Question

Multiply. State any restrictions on the variable.
(2x^2 +15x+18)/(4x^2 -9) x (2x^2 +3x-9)/(x^2 +3x-18)
Simplify the rational expression.
What are the restrictions in the variable? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
1)x(=/)____Type an integer or a simplified fraction. Use a comma to separate answers as needed.
2)There are no restrictions on x.

Answer 1

To multiply the rational expressions

\[
\frac{2x^2 + 15x + 18}{4x^2 - 9} \times \frac{2x^2 + 3x - 9}{x^2 + 3x - 18},
\]

we first factor each expression where possible.

### Step 1: Factor each expression.

1. **Factoring \(2x^2 + 15x + 18\)**:
To factor this, we look for two numbers that multiply to \(2 \times 18 = 36\) and add up to \(15\). The numbers 12 and 3 work:
\[
2x^2 + 15x + 18 = (2x + 3)(x + 6).
\]

2. **Factoring \(4x^2 - 9\)**:
This is a difference of squares:
\[
4x^2 - 9 = (2x - 3)(2x + 3).
\]

3. **Factoring \(2x^2 + 3x - 9\)**:
We look for two numbers that multiply to \(-18\) (2 times \(-9\)) and add to \(3\). The numbers 6 and -3 work:
\[
2x^2 + 3x - 9 = (2x - 3)(x + 3).
\]

4. **Factoring \(x^2 + 3x - 18\)**:
We look for two numbers that multiply to \(-18\) and add to \(3\). The numbers 6 and -3 work:
\[
x^2 + 3x - 18 = (x + 6)(x - 3).
\]

### Step 2: Substitute the factors back into the expression.

Substituting all factored parts, we have:

\[
\frac{(2x + 3)(x + 6)}{(2x - 3)(2x + 3)} \times \frac{(2x - 3)(x + 3)}{(x + 6)(x - 3)}.
\]

### Step 3: Cancel common factors.

Now, we can cancel common factors:

- \(2x + 3\) (from numerator and denominator)
- \(2x - 3\) (from numerator and denominator)
- \(x + 6\) (from numerator and denominator)

After canceling, we have:

\[
\frac{x + 3}{x - 3}.
\]

### Step 4: Identify restrictions on the variable.

Now we need to identify the restrictions from the original expressions, where we set the denominators to zero:

1. From \(4x^2 - 9 = 0\):
\[
4x^2 - 9 = 0 \Rightarrow (2x - 3)(2x + 3) = 0 \Rightarrow x = \frac{3}{2}, x = -\frac{3}{2}.
\]

2. From \(x^2 + 3x - 18 = 0\):
\[
(x + 6)(x - 3) = 0 \Rightarrow x = -6, x = 3.
\]

### Summary of Restrictions:

The restrictions on \(x\) are:

- \(x \neq \frac{3}{2}\)
- \(x \neq -\frac{3}{2}\)
- \(x \neq -6\)
- \(x \neq 3\)

### Final Output

1) \( x \neq \frac{3}{2}, -\frac{3}{2}, -6, 3 \)
2) There are no restrictions on \(x\). (This would be incorrect here.)

Thus, the reasonable answer is **1)**: \( \frac{3}{2}, -\frac{3}{2}, -6, 3 \) as restrictions on \( x \).