## Engage NY Eureka Math Algebra 2 Module 1 Lesson 11 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 11 Opening Exercise Answer Key

Find all solutions to the equation (x2 + 5x + 6)(x2 – 3x – 4) = 0.
The main point of this opening exercise is for students to recognize and then formalize that the statement “If ab = 0, then a = 0 or b = 0” applies not only when a and b are numbers or linear functions (which we used when solving a quadratic equation), but also applies to cases where a and b are polynomial functions of any degree.

In small groups, let students discuss ways to solve this equation. Walk around the room and offer advice such as, “Have you considered factoring each quadratic expression? What do you get?” As soon as one group factors both quadratic expressions, or when three minutes have passed, show, or let that group show, the factorization on the board. → What are the solutions to this equation?
– 2, – 3, 4, – 1

→ Why?
If x is any number other than – 2, – 3, 4, or – 1, then each factor is a nonzero number that is x + 2 ≠ 0, x + 3 ≠ 0, etc. However, the multiplication of four nonzero numbers is nonzero, so that value of x cannot be a solution. Therefore, the only possible solutions are – 2, – 3, 4, and – 1. It is easy to confirm that these are indeed solutions by substituting them each into the equation individually.

→ Why are these numbers also solutions to the original equation?
Because the expression (x + 2)(x + 3)(x – 4)(x + 1) is equivalent to (x2 + 5x + 6)(x2 – 3x – 4).
Now let’s study the solutions to x2 + 5x + 6 = 0 and x2 – 3x – 4 = 0 separately.

→ What are the solutions to x2 + 5x + 6 = 0?
– 2, – 3

→ What are the solutions to x2 – 3x – 4 = 0?
4, – 1

→ Relate the solutions of the equation (x2 + 5x + 6)(x2 – 3x – 4) = 0 to the solutions of the compound statement,”x2 + 5x + 6 = 0 or x2 – 3x – 4 = 0.”
They are the same.

→ Given two polynomial functions p and q of any degree, the solution set of the equation p(x) q(x) = 0 is the union of the solution set of p(x) = 0 and the solution set of q(x) = 0. Let’s think about why.

### Eureka Math Algebra 2 Module 1 Lesson 11 Example Answer Key

Example 1.
Suppose we know that the polynomial equation 4x3 – 12x2 + 3x + 5 = 0 has three real solutions and that one of the factors of 4x3 – 12x2 + 3x + 5 is (x – 1). How can we find all three solutions to the given equation?
Steer the discussion to help students conjecture that the polynomial 4x3 – 12x2 + 3x + 5 must be the product of (x – 1) and some quadratic polynomial.

Since (x – 1) is a factor, and we know how to divide polynomials, we can find the quadratic polynomial by dividing:
$$\frac{4 x^{3}-12 x^{2}+3 x+5}{x-1}$$ = 4x2 – 8x – 5.

Now we know that 4x3 – 12x2 + 3x + 5 = (x – 1) (4x2 – 8x – 5), and we also know that 4x2 – 8x – 5 is a quadratic polynomial that has linear factors (2x + 1) and (2x – 5).

Therefore, 4x3 – 12x2 + 3x + 5 = 0 has the same solutions as (x – 1)(4x2 – 8x – 5) = 0, which has the same solutions as
(x – 1)(2x + 1)(2x – 5) = 0.
In this factored form, the solutions of f(x) = 0 are readily apparent: –$$\frac{1}{2}$$, 1, and $$\frac{5}{2}$$.

### Eureka Math Algebra 2 Module 1 Lesson 11 Exercise Answer Key

Exercise 1.
Find the solutions of (x2 – 9) (x2 – 16) = 0.
The solutions to (x2 – 9)(x2 – 16) are the solutions of x2 – 9 = 0 combined with the solutions of x2 – 16 = 0.
These solutions are – 3, 3, – 4, and 4.

Exercise 2.
Find the zeros of the following polynomial functions, with their multiplicities.
a. f(x) = (x+ 1) (x – 1) (x2 + 1)
– 1 with multiplicity 1
1 with multiplicity 1

b. g(x) = (x – 4)3 (x – 2)8
4 with multiplicity 3
2 with multiplicity 8

c. h(x) = (2x – 3)5
$$\frac{3}{2}$$ with multiplicity 5

d. k(x) = (3x + 4)100 (x – 17)4
– $$\frac{4}{3}$$ with multiplicity 100
17 with multiplicity 4

Exercise 3.
Find a polynomial function that has the following zeros and multiplicities. What is the degree of your polynomial?

 Zero Multiplicity 2 3 – 4 1 6 6 -8 10

p(x) = (x – 2)3 (x + 4) (x – 6)6 (x + 8)10
The degree of p is 20.

Exercise 4.
Is there more than one polynomial function that has the same zeros and multiplicities as the one you found in Exercise 3?
Yes. Consider (x) = (x2 + 5) (x – 2)3 (x + 4) (x – 6)6 (x + 8)10. Since there are no real solutions to x2 + 5 = 0, adding this factor does not produce a new zero. Thus p and q have the same zeros and multiplicities but are different functions.

