## Engage NY Eureka Math Algebra 2 Module 1 Lesson 7 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 7 Opening Exercise Answer Key

a. How are these two equations related?

\(\frac{x^{2}-1}{x+1}\) = x – 1 and x^{2} – 1 = (x + 1)(x – 1)

Answer:

They represent the same relationship between the expressions x^{2} – 1, x – 1, and x + 1 as long as x â‰ – 1. One shows the relationship as division and the other as multiplication.

b. Explain the relationship between the polynomial identities

x^{2} – 1 = (x + 1)(x – 1) and x^{2} – a^{2} = (x – a)(x + a).

Answer:

The expression x^{2} – 1 is of the form x^{2} – a^{2}, with a = 1. Note that this works with a = – 1 as well.

### Eureka Math Algebra 2 Module 1 Lesson 7 Exercise Answer Key

Exercise 1.

Compute the following products using the identity x^{2} – a^{2} = (x – a)(x + a) Show your steps.

a. 6 âˆ™ 8

Answer:

6 âˆ™ 8 = (7 – 1) (7 + 1)

= 7^{2} – 1^{2}

=49 – 1

=48

b. 11 âˆ™ 19

Answer:

11 âˆ™ 19 = (15 – 4) (15 + 4)

= 15^{2} – 4^{2}

= 225 – 16

= 209

c. 23 âˆ™ 17

Answer:

23 âˆ™ 17 = (20 + 3) (20 – 3)

= 20^{2} – 3^{2}

= 400 – 9

= 391

d. 34 âˆ™ 26

Answer:

34 âˆ™ 26 = (30 + 4)(30 – 4)

= 30^{2} – 4^{2}

= 900 – 16

= 884

Exercises 2

Find two additional factors of 2^{100} – 1.

Answer:

2^{100} – 1 = (2^{5})^{20} – 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2^{100}– 1 = (2^{2})^{50} – 1

= 32^{20} – 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4^{50} – 1

= (32 – 1)(32^{19} + 32^{18} + ……… + 32 + 1)Â Â Â Â Â Â Â Â Â Â Â Â Â = (4 – 1)(4^{49} + 4^{48} + ……… + 4 + 1)

Thus 31 is a factor and so is 3.

Exercises 3.

Show that 8^{3} – 1 is divisible by 7.

Answer:

8^{3} – 1 = (8 – 1)(8^{2<s/up> + 8 + 1) = 7 âˆ™ 73}

### Eureka Math Algebra 2 Module 1 Lesson 7 Problem Set Answer Key

Question 1.

Using an appropriate polynomial identity, quickly compute the following products. Show each step. Be sure to state your values for x and a.

a. 41 âˆ™ 19

Answer:

a = 11, x = 30

(x + a) (x – a) = x^{2} – a^{2}

(30 + 11) (30 – 11) = 30^{2} – 11^{2}

= 900 – 121

= 779

b. 993 âˆ™ 1,007

Answer:

a = 7, x = 100

(x – a) (x + a) = x^{2} – a^{2}

(1000 – 7) (1000 + 7) = 1000^{2} – 7^{2}

= 1000 000 – 49

= 999 951

c. 13 âˆ™ 187

Answer:

a = 13, x = 200

(x – a) (x + a) = x^{2} – a^{2}

(200 – 13) (200 + 13) = 200^{2} – 13^{2}

= 40000 – 169

= 39831

d. 29 âˆ™ 51

Answer:

a = 11, x = 40

(x – a) (x + a) = x^{2} – a^{2}

(40 – 11) (40 + 11) = 40^{2} – 11^{2}

= 1600 – 121

= 1479

e. 125 âˆ™ 75

Answer:

a = 25, x = 100

(x – a) (x + a) = x^{2} – a^{2}

(100 – 25) (100 + 25) = 100^{2} – 25^{2}

= 10000 – 625

= 9375

Question 2.

Give the general steps you take to determine and when asked to compute a product such as those in Problem 1.

Answer:

The number x is the mean (average is also acceptable) of the two factors, and a is the positive difference between x and either factor.

Question 3.

Why is 17 âˆ™ 23 easier to compute than 17 âˆ™ 22?

Answer:

The mean of 17 and 22 is 19.5 whereas the mean of 17 and 23 is the integer 20. I know that the square of 20 is 400 and the square of 3 is 9. However, I cannot quickly compute the squares of 19.5 and 2.5.

Question 4.

