## Engage NY Eureka Math Algebra 2 Module 1 Lesson 19 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 19 Exercise Answer Key

Exercise 1.

Consider the polynomial function f(x) = 3x^{2} + 8x – 4.

a. Divide f by x – 2.

Answer:

Answer:

b. Find f(2)

Answer:

f(2) = 24

Exercise 2.

Consider the polynomial function g(x) = x^{3} – 3x^{2} + 6x + 8.

a. Divide g by x + 1.

Answer:

b. Find g(-1).

Answer:

g(-1) = – 2

Exercise 3.

Consider the polynomial function h(x) = x^{3} + 2x – 3.

a. Divide h by x – 3.

Answer:

b. Find h(3)

Answer:

h(3) = 30

Exercise 4.

Consider the polynomial P(x) = x^{3} + kx^{2} + x + 6.

a. Find the value of k so that x + 1 is a factor of P.

Answer:

In order for x + 1 to be a factor of P, the remainder must be zero. Hence, since x + 1 = x – ( – 1), we must have P( – 1) = 0 so that 0 = – 1 + k – 1 + 6.

Then k = – 4.

b. Find the other two factors of P for the value of k found in part (a).

Answer:

P(x) = (x + 1)(x^{2} – 5x + 6) = (x + 1) (x – 2) (x – 3)

Exercise 5.

Consider the polynomial P(x) = x^{4} + 3x^{3} – 28x^{2} – 36x + 144.

a. Is 1 a zero of the polynomial P?

Answer:

No

b. Is x + 3 one of the factors of P?

Answer:

Yes; P(- 3) = 81 – 81 – 252 + 108 + 144 = 0.

c. The graph of P is shown to the right. What are the zeros of P?

Answer:

Approximately – 6, – 3, 2, and 4.

d. Write the equation of P in factored form.

Answer:

P(x) = (x + 6) (x + 3) (x – 2) (x – 4)

Exercise 6.

Consider the graph of a degree 5 polynomial shown to the right, with x-intercepts – 4, – 2, 1, 3, and 5.

a. Write a formula for a possible polynomial function that the graph represents using c as the constant factor.

Answer:

P(x) = c(x + 4) (x + 2) (x – 1) (x – 3) (x – 5)

b. Suppose the y-intercept is – 4. Find the value of c so that the graph f P has y-intercept – 4.

Answer:

P(x) = \(\frac{1}{30}\) (x + 4) (x + 2) (x – 1) (x – 3) (x – 5)

### Eureka Math Algebra 2 Module 1 Lesson 19 Problem Set Answer Key

Question 1.

Use the remainder theorem to find the remainder for each of the following divisions.

a. \(\frac{\left(x^{2}+3 x+1\right)}{(x+2)}\)

Answer:

– 1

b. \(\frac{x^{3}-6 x^{2}-7 x+9}{(x-3)}\)

Answer:

– 39

c. \(\frac{32 x^{4}+24 x^{3}-12 x^{2}+2 x+1}{(x+1)}\)

Answer:

– 5

d. \(\frac{32 x^{4}+24 x^{3}-12 x^{2}+2 x+1}{(2 x-1)}\)

Hint for part (d): Can you rewrite the division expression so that the divisor is in the form (x – c) for some constant c?

Answer:

4

Question 2.

Consider the polynomial P(x) = x^{3} + 6x^{2} – 8x – 1. Find P(4) in two ways.

Answer:

P(4) = 4^{3} + 6(4)^{2} – 8(4) – 1 = 127 \(\frac{x^{3}+6 x^{2}-8 x-1}{x-4}\) has a remainder of 127, so P(4) = 127.

Question 3.

Consider the polynomial function P(x) = 2x^{4} + 3x^{2} + 12.

a. Divide P by x + 2, and rewrite P in the form (divisor) (quotient) + remainder.

Answer:

P(x) = (x + 2) (2x^{3} – 4x^{2} + 11x – 22) + 56

b. Find P(- 2).

Answer:

P(- 2) = (- 2 + 2) (q(- 2)) + 56 = 56

Question 4.

Consider the polynomial function P(x) = x^{3} + 42.

a. Divide P by x – 4, and rewrite P In the form (divisor)(quotient) + remainder.

Answer:

P(x) = (x – 4) (x^{2} + 4x + 16) + 106

b. Find P(4).

Answer:

P(4) = (4 – 4) (q(4)) + 106 = 106

Question 5.

Explain why for a polynomial function P, P(a) is equal to the remainder of the quotient of P and x – a.

Answer:

The polynomial P can be rewritten in the form P(x) = (x – a) (q(x)) + r, where q(x) is the quotient function and r is the remainder. Then P(a) = (a – a) (q(a)) + r. Therefore, P(a) = r.

Question 6.

Is x – 5 a factor of the function f(x) = x^{3} + x^{2} – 27x – 15? Show work supporting your answer.

Answer:

Yes, because f(5) = 0 means that dividing by x – 5 leaves a remainder of 0.

Question 7.

Is x + 1 a factor of the function f(x) = 2x^{5} – 4x^{4} + 9x^{3} – x + 13? Show work supporting your answer.

Answer:

No, because f(- 1) = – 1 means that dividing by x + 1 has a remainder of – 1.

Question 8.

A polynomial function p has zeros of 2, 2, -3, -3, 3, and 4. Find a possible formula for P, and state its degree. Why is the degree of the polynomial not 3?

Answer:

0ne solution is P(x) = (x – 2)^{2} (x + 3)^{3} (x – 4). The degree of P is 6. This is not a degree 3 polynomial function because the factor (x – 2) appears twice, and the factor (x + 3) appears 3 times, while the factor (x – 4) appears once.

