## 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 – 16x^{4} as the product of two factors.

Answer:

9 – 16x^{4} = (3)^{2} – (4x^{2})^{2}

= (3 – 4x^{2}) (3 + 4x^{2})

Example 2.

Factor 4x^{2}y^{4} – 25x^{4}z^{6}.

Answer:

4x^{2}y^{4} – 25x^{4}z^{6} = (2xy^{2})^{2} – (5x^{2}z^{3})^{2}

= (2xy^{2} + 5x^{2}z^{3}) (2xy^{2} – 5x^{2}z^{3})

= [x(2y^{2} + 5xz^{3})] [x(2y^{2} – 5xz^{3})]

= x^{2}(2y^{2} + 5xz^{3}) (2y^{2} – 5xz^{3})

### 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. x^{2} – 1

Answer:

(x – 1) (x + 1)

b. 9x^{2} – 1

Answer:

(3x – 1) (3x + 1)

c. x^{2} + 8x + 15

Answer:

(x + 5) (x + 3)

d. 4x^{2} + 16x + 15

Answer:

(2x + 5) (2x + 3)

e. x^{2} – y^{2}

Answer:

(x – y) (x + y)

f. x^{4} – y^{4}

Answer:

(x^{2} – y^{2}) (x^{2} + y^{2})

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

Exercise 1.

Factor the following expressions:

a. 4x^{2} + 4x – 63

Answer:

4x^{2} + 4x – 63 = (2x)^{2} + 2(2x) – 63

= (2x + 9) (2x – 7)

b. 12y^{2} – 24y – 15

Answer:

12y^{2} – 24y- 15 = 3(4y^{2} – 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.

a^{3} + 27

Answer:

(a + 3) (a^{2} – 3a + 9)

Exercise 3.

x^{3} – 64

Answer:

(x – 4) (x^{2} + 4x + 16)

Exercise 4.

2x^{3} + 128

Answer:

2(x^{3} + 64) = 2(x + 4)(x^{2} – 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. 25x^{2} – 25x – 14

Answer:

(5x – 7) (5x + 2)

b. 9x^{2}y^{2} – 18xy + 8

Answer:

(3xy – 4)(3xy – 2)

c. 45y^{2} + 15y – 10

Answer:

5(3y + 2)(3y – 1)

d. y^{6} – y^{3} – 6

Answer:

(y^{3} – 3) (y^{3} + 2)

e. x^{3} – 125

Answer:

(x – 5) (x^{2} + 5x + 25)

f. 2x^{4} – 16x

Answer:

2x(x – 2) (x^{2} + 2x + 4)

g. 9x^{2} – 25y^{4}z^{6}

Answer:

(3x – 5y^{2} z^{3}) (3x + 5y^{2}z^{3})

h. 36x^{6}y^{4}z^{2} – 25x^{2}z^{10}

Answer:

x^{2}z^{2} (6x^{2}y^{2} – 5z^{4}) (6x^{2}y^{2} + 5z^{4})

i. 4x^{2} + 9

Answer:

Cannot be factored.

j. x^{4} – 36

Answer:

(x – √6) (x + √6) (x^{2} + 6)

k. 1 + 27x^{9}

Answer:

(1 + 3x^{3}) (1 – 3x^{3} + 9x^{6})

I. x^{3}y^{6} + 8z^{3}

Answer:

(xy^{2} + 2z) (x^{2}y^{4} – 2xy^{2}z + 4z^{2})

Question 2.

Consider the polynomial expression y^{4} + 4y^{2} + 16.

a. Is y^{4} + 4y^{2} + 16 factorable using the methods we have seen so far?

Answer:

No. This will not factor into the form (y^{2} + a) (y^{2} + b) using any of our previous methods.

b. Factor y^{6} – 64 first as a difference of cubes, and then factor completely: (y^{2})^{3} – 4^{3}.

Answer:

y^{6} – 64 = (y^{2} – 4) (y^{4} + 4y^{2} + 16)

=(y – 2) (y + 2) (y^{4} + 4y^{2} + 16)

c. Factor y6 – 64 first as a difference of squares, and then factor completely: (y^{3})^{2} – 8^{2}.

Answer:

y^{6} – 64 = (y^{3} – 8) (y^{3} + 8)

=(y – 2) (y^{2} + 2y + 4) (y + 2) (y^{2} – 2y + 4)

=(y – 2) (y + 2) (y^{2} – 2y + 4) (y^{2} + 2y + 4)

d. Explain how your answers to parts (b) and (c) provide a factorization of y^{4} + 4y^{2} + 16.

Answer:

Since y^{6} – 64 can be factored two different ways, those factorizations are equal. Thus we have

(y – 2) (y + 2) (y^{4} + 4y^{2} + 16) = (y – 2) (y + 2) (y^{2} – 2y + 4) (y^{2} + 2y + 4).

If we specify that y ≠ 2 and y ≠ – 2, we can cancel the common terms from both sides:

(y^{4} + 4y^{2} + 16) = (y^{2} – 2y + 4) (y^{2} + 2y + 4).

Multiplying this out, we see that

(y^{2} – 2y + 4) (y^{2} + 2y + 4) = y^{4} + 2y^{3} + 4y^{2} – 2y^{3} – 4y^{2} – 8y + 4y^{2} + 8y + 16

= y^{4} + 4y^{2} + 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:

x^{14}y^{4} – 225z^{10}

b. Difference of cubes

Answer:

27x^{9}y^{6} – 1

c. Sum of cubes

Answer:

x^{6}z^{3} + 64y^{12}

### 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: 4x^{2} – 9a^{6}

Answer:

(2x – 3a^{3}) (2x + 3a^{3}) = 4x^{2} + 6a^{3}x – 6a^{3}x – 9a^{6}

= 4x^{2} – 9a^{6}

Question 2.

Factor the following expression, and verify that the factored expression is equivalent to the original: 16x^{2} – 8x – 3

Answer:

(4x – 3) (4x +1 ) = 16x^{2} + 4x – 12x – 3

= 16x^{2} – 8x – 3