 2 Simplify.14

12

14

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.

14

14 1

412

12

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


Find (p4 – 8)(p4 + 8).

(p4 – 8)(p4 + 8) = (p4)2 – (8)2 Find the difference of squares.

= p8 – 64 Simplify.

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


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

multiplying binomials algebra 1 lesson 9-3 (for help, go to lesson 9-2.) find each product. 1.4r(r...

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