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