To multiply the given rational expressions \(\frac{x^2 - 9}{x^2 - 4}\) and \(\frac{4x - 8}{12x + 36}\), we first factor all the expressions involved.
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Factor \(x^2 - 9\) and \(x^2 - 4\):
- \(x^2 - 9\) is a difference of squares: \[ x^2 - 9 = (x - 3)(x + 3) \]
- \(x^2 - 4\) is also a difference of squares: \[ x^2 - 4 = (x - 2)(x + 2) \]
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Factor the numerator and the denominator of the second expression:
- \(4x - 8\) can be factored as: \[ 4x - 8 = 4(x - 2) \]
- \(12x + 36\) can be factored as: \[ 12x + 36 = 12(x + 3) = 12(x + 3) \]
Now we can rewrite the expression as: \[ \frac{(x - 3)(x + 3)}{(x - 2)(x + 2)} \cdot \frac{4(x - 2)}{12(x + 3)} \]
Now, combine the fractions by multiplying the numerators and the denominators: \[ = \frac{(x - 3)(x + 3) \cdot 4(x - 2)}{(x - 2)(x + 2) \cdot 12(x + 3)} \]
Simplifying gives us: \[ = \frac{4(x - 3)(x + 3)(x - 2)}{12(x - 2)(x + 2)(x + 3)} \]
Next, we can cancel the common factors in the numerator and the denominator:
- The factor \(x - 2\) cancels out.
- The factor \(x + 3\) also cancels out.
This leaves us with: \[ = \frac{4(x - 3)}{12(x + 2)} \]
Now we can simplify further: \[ = \frac{4}{12} \cdot \frac{(x - 3)}{(x + 2)} = \frac{1}{3} \cdot \frac{(x - 3)}{(x + 2)} \]
Thus, the simplest form of the product is: \[ \frac{x - 3}{3(x + 2)} \]
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