Eureka Math Algebra 2 Module 1 Lesson 13 Answer Key (2024)

Eureka Math Algebra 2 Module 1 Lesson 13 Example Answer Key

Example 1.
Write 9 – 16x4 as the product of two factors.
Answer:
9 – 16x4 = (3)2 – (4x2)2
= (3 – 4x2) (3 + 4x2)

Example 2.
Factor 4x2y4 – 25x4z6.
Answer:
4x2y4 – 25x4z6 = (2xy2)2 – (5x2z3)2
= (2xy2 + 5x2z3) (2xy2 – 5x2z3)
= [x(2y2 + 5xz3)] [x(2y2 – 5xz3)]
= x2(2y2 + 5xz3) (2y2 – 5xz3)

Eureka Math Algebra 2 Module 1 Lesson 13 Opening Exercise Answer Key

Factor each of the following expressions. What similarities do you notice between the examples in the left column and those on the right?
a. x2 – 1
Answer:
(x – 1) (x + 1)

b. 9x2 – 1
Answer:
(3x – 1) (3x + 1)

c. x2 + 8x + 15
Answer:
(x + 5) (x + 3)

d. 4x2 + 16x + 15
Answer:
(2x + 5) (2x + 3)

e. x2 – y2
Answer:
(x – y) (x + y)

f. x4 – y4
Answer:
(x2 – y2) (x2 + y2)

Eureka Math Algebra 2 Module 1 Lesson 13 Exercise Answer Key

Exercise 1.
Factor the following expressions:
a. 4x2 + 4x – 63
Answer:
4x2 + 4x – 63 = (2x)2 + 2(2x) – 63
= (2x + 9) (2x – 7)

b. 12y2 – 24y – 15
Answer:
12y2 – 24y- 15 = 3(4y2 – 8y – 5)
= 3((2y)2 – 4(2y) – 5)
= 3(2y + 1) (2y – 5)

Factor each of the following, and show that the factored form is equivalent to the original expression.

Exercise 2.
a3 + 27
Answer:
(a + 3) (a2 – 3a + 9)

Exercise 3.
x3 – 64
Answer:
(x – 4) (x2 + 4x + 16)

Exercise 4.
2x3 + 128
Answer:
2(x3 + 64) = 2(x + 4)(x2 – 4x + 16)

Eureka Math Algebra 2 Module 1 Lesson 13 Problem Set Answer Key

Question 1.
If possible, factor the following expressions using the techniques discussed in this lesson.
a. 25x2 – 25x – 14
Answer:
(5x – 7) (5x + 2)

b. 9x2y2 – 18xy + 8
Answer:
(3xy – 4)(3xy – 2)

c. 45y2 + 15y – 10
Answer:
5(3y + 2)(3y – 1)

d. y6 – y3 – 6
Answer:
(y3 – 3) (y3 + 2)

e. x3 – 125
Answer:
(x – 5) (x2 + 5x + 25)

f. 2x4 – 16x
Answer:
2x(x – 2) (x2 + 2x + 4)

g. 9x2 – 25y4z6
Answer:
(3x – 5y2 z3) (3x + 5y2z3)

h. 36x6y4z2 – 25x2z10
Answer:
x2z2 (6x2y2 – 5z4) (6x2y2 + 5z4)

i. 4x2 + 9
Answer:
Cannot be factored.

j. x4 – 36
Answer:
(x – √6) (x + √6) (x2 + 6)

k. 1 + 27x9
Answer:
(1 + 3x3) (1 – 3x3 + 9x6)

I. x3y6 + 8z3
Answer:
(xy2 + 2z) (x2y4 – 2xy2z + 4z2)

Question 2.
Consider the polynomial expression y4 + 4y2 + 16.
a. Is y4 + 4y2 + 16 factorable using the methods we have seen so far?
Answer:
No. This will not factor into the form (y2 + a) (y2 + b) using any of our previous methods.

b. Factor y6 – 64 first as a difference of cubes, and then factor completely: (y2)3 – 43.
Answer:
y6 – 64 = (y2 – 4) (y4 + 4y2 + 16)
=(y – 2) (y + 2) (y4 + 4y2 + 16)

c. Factor y6 – 64 first as a difference of squares, and then factor completely: (y3)2 – 82.
Answer:
y6 – 64 = (y3 – 8) (y3 + 8)
=(y – 2) (y2 + 2y + 4) (y + 2) (y2 – 2y + 4)
=(y – 2) (y + 2) (y2 – 2y + 4) (y2 + 2y + 4)

d. Explain how your answers to parts (b) and (c) provide a factorization of y4 + 4y2 + 16.
Answer:
Since y6 – 64 can be factored two different ways, those factorizations are equal. Thus we have
(y – 2) (y + 2) (y4 + 4y2 + 16) = (y – 2) (y + 2) (y2 – 2y + 4) (y2 + 2y + 4).
If we specify that y ≠ 2 and y ≠ – 2, we can cancel the common terms from both sides:
(y4 + 4y2 + 16) = (y2 – 2y + 4) (y2 + 2y + 4).
Multiplying this out, we see that
(y2 – 2y + 4) (y2 + 2y + 4) = y4 + 2y3 + 4y2 – 2y3 – 4y2 – 8y + 4y2 + 8y + 16
= y4 + 4y2 + 16
for every value of y.

e. If a polynomial can be factored as either a difference of squares or a difference of cubes, which formula should you apply first, and why?
Answer:
Based on this example, a polynomial should first be factored as a difference of squares and then as a difference of cubes. This will produce factors of lower degree.

Question 3.
Create expressions that have a structure that allows them to be factored using the specified identity. Be creative, and produce challenging problems!
a. Difference of squares
Answer:
x14y4 – 225z10

b. Difference of cubes
Answer:
27x9y6 – 1

c. Sum of cubes
Answer:
x6z3 + 64y12

Eureka Math Algebra 2 Module 1 Lesson 13 Exit Ticket Answer Key

Question 1.
Factor the following expression, and verify that the factored expression is equivalent to the original: 4x2 – 9a6
Answer:
(2x – 3a3) (2x + 3a3) = 4x2 + 6a3x – 6a3x – 9a6
= 4x2 – 9a6

Question 2.
Factor the following expression, and verify that the factored expression is equivalent to the original: 16x2 – 8x – 3
Answer:
(4x – 3) (4x +1 ) = 16x2 + 4x – 12x – 3
= 16x2 – 8x – 3

Eureka Math Algebra 2 Module 1 Lesson 13 Answer Key (2024)
Top Articles
Latest Posts
Article information

Author: Zonia Mosciski DO

Last Updated:

Views: 5407

Rating: 4 / 5 (51 voted)

Reviews: 90% of readers found this page helpful

Author information

Name: Zonia Mosciski DO

Birthday: 1996-05-16

Address: Suite 228 919 Deana Ford, Lake Meridithberg, NE 60017-4257

Phone: +2613987384138

Job: Chief Retail Officer

Hobby: Tai chi, Dowsing, Poi, Letterboxing, Watching movies, Video gaming, Singing

Introduction: My name is Zonia Mosciski DO, I am a enchanting, joyous, lovely, successful, hilarious, tender, outstanding person who loves writing and wants to share my knowledge and understanding with you.