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ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
(For help, go to Lessons 8–4 and 9-3.)
Simplify.
1. (7x)2
2. (3v)2
3. (–4c)2
4. (5g3)2
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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ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
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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ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
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
5-6
ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
a. Find (y + 11)2.
(y + 11)2 = y2 + 2y(11) + 72
= y2 + 22y + 121
Square the binomial.
Simplify.
b. Find (3w – 6)2.
(3w – 6)2 = (3w)2 –2(3w)(6) + 62
= 9w2 – 36w + 36
Square the binomial.
Simplify.
5-6
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. Since WW is 1 of the outcomes, the probability that a guinea
4
pig has white fur is 1.
4
B
B
W
BB BW
W BW WW
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ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
(continued)
You can model the probabilities found in the Punnett square with the
expression ( 1 B + 1 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 B2 + 2 BW + 4 W 2
Square the binomial.
Simplify.
The expressions 1 B2 and 1 W 2 indicate the probability that offspring will
4
4
1
have either two dominant genes or two recessive genes is . The
4
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.
5-6
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.
5-6
ALGEBRA 1 LESSON 9-4
Multiplying Special Cases
Find each square.
1. (y + 9)2
2. (2h – 7)2
4h2 – 28h + 49
y2 + 18y + 81
3. 412
4. 292
1681
5. Find (p3 – 7)(p3 + 7).
841
6. Find 32 • 28.
p6 – 49
896
5-6