# Eureka Math Algebra 2 Module 1 Lesson 13 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 13 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 13 Example Answer Key

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

Example 2.
Factor 4x2y4 – 25x4z6.
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
(x – 1) (x + 1)

b. 9x2 – 1
(3x – 1) (3x + 1)

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

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

e. x2 – y2
(x – y) (x + y)

f. x4 – y4
(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
4x2 + 4x – 63 = (2x)2 + 2(2x) – 63
= (2x + 9) (2x – 7)

b. 12y2 – 24y – 15
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
(a + 3) (a2 – 3a + 9)

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

Exercise 4.
2x3 + 128
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
(5x – 7) (5x + 2)

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

c. 45y2 + 15y – 10
5(3y + 2)(3y – 1)

d. y6 – y3 – 6
(y3 – 3) (y3 + 2)

e. x3 – 125
(x – 5) (x2 + 5x + 25)

f. 2x4 – 16x
2x(x – 2) (x2 + 2x + 4)

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

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

i. 4x2 + 9
Cannot be factored.

j. x4 – 36
(x – √6) (x + √6) (x2 + 6)

k. 1 + 27x9
(1 + 3x3) (1 – 3x3 + 9x6)

I. x3y6 + 8z3
(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?
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.
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.
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.
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?
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
x14y4 – 225z10

b. Difference of cubes
27x9y6 – 1

c. Sum of cubes
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