AnswersINMT3 Integrated Math 3 Sem 1 26-27Solving Polynomial Equations using Technology

Solving Polynomial Equations using Technology — Unit test Answers

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1
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The polynomial function f(x) = 3x5 – 2x2 + 7x models the motion of a roller coaster. The roots of the function represent when the roller coaster is at ground level. Which answer choice represents all potential values of when the roller coaster is at ground level? Begin by factoring x to create a constant term.

A
0, , 1, , 3
Option A
B
0, , 1, , 7
Option B
C
, 1, , 3
Option C
D
, 1, , 7
Option D
2
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Which statement about the polynomial function g(x) is true?

I
If all rational roots of g(x) = 0 are integers, the leading coefficient of g(x) must be 1.
I
If all roots of g(x) = 0 are integers, the leading coefficient of g(x) must be 1.
I
If the leading coefficient of g(x) is 1, all rational roots of g(x) = 0 must be integers.
I
If the leading coefficient of g(x) is 1, all roots of g(x) = 0 must be integers.
3

Anja divides (8x3 – 36x2 + 54x – 27) by (2x – 3) as shown below. What error does Anja make?

Question illustration
S
She makes a subtraction error.
S
She makes an error choosing the x-term in the quotient.
S
She makes a multiplication error.
S
She makes an error rewriting the problem in long division.
4

A polynomial function has a root of –7 with multiplicity 2, a root of –1 with multiplicity 1, a root of 2 with multiplicity 4, and a root of 4 with multiplicity 1. If the function has a positive leading coefficient and is of even degree, which statement about the graph is true?

A
The graph of the function is positive on (2, 4).
B
The graph of the function is negative on (4, ).
Option B
C
The graph of the function is positive on (, –7).
Option C
D
The graph of the function is negative on (–7, –1).
5

What are the roots of the polynomial equation ? Use a graphing calculator and a system of equations.

Question illustration
A
–24, –3, 12
B
–12, 3, 24
C
–4, –2, 3
D
–3, 2, 4
6

Let a and b be real numbers where a b 0. Which of the following functions could represent the graph below?

Question illustration
A
f(x) = x(x – a)2(x – b)4
B
f(x) = x(x – a)3(x – b)2
C
f(x) = x4(x – a)(x – b)2
D
f(x) = x2(x – a)5(x – b)
7

What is the completely factored form of ?

Question illustration
A
(2x - 5)(3x + 1)
B
(2x + 5)(3x - 1)
C
(2x - 1)(3x - 5)
D
(2x + 1)(3x + 5)
8

Which second degree polynomial function f(x) has a lead coefficient of 3 and roots 4 and 1?

A
f(x) = 3x2 + 5x + 4
B
f(x) = 3x2 + 15x + 12
C
f(x) = 3x2 – 5x + 4
D
f(x) = 3x2 – 15x + 12
9

What are the roots of the polynomial equation ? Use a graphing calculator and a system of equations.

Question illustration
A
–2, –1, 0, 2
B
–2, 0, 1, 2
C
–1, 0
D
0, 1
10

must

f
f(k) = 0
f
f(–k) = 0
A
A root of f(x) is x = k.
A
A y intercept of f(x) is x = –k.
11

The side length, s, of a cube is 3x + 2y. If V = s3, what is the volume of the cube?

3
3x3 + 18x2y + 36xy2 + 8y3
2
27x3 + 54x2y + 18xy2 + 2y3
2
27x3 + 18x2y + 12xy2 + 2y3
2
27x3 + 54x2y + 36xy2 + 8y3
14

Which polynomial function could be represented by the graph below?

Question illustration
f
f(x) = x2 + x – 2
f
f(x) = 2x2 + 2x – 4
f
f(x) = x2 – x – 2
f
f(x) = 2x2 –2x – 4
16

If f(–5) = 0, what are all the factors of the function ? Use the Remainder Theorem.

Question illustration
(
(x – 2)(x + 5)(x – 3)
(
(x + 2)(x – 5)(x + 3)
(
(x – 2)(x + 5)
(
(x + 2)(x – 5)
17

Which second degree polynomial function has a leading coefficient of 2 and roots –3 and 5?

f
f(x) = 2x2 + 4x – 30
f
f(x) = 2x2 + 2x – 15
f
f(x) = 2x2 – 4x – 30
f
f(x) = 2x2 – 2x – 15
18

Which polynomial function has x-intercepts –1, 0, and 2 and passes through the point (1, –6)?

f
f(x) = x3 – x2 – 2x
f
f(x) = 3x3 – 3x2 – 6x
f
f(x) = x3 + x2 – 2x
f
f(x) = 3x3 + 3x2 – 6x
19

Which of the following is the complete list of roots for the polynomial function ?

Question illustration
–
–5, 3
–
–5, 3, –4 + i, –4 – i
–
–5, 3, –4 + i, 4 + i
–
–4 + i, –4 – i
20

best

T
The equation is quadratic in form because it is a fifth-degree polynomial.
T
The equation is quadratic in form because the difference of the exponent of the lead term and the exponent of the middle term is 2.
T
The equation is not quadratic in form because it cannot be rewritten as a second-degree polynomial.
T
The equation is not quadratic in form because the exponent of the lead term is not the square of the exponent of the middle term.

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