AnswersAlgebra 2 - Semester 1 PathwaysThe Rational Roots Theorem

Quadratic in Form Polynomials Answers

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Select all of the zeroes of the function.

A
Option
Option A
B
i
Option B
C
Option
Option C
D
i
Option D
4

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
5

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
6

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
8

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

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

A
x = ± i
B
x = ±1
C
Option
Option C
D
Option
Option D
14

Select the solution(s) of the original equation.

A
Option
Option A
B
Option
Option B
C
Option
Option C
D
Option
Option D
15

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.

16

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

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