Question

Which of the following is a rational expression whose domain has a restriction of x

=−1?
A.
x
3

x+1
​
B.
2x+2
x
​

​
C.
6x+6
x
2
+6x+5
​
D.
x
6x
3
+60
​

Answer 1

Are you sure

Answer 2

To determine which rational expression has a restriction of \( x \neq -1 \) in its domain, we need to identify expressions where the denominator equals zero when \( x = -1 \).

Let's analyze each option:

A.
\[
\frac{x^3}{x + 1}
\]
The denominator is \( x + 1 \). Setting this equal to zero:
\[
x + 1 = 0 \implies x = -1
\]
Thus, this expression has a restriction of \( x \neq -1 \).

B.
\[
\frac{2x + 2}{x}
\]
The denominator is \( x \). Setting this equal to zero:
\[
x = 0
\]
This expression has a restriction of \( x \neq 0 \).

C.
\[
\frac{6x + 6}{x^2 + 6x + 5}
\]
The denominator is \( x^2 + 6x + 5 \). Factoring the quadratic:
\[
x^2 + 6x + 5 = (x + 1)(x + 5)
\]
Setting this equal to zero:
\[
(x + 1)(x + 5) = 0 \implies x = -1 \text{ or } x = -5
\]
This expression has a restriction of \( x \neq -1 \) and \( x \neq -5 \).

D.
\[
\frac{x}{6x^3 + 60}
\]
The denominator is \( 6x^3 + 60 \). Setting this equal to zero:
\[
6x^3 + 60 = 0 \implies 6x^3 = -60 \implies x^3 = -10 \implies x = -\sqrt[3]{10}
\]
Thus, there is no restriction at \( x = -1 \).

### Conclusion:
Both options A and C have restrictions of \( x \neq -1 \) in their domain. However, the question looks for only one option. Since the explicit mention of the restriction is required and since both A and C fulfill that requirement, the best fit for the answer that represents a simplistic understanding would be option A.

Therefore, the answer is:
**A. \( \frac{x^3}{x + 1} \)**.

Answer 3

Let's reevaluate the options to confirm the rational expressions and their respective restrictions clearly:

### Options:
A.
\[
\frac{x^3}{x + 1}
\]
**Denominator:** \( x + 1 \)
Setting \( x + 1 = 0 \):
\[
x = -1
\]
**Restriction:** \( x \neq -1 \)

---

B.
\[
\frac{2x + 2}{x}
\]
**Denominator:** \( x \)
Setting \( x = 0 \):
\[
x = 0
\]
**Restriction:** \( x \neq 0 \)

---

C.
\[
\frac{6x + 6}{x^2 + 6x + 5}
\]
**Denominator:** \( x^2 + 6x + 5 \)
Factoring gives:
\[
x^2 + 6x + 5 = (x + 1)(x + 5)
\]
Setting \( (x + 1)(x + 5) = 0 \):
\[
x = -1 \text{ or } x = -5
\]
**Restrictions:** \( x \neq -1 \) and \( x \neq -5 \)

---

D.
\[
\frac{x}{6x^3 + 60}
\]
**Denominator:** \( 6x^3 + 60 \)
Setting \( 6x^3 + 60 = 0 \):
\[
6x^3 = -60 \implies x^3 = -10 \implies x = -\sqrt[3]{10}
\]
**Restriction:** \( x \neq -\sqrt[3]{10} \)

---

### Summary of Restrictions:
- **A:** \( x \neq -1 \)
- **B:** \( x \neq 0 \)
- **C:** \( x \neq -1, -5 \)
- **D:** \( x \neq -\sqrt[3]{10} \)

---

### Conclusion:
Both **A** and **C** contain the restriction \( x \neq -1 \), but since the question asks for a rational expression with this specific restriction and no other restrictions, the most straightforward choice is:

**A. \( \frac{x^3}{x + 1} \)**, since it directly restricts \( x \) only at \(-1\) and does not impose any other restrictions.

Thank you for your patience, and I hope this clears up any confusion!