To multiply the given rational expressions, we first need to factor everything we can.
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Factor the expressions:
\[ \frac{x^2 - 9x + 14}{x^2 - 4} \cdot \frac{x^2 + 3x + 2}{x^2 - 6x - 7} \]
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For \(x^2 - 9x + 14\):
- This factors to \((x - 7)(x - 2)\).
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For \(x^2 - 4\) (the difference of squares):
- This factors to \((x - 2)(x + 2)\).
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For \(x^2 + 3x + 2\):
- This factors to \((x + 2)(x + 1)\).
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For \(x^2 - 6x - 7\):
- This factors to \((x - 7)(x + 1)\).
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So the expression becomes:
\[ \frac{(x - 7)(x - 2)}{(x - 2)(x + 2)} \cdot \frac{(x + 2)(x + 1)}{(x - 7)(x + 1)} \]
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Now, we can combine and simplify it:
\[ = \frac{(x - 7)(x - 2)(x + 2)(x + 1)}{(x - 2)(x + 2)(x - 7)(x + 1)} \]
Now, we can cancel the common factors in the numerator and the denominator.
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Canceling terms:
- \((x - 7)\) cancels with \((x - 7)\),
- \((x - 2)\) cancels with \((x - 2)\),
- \((x + 2)\) cancels with \((x + 2)\),
- \((x + 1)\) cancels with \((x + 1)\).
After canceling, we have:
\[ = 1 \]
- Restrictions on the variable:
The restrictions come from the factors that we canceled and must not equal zero. We need to find where the original expressions are undefined, which occurs when the denominator is zero:
- From \(x - 2 = 0\) → \(x = 2\)
- From \(x + 2 = 0\) → \(x = -2\)
- From \(x - 7 = 0\) → \(x = 7\)
- From \(x + 1 = 0\) → \(x = -1\)
Thus, the restrictions on \(x\) are:
- \(x \neq 2, -2, 7, -1\)
So the final answers are:
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\(x \neq 2, -2, 7, -1\) (use a comma to separate answers as needed)
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There are no restrictions on the variable \(x\).
1 answer