Rewrite the following differences of squares as a product of two integers.

a. 81 – 1

Answer:

81 – 1 = 9^{2} – 1^{2}

=(9 – 1) (9 + 1)

= 8 âˆ™ 10

b. 400 – 121

Answer:

400 – 121 = 20^{2} – 11^{2}

= (20 – 11)(20 + 11)

= 9 âˆ™ 31

Question 5.

Quickly compute the following differences of squares.

a. 64^{2} – 14^{2}

Answer:

64^{2} – 14^{2} = (64 – 14) (64 + 14)

= 50 âˆ™ 78

= 3900

b. 112^{2} – 88^{2}

Answer:

112^{2} – 88^{2} = (112 – 88)(112 + 88)

= 24 âˆ™ 200

= 4800

c. 785^{2} – 215^{2}

Answer:

785^{2} – 215^{2} = (785 – 215) (785 + 215)

= 570 âˆ™ 1000

= 570000

Question 6.

Is 323 prime? Use the fact that 18^{2} = 324and an identity to support your answer.

Answer:

No, 323 is not prime because it is equal to 18^{2} – 1. Therefore, 323 = (18 – 1) (18 + 1).

Note: This problem can also be solved through factoring.

Question 7.

The number 2^{3} – 1 is prime and so are 2^{5} – 1 and 2^{7} – 1. Does that mean 2^{9} – 1 is prime? Explain why or why not.

Answer:

2^{9} – 1 = (2^{3})^{3} – 1

= (2^{3} – 1) ((2^{3})^{2} + 2^{3}(1) + 1^{2})

The factors are 7 and 73. As such, 2^{9} – 1 is not prime.

Question 8.

Show that 9, 999, 999, 991 is not prime without using a calculator or computer.

Answer:

Note that 9, 999, 999, 991 = 10 000 000 000 – 9 Since 10^{10} is the square of 10^{5}, 10, 000, 000, 000 is the square of 100 000. Since 9 is the square of 3, 9 999 999 991 = 100 000^{2} – 3^{2}, which is divisible by 100 000 – 3 and by 100 000 + 3.

Question 9.

Show that 999, 973 is not prime without using a calculator or computer.

Answer:

Note that 999 973 = 1 000 000 – 27. Since 27 = 3^{3} and 1 000 000 = 100^{3} are both perfect cubes, we have 999 973 = 100^{3} – 3^{3} Therefore, we know that 999, 973 is divisible by 100 – 3 = 97.

Question 10.

Find a value of so that the expression b^{n} – 1 is always divisible by 5 for any positive integer n. Explain why your value of b works for any positive integer n.

Answer:

There are many correct answers. If b = 6, then the expression 6^{n} – 1 will always be divisible by 5 because 5 = 6 – 1. This will work for any value of b that is one more than a multiple of 5, such as b = 101 or b = 11.

Question 11.

Find a value of b so that the expression b^{n} – 1 is always divisible by 7 for any positive integer n. Explain why your value of b works for any positive integer n.

Answer:

There are many correct answers. If b = 8, then the expression 8^{n } – 1 will always be divisible by 7 because 7 = 8 – 1. This will work for any value of b that is one more than a multiple of such as b = 50 or b = 15.

Question 12.

Find a value of b so that the expression b^{n} – 1 is divisible by both 7 and 9 for any positive integer n. Explain why your value of b works for any positive integer n.

Answer:

There are multiple correct answers, but one simple answer is b = 64. Since 64 = 8^{2}, 64^{n} – 1 = (8^{2})^{n} – 1 has a factor of 8^{2} – 1, which factors into (8 – 1) (8 + 1) = 7 âˆ™ 9.

### Eureka Math Algebra 2 Module 1 Lesson 7 Exit Ticket Answer Key

Question 1.

Explain how you could use the patterns in this lesson to quickly compute (57)(43).

Answer:

Subtract 49 from 2,500. That would be 2,451. You can use the identity x^{2} – a^{2} = (x + a)(x – a). In this case, x = 50 and a = 7.

Question 2.

Jessica believes that 10 – 1 is divisible by 9. Support or refute her claim using your work In this lesson.

Answer:

Since we recognize that 9 = 10 – 1, then \(\frac{10^{3}-1}{9}\) fits the pattern of \(\frac{x^{3}-a^{3}}{x-a}\) where x = 10 and a = 1. Therefore,

\(\frac{10^{3}-1}{9}=\frac{10^{3}-1}{10-1}\) = 10^{2} + 10 + 1 = 111,

and Jessica is correct.