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.
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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: \[ 2x^2 + 15x + 18 = (2x + 3)(x + 6). \]
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Factoring \(4x^2 - 9\): This is a difference of squares: \[ 4x^2 - 9 = (2x - 3)(2x + 3). \]
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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: \[ 2x^2 + 3x - 9 = (2x - 3)(x + 3). \]
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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:
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From \(4x^2 - 9 = 0\): \[ 4x^2 - 9 = 0 \Rightarrow (2x - 3)(2x + 3) = 0 \Rightarrow x = \frac{3}{2}, x = -\frac{3}{2}. \]
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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
- \( x \neq \frac{3}{2}, -\frac{3}{2}, -6, 3 \)
- 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 \( x \).
1 answer