## Engage NY Eureka Math Algebra 2 Module 1 Lesson 39 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 39 Opening Exercise Answer Key

Rewrite each expression as a polynomial in standard form.

a. (x + i)(x – i)

Answer:

(x + i)(x – i) = x^{2} + ix – ix – i^{2}

= x^{2} – i^{2}

= x^{2} – (-1)

= x^{2} + 1

b. (x + 5i)(x – 5i)

Answer:

(x + 5i)(x – 5i) =x^{2} + 5ix – 5ix – 25i^{2}

= x^{2} – 25i^{2}

= x^{2} – 25(- 1)

=x^{2} + 25

c. (x – (2 + i))(x – (2 – i))

Answer:

(x – (2 + i))(x – (2 – 1)) = x^{2} – (2 + i)x – (2 – i)x + [(2 + i)(2 – t)]

= x^{2} – 2x – ix – 2x + ix + [4 – i^{2}]

= x^{2} – 4x + [4-(-1)]

= x^{2} – 4x + 5

### Eureka Math Algebra 2 Module 1 Lesson 39 Exercise Answer Key

Factor the following polynomial expression into products of linear terms.

Exercise 1.

x^{2} + 9

Answer:

x^{2} + 9 = (x + 3i) (x – 3i)

Exercise 2.

x^{2} + 5

Answer:

x^{2} + 5 = (x + iâˆš5) (x – iâˆš5)

Exercise 3.

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

a. What are the solutions to x^{4} – 3x^{2} – 4 = 0?

Answer:

x^{4} – 3x^{2} – 4 = 0

(x^{2})^{2} – 3x^{2} – 4 = 0

(x^{2} + 1)(x^{2} -4) = 0

(x + i) (x – i) (x + 2) (x – 2) = 0

The solutions are -i, i, -2, and 2.

b. How many x-intercepts does the graph of the equation y = x^{4} – 3x^{2} – 4 have? What are the coordinates of the x-intercepts?

Answer:

The graph of y = x^{4} – 3x^{2} – 4 has two x-intercepts: (-2, 0) and (2, 0).

c. Are solutions to the polynomial equation P(x) = 0 the same as the x-lntercepts of the graph of y = Justify your reasoning.

Answer:

No. Only the real solutions to the equation are x-intercepts of the graph. By comparing the graph of the polynomial in part (b) to the equationâ€™s solutions from part (c), you can see that only the real number solutions to the equation correspond to the x-intercepts in the Cartesian plane.

Exercise 4.

Write a polynomial P with the lowest possible degree that has the given solutions. Explain how you generated each answer.

a. – 2, 3, – 4i, 4i

Answer:

The polynomial P has two real zeros and two complex zeros. Since the two complex zeros are members of a conjugate pair, P may have as few as four total factors. Therefore, P has degree at least 4.

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

= (x^{2} – x – 6) ( x^{2} – 16i^{2})

= (x^{2} – x – 6) (x^{2} + 16)

= x^{4} – x^{3} – 6x^{2} + 16x^{2}– 1 6x – 96

= x^{4} – x^{3} + 10x^{2} – 16x -96

b. – 1, 3i

Answer:

The polynomial P has one real zero and two complex zeros because complex zeros come in pairs. Since 3 i and – 3i form a conjugate pair, P has at least three total factors. Therefore, P has degree at least 3.

P(x) = (x + 1) (x – 3i) (x + 3i)

= (x + 1) (x^{2} – 9i^{2})

= (x + 1) (x^{2} + 9)

= x^{3} + x^{2} + 9x + 9

c. 0, 2, 1+i, 1-i

Answer:

Since 1 + i and 1 – i are complex conjugates, P is at least a 4th degree polynomial.

P(x) = x(x – 2) (x – (1 + i)) (x – (1 – i))

= x(x – 2) [(x – 1) -i [(x – 1) + i]

= x(x -2) [(x – 1)^{2} – i^{2}]

= x(x – 2) [(x^{2} – 2x + 1) + 1]

= x(x – 2) (x^{2} – 2x + 2)

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

= x (x^{3} – 4x^{2} + 6x – 4)

= x^{4} – 4x^{3} + 6x^{2} – 4x

d. âˆš2, -âˆš2, 3, 1 + 2i

Answer:

Since 1 + 2i is a complex solution to P(x) = 0, its conjugate, 1 – 2i, must also be a complex solution. Thus, P is at least a fifth-degree polynomial.