Question 9.

Consider the polynomial function P(x) = x^{3} – 8x^{2} – 29x + 180.

a. Verify that P(9) = 0. Since P(9) = 0 what must one of the factors of P be?

Answer:

P(9)= 9^{3} – 8(9^{2}) – 29(9) + 180 0; x – 9

b. Find the remaining two factors of P.

Answer:

P(x) = (x – 9) (x – 4) (x + 5)

c. State the zeros of P.

Answer:

x = 9, 4, – 5

d. Sketch the graph of P.

Answer:

Question 10.

Consider the polynomial function P(x) = 2x^{3} + 3x^{2} – 2x – 3.

a. Verify that P(-1) = 0. Since P(-1) = 0, what must one of the factors of P be?

Answer:

P(- 1) = 2(- 1)^{3} + 3(- 1)^{2} – 2(- 1) – 3 = 0; x + 1

b. Find the remaining two factors of P.

Answer:

P(x) = (x + 1) (x – 1) (2x + 3)

c. State the zeros of P.

Answer:

x = – 1, 1, –\(\frac{3}{2}\)

d. Sketch the graph of P.

Answer:

Question 11.

The graph to the right is of a third-degree polynomial function f.

a. State the zeros off.

Answer:

x = – 10, – 1, 2

b. Write a formula for f in factored form using c for the constant factor.

Answer:

f(x) = c (x + 10) (x + 1) (x – 2)

c. Use the fact that f(- 4) = – 54 to find the constant factor c.

Answer:

– 54 = c(- 4 + 10) (- 4 + 1) (- 4 – 2)

c = – \(\frac{1}{2}\)

f(x) = – \(\frac{1}{2}\) (x + 10) (x + 1) (x – 2)

d. Verify your equation by using the fact that f(1) = 11.

Answer:

f(1) = – \(\frac{1}{2}\) (1 + 10)(1 + 1) (1 – 2) = – \(\frac{1}{2}\) (11) (2) (- 1) = 11

Question 12.

FInd the value of k so that \(\frac{x^{3}-k x^{2}+2}{x-1}\) has remainder 8.

Answer:

k = – 5

Question 13.

Find the value k so that \(\frac{k x^{3}+x-k}{x+2}\) has remainder 16.

Answer:

k = – 2

Question 14.

Show that x^{51} – 21x + 20 is divisible by x – 1.

Answer:

Let P(x) = x^{51} – 21x + 20.

Then P(1) = 1^{51} – 21(1) + 20 = 1 – 21 + 20 = 0.

Since P(1) = 0, the remainder of the quotient (x^{51} – 21x + 20) ÷ (x – 1) is 0.

Therefore, x^{51} – 21 x + 20 is divisible by x – 1.

Question 15.

Show that x + 1 Is a factor of 19x^{42} + 18x – 1.

Answer:

LetP(x) = 19x^{42} + 18x -1.

Then P(- 1) = 19(-1)^{42} + 18(- 1) – 1 = 19 – 18 – 1 = 0.

Since P(- 1) = 0, x + 1 must be a factor of P.

Write a polynomial function that meets the stated conditions.

Question 16.

The zeros are – 2 and 1.

Answer:

p(x) = (x + 2)(x – 1) or, equivalently, p(x) = x<sup2</sup + x – 2

Question 17.

The zeros are – 1, 2, and 7.

Answer:

p(x) = (x + 1) (x – 2) (x – 7) or, equivalently, p(x) = x<sup3</sup – 8x<sup2</sup + 5x + 14

Question 18.

The zeros are – \(\frac{1}{2}\) and –\(\frac{3}{4}\).

Answer:

p(x) = (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\)) or, equivalently, p(x) = x^{2} – \(\frac{x}{4}\) – \(\frac{3}{8}\)

Question 19.

The zeros are – \(\frac{2}{3}\) and 5, and the constant term of the polynomial is – 1o.

Answer:

p(x) = (x – 5)(3x + 2) or, equivalently, p(x) = 3x^{2} – 13x – 10

Question 20.

The zeros are 2 and –\(\frac{3}{2}\), the polynomial has degree 3, and there are no other zeros.

Answer:

p(x) = (x – 2)^{2} (2x + 3) or, equivalently, p(x) = (x – 2)(2x + 3)^{2}

### Eureka Math Algebra 2 Module 1 Lesson 19 Exit Ticket Answer Key

Consider polynomial P(x) = x^{3} + x^{2} – 10x – 10.

Question 1.

Is x + 1 one of the factors of P? Explain.

Answer:

P(- 1) = (- 1)^{3} + (- 1)^{2} – 10 (- 1) – 10

= – 1 + 1 + 10 – 10 = 0

Yes, x + 1 isa factor of P because P (-1) = 0. Or, using factoring by grouping, we have

P(x) = x^{2}(x + 1) – 10 (x + 1) = (x + 1) (x^{2} – 10).

Question 2.

The graph shown has x-intercepts at √10, – 1 and – √10 Could this be the graph of P(x) = x^{3} + x^{2} – 10x – 10? Explain how you know.

Answer:

Yes, this could be the graph of P. Since this graph has x-intercepts at √10 – 1, and – √10, the factor theorem says that (x – √10), (x – 1), and (x + √10) are all factors of the equation that goes with this graph. Since (x – √10) (x + √10) (x – 1) = x^{2} + x^{2} – 10x – 10, the graph shown is quite likely to be the graph of P.