Quadratic in Form Polynomials Answers

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Use substitution to write an equivalent quadratic equation.(3x + 2)2 + 7(3x + 2) – 8 = 0

A
u² + 7u - 8 = 0, where u = (3x + 2)²
B
u² + 7u - 8 = 0, where u = 3x + 2
C
u² + 7u - 8 = 0, where u = 7(3x + 2)
D
u² + u - 8 = 0
3

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
4

Select all of the zeroes of the function.

A
mr001-1.jpg
Option A
B
i
Option B
C
mr001-3.jpg
Option C
D
i
Option D
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
10

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

Answer not available
11

Solve the equation 6(2x + 4)2 = (2x + 4) + 2.

A
mc001-1.jpg
Option A
B
mc001-2.jpg
Option B
C
mc001-3.jpg
Option C
D
mc001-4.jpg
Option D
12

Select the solution(s) of the original equation.

A
mr001-1.jpg
Option A
B
mr001-2.jpg
Option B
C
mr001-3.jpg
Option C
D
mr001-4.jpg
Option D
14

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

Answer:

The equation is not quadratic in form because the square of x^3 is x^6, not x^9.

15

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
16

Factor the equation.

A
(u + 2)(u – 1) = 0
B
(u + 1)(u – 2) = 0
C
(u – 2)(u – 1) = 0
17

What are the solutions of the original equation?

A
x = 1 or x = –2
B
x = –1 or x = 2
C
x = –5 or x = –2
D
x = 4 or x = –5

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