P(x) = (x – âˆš2) (x + âˆš2) (x – 3) (x – (1 + 2i)) (x – (1 – 2i))

= (x^{2} – 2) (x – 3) [(x – 1) – 2i] [(x – 1) + 2i]

= (x^{2} – 2) (x – 3) [(x – 1)^{2} – 4i^{2}]

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

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

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

= x^{5} – 5x^{4} + 9x^{3} – 5x^{2} – 22x + 30

e. 2i, 3 – i

Answer:

The complex conjugates of 2i and 3 – i are -2i and 3 + i, respectively. So, P is at least a fourth-degree polynomial.

P(x) = (x – 2i) (x + 2i)( x – (3 – i))(x – (3 + i))

= (x^{2} – 4i^{2}) [(x – 3) + i] [(x – 3) – i]

= (x^{2} + 4) [(x – 3)^{2} – i^{2}]

= (x^{2} + 4) [(x^{2} – 6x + 9) + 1]

= (x^{2} + 4) (x^{2} – 6x + 10)

= x^{4} – 6x^{3} + 14x^{2} – 24x + 40

### Eureka Math Algebra 2 Module 1 Lesson 39 Problem Set Answer Key

Question 1.

Rewrite each expression in standard form.

a. (x + 3i) (x – 3i)

Answer:

x^{2} + 3^{2} = x^{2} + 9

b. (x – a + bi) (x – (a + bi))

Answer:

(x – a + bi) (x – (a + bi)) = ((x – a) + bi) ((x – a) – bi)

= (x – a)^{2} + b^{2}

= x^{2} – 2ax + a^{2} + b^{2}

c. (x + 2i)(x – i)(x + i)(x – 2i)

Answer:

(x + 2i) (x – 2i) (x + i) (x – i) = (x^{2} + 2^{2})( x^{2} + 1^{2})

= (x^{2} + 4) (x^{2} + 1)

= x^{4} + 5x^{2} + 4

d. (x + i)^{2} âˆ™ (x – i)^{2}

Answer:

(x + i)(x – i) âˆ™ (x + i) (x – i) = (x^{2} + 1) (x^{2} + 1)

= x^{4} + 2x^{2} + 1

Question 2.

Suppose in Problem 1 that you had no access to paper, writing utensils, or technology. How do you know that the expressions in parts (a)-(d) are polynomials with real coefficients?

Answer:

In part (a), the identity (x + ai) (x – ai) = x^{2} + a^{2} can be applied. Since the number a is real, the resulting polynomial will have real coefficients. The remaining three expressions can all be rearranged to take advantage of the conjugate pairs identity. In parts (c) and (d), regrouping terms will produce products of polynomial expressions with real coefficients, which will again have real coefficients.

Question 3.

Write a polynomial equation of degree 4 in standard form that has the solutions i, – i, 1, – 1.

Answer:

The first step is writing the equation in factored form:

(x + 1) (x – i) (x + 1) (x – 1) = 0.

Then, use the commutative property to rearrange terms and apply the difference of squares formula:

(x + i) (x – i) (x + 1) (x – 1) = (x^{2} + 1) (x^{2} – 1)

= x^{4} – 1.

So, the standard form of the equation Â¡s

x^{4} – 1 = 0.

Question 4.

Explain the difference between x-intercepts and solutions to an equation. Give an example of a polynomial with real coefficients that has twice as many solutions as x-intercepts. Write it in standard form.

Answer:

The x-intercepts are the real solutions to a polynomial equation with real coefficients. The solutions to an equation can be real or not reaL The previous problem is an example of a polynomial with twice as many solutions than x intercepts. Or, we could consider the equation x^{4} – 6x^{3} + 13x^{2} – 12x + 4 = 0, which has zeros of multiplicity 2 at both 1 and 2.

Question 5.

