AnswersSLPS-1200710-Mathematics for College Algebra - S2 - GFSynthetic Division and the Remainder Theorem

Synthetic Division and the Remainder Theorem — Unit test Answers

25 verified answers
1
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What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i?

A
f(x) = x2 – 2x + 2
B
f(x) = x3 – x2 + 4x – 2
C
f(x) = x3 – 3x2 + 4x – 2
D
f(x) = x2 – x + 2
2
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What is the factored form of ?

Question illustration
A
mc023-2.jpg
Option A
B
mc023-3.jpg
Option B
C
mc023-4.jpg
Option C
D
mc023-5.jpg
Option D
5

What are the possible degrees for the polynomial function?

Question illustration
A
degrees of 6 or greater
B
even degrees of 6 or greater
C
degrees of 5 or greater
D
odd degrees of 5 or greater
6

A polynomial function has a root of –6 with multiplicity 1, a root of –2 with multiplicity 3, a root of 0 with multiplicity 2, and a root of 4 with multiplicity 3. If the function has a positive leading coefficient and is of odd degree, which statement about the graph is true?

A
The graph of the function is positive on (–6, –2).
B
The graph of the function is negative on (, 0).
Option B
C
The graph of the function is positive on (–2, 4).
D
The graph of the function is negative on (4, ).
Option D
7

What is the completely factored form of f(x) = 6x3 – 13x2 – 4x + 15?

A
(x + 1)(6x2 – 19x + 15)
B
(x + 1)2(2x – 3)
C
(x + 1)(3x – 2)(5x – 3)
D
(x + 1)(2x – 3)(3x – 5)
8

Use synthetic division to solve (x4 – 1) ÷ (x – 1). What is the quotient?

A
x cubed minus x squared + x minus 1
Option A
B
x3
C
x cubed + x squared + x + 1
Option C
D
x3 – 2
10

Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots and 2?

Question illustration
A
f (x) = 3 x cubed minus 6 x squared minus 15 x + 30
Option A
B
f (x) = x cubed minus 2 x squared minus 5 x + 10
Option B
C
f (x) = 3 x squared minus 21 x + 30
Option C
D
f (x) = x squared minus 7 x + 10
Option D
13

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.

A
x = StartFraction negative 4 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
Option A
B
x = StartFraction negative 7 plus-or-minus StartRoot 5 EndRoot Over 2 EndFraction
Option B
C
x = –7 and x = –2
D
x = –1 and x = 4
14

What are the solutions of the equation (x – 3)2 + 2(x – 3) – 8 = 0? Use u substitution to solve.

A
x = –5 and x = 1
B
x = –1 and x = 5
C
x = –1 and x = –7
D
x = 1 and x = 7
15

Which polynomial function has a leading coefficient of 1, roots –2 and 7 with multiplicity 1, and root 5 with multiplicity 2?

A
f(x) = 2(x + 7)(x + 5)(x – 2)
B
f(x) = 2(x – 7)(x – 5)(x + 2)
C
f(x) = (x + 7)(x + 5)(x + 5)(x – 2)
D
f(x) = (x – 7)(x – 5)(x – 5)(x + 2)
17

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

Question illustration
A
f(x) = x2(x – a)2(x – b)4(x – c)
B
f(x) = x3(x – a)3(x – b)(x – c)2
C
f(x) = x4(x – a)(x – b)3(x – c)3
D
f(x) = (x – a)2(x – b)(x – c)6
18

Which statement best describes the equation (x + 5)2 + 4(x + 5) + 12 = 0?

A
The equation is quadratic in form because it can be rewritten as a quadratic equation with u substitution u = (x + 5).
B
The equation is quadratic in form because when it is expanded, it is a fourth-degree polynomial.
C
The equation is not quadratic in form because it cannot be solved by using the quadratic formula.
D
The equation is not quadratic in form because there is no real solution.
19

If (x – 2k) is a factor of f(x), which of the following must be true?

A
f(2k) = 0
B
f(–2k) = 0
C
A root of f(x) is x = –2k.
D
A y intercept of f(x) is x = 2k.
20

If f(–2) = 0, what are all the factors of the function ? Use the Remainder Theorem.

Question illustration
A
(x + 2)(x + 60)
B
(x – 2)(x – 60)
C
(x – 10)(x + 2)(x + 6)
D
(x + 10)(x – 2)(x – 6)

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