AnswersMVP MSD Common Core Algebra II A - MA3111 ICWriting Polynomial Functions from Complex Roots

Quadratic in Form Polynomials Answers

17 verified answers2 views
5

Rewrite the equation in terms of u.

A
(u2 + 3) + u – 2 = 0
B
u2 + u – 2 = 0
C
(u2 + 9) + u – 2 = 0
D
u2 + u + 1 = 0
9

Factor the equation. ✔ (u + 2)(u – 1) = 0(u + 1)(u – 2) = 0(u – 2)(u – 1) = 0

A
✔ (u + 2)(u – 1) = 0
B
(u + 1)(u – 2) = 0
C
(u – 2)(u – 1) = 0
11

Examine the criteria below. Check which of these you included in your response.

A
The equation is not quadratic in form.
B
The variable part of the first term is not the square of the variable part of thesecond term.
C
The result of squaring x cubed is x to the 6th power, not the 9th power.
D
x9 (x3)2
Option D
13

Rewrite the equation in terms of u. u² - 17u - 16 = 017u² + u + 16 = 0-u² + 17u + 16 = 0✔ u² - 17u + 16 = 0

A
u² - 17u - 16 = 0
B
17u² + u + 16 = 0
C
-u² + 17u + 16 = 0
D
✔ u² - 17u + 16 = 0
16

Factor. (u + 16)(u + 1) = 0✔ (u - 16)(u - 1) = 0(u + 17)(u - 1) = 0

A
(u + 16)(u + 1) = 0
B
✔ (u - 16)(u - 1) = 0
C
(u + 17)(u - 1) = 0
17

Select all of the solutions to the original equation.

A
x = 4
B
x = - 16
C
x = - 4
D
x = 16
E
x = - 8
F
x = - 1
G
x = 1
H
x = 8
24

What are the solutions of 4x4 – 4x2 = 8?

A
x = ± i
B
x = ±1
C
Option
Option C
D
Option
Option D
25

Is the equation x9 – 5x3 + 6 = 0 quadratic in form? Explain why or why not.

Answer:

No. It is not quadratic in form because (x³)² = x⁶, not x⁹, so the variable part of the first term is not the square of the variable part of the second term.

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