Find the solutions to x^{4} – 5x^{2} – 36 = 0 and the x-intercepts of the graph of y = x^{4} – 5x^{2} – 36.

Answer:

(x^{2} + 4) (x^{2} – 9) = 0

(x + 2i) (x – 2i) (x + 3) (x – 3)= 0

Since the solutions are 2i, -21, 3, and -3, and only real solutions to the equation are x-intercepts of the graph, the x-intercepts are 3 and – 3.

Question 6.

Find the solutions to 2x^{4} – 24x^{2} + 40 = 0 and the x-intercepts of the graph of y = 2x^{4} – 24x^{2} + 40.

Answer:

2(x^{4} – 12x^{2} + 20) = 0

2(x^{2} – 10) (x^{2} – 2) = 0

Since all of the solutions âˆš10, – âˆš10, âˆš2, and -âˆš2 are real numbers, the x-intercepts of the graph are

âˆš10, -âˆš10, âˆš2, and -âˆš2.

Question 7.

Find the solutions to x^{4} – 64 = 0 and the x-intercepts of the graph of y = x^{4} – 64.

Answer:

(x^{2} + 8) (x^{2} – 8) = 0

(x + âˆš8i) (x – âˆš8i) (x + âˆš8) (x – âˆš8) = 0

The x-intercepts are 2âˆš2 and – 2âˆš2.

Question 8.

Use the fact that x^{4} + 64 = (x^{2} – 4x + 8) (x^{2} + 4x + 8) to explain how you know that the graph of y = x^{4} + 64 has no x-intercepts. You need not find the solutions.

Answer:

The x-intercepts of y = x^{4} + 64 are solutions to (x^{2} – 4x + 8) (x^{2} + 4x + 8) = 0. Both x^{2} – 4x + 8 = 0 and x^{2} + 4x + 8 = 0 have negative discriminant values of -16, so the equations x^{2} – 4x + 8 = 0 and x^{2} + 4x + 8 = 0 have no real solutions. Thus, the equation x^{4} + 64 = 0 has no real solutions, and the graph of y = x^{4} + 64 has no x-intercepts.

Since x^{4} + 64 = 0 has no real solutions, the graph of y = x^{4} + 64 has no x-intercepts.

### Eureka Math Algebra 2 Module 1 Lesson 39 Exit Ticket Answer Key

Question 1.

Solve the quadratic equation x^{2} + 9 = 0. What are the x-lntercepts of the graph of the function f(x) = x^{2} + 9?

Answer:

x^{2} + 9 = 0

x = âˆš-9 or x = -âˆš-9

x = 3âˆš-1 or x = – 3âˆš-1

x = 3i or x = – 3i

The x-intercepts of the graph of the function f(x) = x^{2} + 9 are the real solutions to the equation x^{2} + 9 = 0.

However, since both solutions to x^{2} + 9 = 0 are not real, the function f(x) = x^{2} + 9 does not have any x intercepts.

Question 2.

Find the solutions to 2x^{5} – 5x^{3} – 3x = 0. What are the x-intercepts of the graph of the function f(x) = 2x^{5} – 5x^{3} – 3x?

Answer:

(2x^{4} – 5x^{2} – 3

x(x^{2} – 3) (2x^{2} + 1) = 0

x(x + âˆš3) (x – âˆš3) (2x^{2} + 1) = 0

x(x + âˆš3) (x – âˆš3) (x + \(\frac{i \sqrt{2}}{2}\)) (x – \(\frac{i \sqrt{2}}{2}\)) = 0

Thus, x = 0, x = -âˆš3, x = âˆš3, x = –\(\frac{i \sqrt{2}}{2}\), or x = \(\frac{i \sqrt{2}}{2}\)

The solution are, 0, âˆš3, -âˆš3, \(\frac{i \sqrt{2}}{2}\) and –\(\frac{i \sqrt{2}}{2}\)

The x – intercepts of the graph of the function f(x) = 2x^{5} – 5x^{3} – 3x are the real solutions to the equation 2x^{5} – 5x^{3} – 3x = 0, so the x-intercepts are 0, âˆš3, and -âˆš3.