Exercise 5.
Can you find a rule that relates the multiplicities of the zeros to the degree of the polynomial function?
Yes. If p is a polynomial function of degree n, then the sum of the multiplicities of all of the zeros is less than or equal to n. If p can be factored into linear terms, then the sum of the multiplicities of all of the zeros is exactly equal to n.

### Eureka Math Algebra 2 Module 1 Lesson 11 Problem Set Answer Key

For Problems 1 – 4, find all solutions to the given equations.

Question 1.
(x – 3) (x + 2) = 0
3, – 2

Question 2.
(x – 5) (x + 2) (x + 3) = 0
5, – 2, – 3

Question 3.
(2x – 4) (x + 5) = 0
2, – 5

Question 4.
(2x – 2) (3x + 1) (x – 1) = 0
1, – $$\frac{1}{3}$$, 1

Question 5.
Find four solutions to the equation (x2 – 9) (x4 – 16) = 0.
2, – 2, 3, – 3

Question 6.
Find the zeros with multiplicity for the function p(x) = (x3 – 8) (x5 – 4x3).
We can factor p to give p(x) = x3 (x – 2) (x2 + 2x + 4) (x – 2) (x + 2) = x3 (x – 2)2 (x + 2) (x2 + 2x + 4).
Then, 0 is a zero of multiplicity 3, – 2 is a zero of multiplicity 1, and 2 is a zero of multiplicity 2.

Question 7.
Find two different polynomial functions that have zeros at 1, 3, and 5 of multiplicity 1.
p(x) = (x – 1)(x – 3)(x – 5) and q(x) = (x2 + 1)(x – 1)(x – 3)(x – 5)

Question 8.
Find two different polynomial functions that have a zero at 2 of multIplicity 5 and a zero at – 4 of multiplIcity 3.
p(x) = (x – 2)5 (x + 4)3 and q(x) = (x2 + 1)(x – 2)5(x + 4)3

Question 9.
Find three solutions to the equation (x2 – 9)(x3 – 8) = 0.
From Lesson 6, we know that (x – 2) is a factor of (x3 – 8), so three solutions are 3, – 3, and 2.

Question 10.
Find two solutions to the equation (x3 – 64)(x5 – 1) = 0.
From Lesson 6, we know that (x – 4) is a factor of (x3 – 64), and (x – 1) is a factor of (x5 – 1), so two solutions are 1 and 4.

Question 11.
If p, q, r, s are nonzero numbers, find the solutions to the equation (px + q) (rx + s) = 0 in terms of p, q, r, s.
Setting each factor equal to zero gives solutions – $$\frac{q}{p}$$ and $$\frac{s}{r}$$.

Use the identity a2 – b2 = (a – b)(a + b) to solve the equations given in Problems 12 – 13.

Question 12.
(3x – 2)2 = (5x + 1)2
Using algebra, we have (3x – 2)2 – (5x + 1)2 = 0. Applying the difference of squares formula, we have
((3x – 2) – (5x + 1))((3x – 2) + (5x + 1)) = 0. Combining like terms gives (- 2x – 3)(8x – 1) = 0, so the solutions are – $$\frac{3}{2}$$ and $$\frac{1}{8}$$.

Question 13.
(x + 7)2 = (2x + 4)2
Using algebra, we have (x + 7)2 – (2x + 4)2 = 0. Then ((x + 7) – (2x + 4)) ((x + 7) + (2x + 4)) = 0,so we have (- x + 3) (3x + 11) = 0. Thus the solutions are –$$\frac{11}{3}$$ and 3.

Question 14.
Consider the polynomial function P(x) = x3 + 2x2 + 2x – 5.
a. Divide P by the divisor (x – 1) and rewrite in the form P(x) = (divisor) (quotient) + remainder.
P(x) = (x – 1) (x + 3x + 5) + 0

b. Evaluate P(1).
P(1) = 0

Question 15.
Consider the polynomial function Q(x) = x6 – 3x5 + 4x3 – 12x2 + x – 3.
a. Divide Q by the divisor (x – 3) and rewrite in the form Q(x) = (divisor)(quotient) + remainder.
Q(x) = (x – 3) (x5 + 4x2 + 1) + 0

b. Evaluate Q(3).
Q(3) = 0

Question 16.
Consider the polynomial function R(x) = x4 + 2x3 – 2x2 – 3x + 2.
a. Divide R by the divisor (x + 2) and rewrite in the form R(x) = (divisor) (quotient) + remainder.
R(x) = (x + 2) (x3 – 2x + 1) + 0

b. Evaluate R(- 2).
R(- 2) = 0

Question 17.
Consider the polynomial function S(x) = x7 + x6 – x5 – x4 + x3 + x2 – x – 1.
a. Divide S by the divisor (x + 1) and rewrite in the form S(x) = (divisor)(quotient) + remainder.
S(x) = (x + 1) (x6 – x4 + x2 – 1) + 0

b. Evaluate S(- 1).
S(- 1) = 0

Question 18.
Make a conjecture based on the results of Problems 14 – 17.