1. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
(For help, go to Topic 6.)
Simplify.
1. (7x)2 2. (3v)2 3. (–4c)2 4. (5g3)2
Use FOIL 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)
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2. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
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
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3. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
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
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4. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
a. Find (y + 11)2.
(y + 11)2 = y2 + 2y(11) + 72 Square the binomial.
= y2 + 22y Simplify.
b. Find (3w – 6)2.
(3w – 6)2 = (3w)2 –2(3w)(6) + 62 Square the binomial.
= 9w2 – 36w + 36 Simplify.
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5. 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. 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
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6. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
(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
= B 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
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7. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
a. Find 812 using mental math.
812 = (80 + 1)2
= (80 • 1) + 12 Square the binomial.
= = 6561 Simplify.
b. Find 592 using mental math.
592 = (60 – 1)2
= 602 – 2(60 • 1) + 12 Square the binomial.
= 3600 – = 3481 Simplify.
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8. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
Find (p4 – 8)(p4 + 8).
(p4 – 8)(p4 + 8) = (p4)2 – (8)2 Find the difference of squares.
= p8 – 64 Simplify.
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9. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
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.
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10. Multiplying Special Cases
ALGEBRA 1 LESSON 7-3
Find each square.
1. (y + 9)2 2. (2h – 7)2
5. Find (p3 – 7)(p3 + 7). 6. Find 32 • 28.
y2 + 18y + 81
4h2 – 28h + 49
1681
841
p6 – 49
896
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