AnswersAZ-Common Core Algebra II AThe Rational Roots Theorem

The Rational Roots Theorem Answers

28 verified answers6 views
3

p(2) = [___]

Answer:

-16

4

–16–202

A
–16
B
–2
C
0
D
2
9

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
10

No, this polynomial has complex roots.No, this polynomial has irrational roots.No, this polynomial has irrational and complex roots.Yes.

A
No, this polynomial has complex roots.
B
No, this polynomial has irrational roots.
C
No, this polynomial has irrational and complex roots.
D
Yes.
11

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

A
Yes. Because is a root, you can cut squares with sides of in. to make the box
Option A
B
No. This equation has no real solutions.
Option B
C
No. The only real solution is x = 4. It is not possible to cut squares of this size.
Option C
12

–10 –7–531524

A
–10
B
–7
C
–5
D
3
E
15
F
24
14

⇒ 0

Question illustration
Answer:

-16

15

No, this polynomial has complex roots.No, this polynomial has irrational roots.No, this polynomial has irrational and complex roots.✔ Yes.

Answers:
Selection 1:Yes.
Selection 2:No, this polynomial has complex roots.
Selection 3:No, this polynomial has irrational roots.
16

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, potential rational roots must be in the form p/q, where p is a factor of the constant term (-28) and q is a factor of the leading coefficient (6). Since 3 is not a factor of 28 and 4 is not a factor of 6, 3/4 cannot be a rational root. Additionally, substituting 3/4 into the function for x does not result in 0, and dividing the polynomial by (x - 3/4) would result in a non-zero remainder, confirming that 3/4 is not a root.

17

Did you include the following in your response?

A
According to the rational root theorem, potential rational roots must be in theform where p is a factor of the constant term and q is a factor of the leading coefficient.
Option A
B
3 is not a factor of 28, and 4 is not a factor of 6. So, does not satisfy the rational root theorem.
Option B
C
Substituting in for x does not result in 0. The remainder is not 0 when dividing by x –
Option C
18

Using the section of the graph shown, which of these potential roots of the function should you test first?f(x) = 2x3 – 9x2 – 6x + 40

A
–5
B
–2
C
1
D
8
19

Question text not available

Answers:
Selection 1:Yes.
Selection 2:No, this polynomial has complex roots.
Selection 3:No, this polynomial has irrational roots.
21

No, this polynomial has complex roots.No, this polynomial has irrational roots.No, this polynomial has irrational and complex roots.Yes.

A
No, this polynomial has complex roots.
B
No, this polynomial has irrational roots.
C
No, this polynomial has irrational and complex roots.
D
Yes.

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