## Engage NY Eureka Math Algebra 2 Module 1 Lesson 4 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 4 Example Answer Key

Example 1.

If x = 10, then the division 1573 ÷ 13 can be represented using polynomial division.

Answer:

The quotient is x^{2} + 2x + 1

The completed board work for this example should look something like this:

Example 2.

Use the long division algorithm for polynomials to evaluate

\(\frac{2 x^{3}-4 x^{2}+2}{2 x-2}\)

Answer:

The quotient is x^{2} – 2x – 1

### Eureka Math Algebra 2 Module 1 Lesson 4 Opening Exercise Answer Key

Exercise 1.

Use the reverse tabular method to determine the quotient \(\frac{2 x^{3}+11 x^{2}+7 x+10}{x+5}\)

Answer:

Exercise 2.

Use your work from Exercise 1 to write the polynomial 2x^{3} + 11x^{2} + 7x + 10 in factored form, and then multiply the factors to check your work above.

Answer:

(x + 5) (2x^{2} + x + 2)

The product is 2x^{3} + 11x^{2} + 7x + 10

### Eureka Math Algebra 2 Module 1 Lesson 4 Exercise Answer Key

Use the long division algorithm to determine the quotient. For each problem, check your work by using the reverse tabular method.

Exercise 1.

\(\frac{x^{2}+6 x+9}{x+3}\)

Answer:

x + 3

Exercise 2.

\(\frac{7 x^{3}-8 x^{2}-13 x+2}{7 x-1}\)

Answer:

x^{2} – x – 2

Exercise 3.

\(\frac{x^{3}-27}{x-3}\)

Answer:

x^{2} + 3x + 9

Exercise 4.

\(\frac{2 x^{4}+14 x^{3}+x^{2}-21 x-6}{2 x^{2}-3}\)

Answer:

x^{2} + 7x + 2

Exercise 5.

\(\frac{5 x^{4}-6 x^{2}+1}{x^{2}-1}\)

Answer:

5x^{2} – 1

Exercise 6.

\(\frac{x^{6}+4 x^{4}-4 x-1}{x^{3}-1}\)

Answer:

x^{3} + 4x + 1

Exercise 7.

\(\frac{2 x^{7}+x^{5}-4 x^{3}+14 x^{2}-2 x+7}{2 x^{2}+1}\)

Answer:

x^{5} – 2x + 7

Exercise 8.

\(\frac{x^{6}-64}{x+2}\)

Answer:

x^{5} – 2x^{4} + 4x^{3} – 8x^{2} + 16x – 32

### Eureka Math Algebra 2 Module 1 Lesson 4 Problem Set Answer Key

Use the long division algorithm to determine the quotient in problems 1 – 5.

Question 1.

\(\frac{2 x^{3}-13 x^{2}-x+3}{2 x+1}\)

Answer:

x^{2} – 7x + 3

Question 2.

\(\frac{3 x^{3}+4 x^{2}+7 x+22}{x+2}\)

Answer:

3x^{2} – 2x + 11

Question 3.

\(\frac{x^{4}+6 x^{3}-7 x^{2}-24 x+12}{x^{2}-4}\)

Answer:

x^{2} + 6x – 3

Question 4.

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

Answer:

6x^{2} + x – 3

Question 5.

(2x^{3} + 2x^{2} + 2x) ÷ (x^{2} + x + 1)

Answer:

2x

Question 6.

Use long division to find the polynomial, p, that satisfies the equation below.

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

Answer:

p(x) = x^{2} – 2

Question 7.

Given q(x) = 3x^{3} – 4x^{2} + 5x + k

a. Determine the value of k so that 3x – 7 is a factor of the polynomial

Answer:

k = – 28

b. What is the quotient when you divide the polynomial q by 3x – 7?

Answer:

x^{2} + x + 4

Question 8.

In parts (a) – (b) and (d) – (e), use long division to evaluate each quotient. Then, answer the remaining questions.

a. \(\frac{x^{2}-9}{x+3}\)

Answer:

x – 3

b. \(\frac{x^{4}-81}{x+3}\)

Answer:

x^{3} – 3x^{2} + 9x – 27

c. Is x + 3 a factor of x^{3} – 27 Explain your answer using the long division algorithm.

Answer:

No. The remainder is not when you perform long division.

d. \(\frac{x^{3}+27}{x+3}\)

Answer:

x^{2} – 3x + 9

e. \(\frac{x^{5}+243}{x+3}\)

Answer:

x^{4} – 3x^{3} + 9x^{2} – 27x + 81

f. Is x + 3 a factor of x^{2} + 9 Explain your answer using the long division algorithm.

Answer:

No. The remainder is not 0 when you perform long division.

g. For which positive integers n is x + 3 a factor of x^{n} + 3^{n}? Explain your reasoning.

Answer:

Only if n is an odd number. By extending the patterns in parts (a) – (c) and (e), we can generalize that x + 3 divides evenly into x^{n} + 3^{n} for odd powers of n only.

h. If is a positive integer, is x + 3 a factor of x^{n} + 3^{n} Explain your reasoning.

Answer:

Only for even numbers n By extending the patterns in parts (a) – (c), we can generalize that x + 3 will always divide evenly into the dividend.

### Eureka Math Algebra 2 Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.

Write a note to a friend explaining how to use long division to find the quotient.

\(\frac{2 x^{2}-3 x-5}{x+1}\)

Answer:

Set up the divisor outside the division symbol and the dividend underneath it. Then ask yourself what number multiplied by x is 2x^{2} Then multiply that number by x + 1 and record the results underneath 2x^{2} – 3x Subtract these terms and bring down the – 5. Then repeat the process.