AnswersAZ-Common Core Algebra II AThe Rational Roots Theorem

The Rational Roots Theorem Answers

21 verified answers
7

0±2±4 ±9±3±6 ±12

Question illustration
A
0
B
±2
C
±4
D
±9
E
Option
Option E
F
±3
G
±6
H
±12
9

Question text not available

Answer not available
10

–16✔ –202

A
–16
B
–2
C
0
D
2
29

f(x) = 4x2 − 25 g(x) = 4x2 + 25 h(x) = 3x2 − 25

Answers:
Selection 1:Yes.
Selection 2:No, this polynomial has complex roots.
Selection 3:No, this polynomial has irrational roots.
47

–10 –7–531524

A
–10
B
–7
C
–5
D
3
E
15
F
24
49

⇒ 1

Answer not available
50

⇒ -5

Answer not available
51

⇒ -12

Answer not available
52

⇒ 3

Answers:
⇒ 3:answer
⇒ 4:answer
53

⇒ 4

Answer not available
96

Find all real solutions of this equation to answer the question.(6 – 2x)(3 – 2x)x = 40

A
Yes. Because [image 1] is a root, you can cut squares with sides of [image 2] in. to make the box
Option A
B
No. This equation has no real solutions.
C
No. The only real solution is x = 4. It is not possible to cut squares of this size.
98

Yuri thinks that is a root of the following function.q(x) = 6x3 + 19x2 – 15x – 28Explain to Yuri why cannot be a root.

Question illustration
Answer:

According to the Rational Root Theorem, a rational root must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. Since 3 is not a factor of 28 and 4 is not a factor of 6, 3/4 does not satisfy the theorem. Substituting 3/4 into the function gives -833/32, not 0, so it is not a root.

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