Eureka Math Algebra 2 Module 1 Lesson 7 Answer Key

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 x² – 1 = (x + 1)(x – 1)
Answer:
They represent the same relationship between the expressions x² – 1, x – 1, and x + 1 as long as x ≠ – 1. One shows the relationship as division and the other as multiplication.

b. Explain the relationship between the polynomial identities
x² – 1 = (x + 1)(x – 1) and x² – a² = (x – a)(x + a).
Answer:
The expression x² – 1 is of the form x² – a², 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 x² – a² = (x – a)(x + a) Show your steps.
a. 6 ∙ 8
Answer:
6 ∙ 8 = (7 – 1) (7 + 1)
= 7² – 1²
=49 – 1
=48

b. 11 ∙ 19
Answer:
11 ∙ 19 = (15 – 4) (15 + 4)
= 15² – 4²
= 225 – 16
= 209

c. 23 ∙ 17
Answer:
23 ∙ 17 = (20 + 3) (20 – 3)
= 20² – 3²
= 400 – 9
= 391

d. 34 ∙ 26
Answer:
34 ∙ 26 = (30 + 4)(30 – 4)
= 30² – 4²
= 900 – 16
= 884

Exercises 2
Find two additional factors of 2¹⁰⁰ – 1.
Answer:
2¹⁰⁰ – 1 = (2⁵)²⁰ – 1                                                        2¹⁰⁰– 1 = (2²)⁵⁰ – 1
= 32²⁰ – 1                                                                       = 4⁵⁰ – 1
= (32 – 1)(32¹⁹ + 32¹⁸ + ……… + 32 + 1)                         = (4 – 1)(4⁴⁹ + 4⁴⁸ + ……… + 4 + 1)
Thus 31 is a factor and so is 3.

Exercises 3.
Show that 8³ – 1 is divisible by 7.
Answer:
8³ – 1 = (8 – 1)(8^{2<s/up> + 8 + 1) = 7 ∙ 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 ∙ 19
Answer:
a = 11, x = 30
(x + a) (x – a) = x² – a²
(30 + 11) (30 – 11) = 30² – 11²
= 900 – 121
= 779

b. 993 ∙ 1,007
Answer:
a = 7, x = 100
(x – a) (x + a) = x² – a²
(1000 – 7) (1000 + 7) = 1000² – 7²
= 1000 000 – 49
= 999 951

c. 13 ∙ 187
Answer:
a = 13, x = 200
(x – a) (x + a) = x² – a²
(200 – 13) (200 + 13) = 200² – 13²
= 40000 – 169
= 39831

d. 29 ∙ 51
Answer:
a = 11, x = 40
(x – a) (x + a) = x² – a²
(40 – 11) (40 + 11) = 40² – 11²
= 1600 – 121
= 1479

e. 125 ∙ 75
Answer:
a = 25, x = 100
(x – a) (x + a) = x² – a²
(100 – 25) (100 + 25) = 100² – 25²
= 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 ∙ 23 easier to compute than 17 ∙ 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 = 9² – 1²
=(9 – 1) (9 + 1)
= 8 ∙ 10

b. 400 – 121
Answer:
400 – 121 = 20² – 11²
= (20 – 11)(20 + 11)
= 9 ∙ 31

Question 5.
Quickly compute the following differences of squares.
a. 64² – 14²
Answer:
64² – 14² = (64 – 14) (64 + 14)
= 50 ∙ 78
= 3900

b. 112² – 88²
Answer:
112² – 88² = (112 – 88)(112 + 88)
= 24 ∙ 200
= 4800

c. 785² – 215²
Answer:
785² – 215² = (785 – 215) (785 + 215)
= 570 ∙ 1000
= 570000

Question 6.
Is 323 prime? Use the fact that 18² = 324and an identity to support your answer.
Answer:
No, 323 is not prime because it is equal to 18² – 1. Therefore, 323 = (18 – 1) (18 + 1).
Note: This problem can also be solved through factoring.

Question 7.
The number 2³ – 1 is prime and so are 2⁵ – 1 and 2⁷ – 1. Does that mean 2⁹ – 1 is prime? Explain why or why not.
Answer:
2⁹ – 1 = (2³)³ – 1
= (2³ – 1) ((2³)² + 2³(1) + 1²)
The factors are 7 and 73. As such, 2⁹ – 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 10¹⁰ is the square of 10⁵, 10, 000, 000, 000 is the square of 100 000. Since 9 is the square of 3, 9 999 999 991 = 100 000² – 3², 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 = 3³ and 1 000 000 = 100³ are both perfect cubes, we have 999 973 = 100³ – 3³ Therefore, we know that 999, 973 is divisible by 100 – 3 = 97.

Question 10.
Find a value of so that the expression bⁿ – 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 6ⁿ – 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 bⁿ – 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 8ⁿ – 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 bⁿ – 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 = 8², 64ⁿ – 1 = (8²)ⁿ – 1 has a factor of 8² – 1, which factors into (8 – 1) (8 + 1) = 7 ∙ 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 x² – a² = (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,
\(\frac{10^{3}-1}{9}=\frac{10^{3}-1}{10-1}\) = 10² + 10 + 1 = 111,
and Jessica is correct.