To find the quotient of the two fractions
\[ \frac{4x^2 - 9}{6x^2 + 13x + 6} \div \frac{4x^2 - 1}{6x^2 + x - 2}, \]
we start by multiplying by the reciprocal of the second fraction:
\[ \frac{4x^2 - 9}{6x^2 + 13x + 6} \times \frac{6x^2 + x - 2}{4x^2 - 1}. \]
Next, we can factor the expressions in each fraction where applicable.
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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 \(6x^2 + 13x + 6\):
- We look for two numbers that multiply to \(6 \cdot 6 = 36\) and add to \(13\). The numbers are \(9\) and \(4\): \[ 6x^2 + 9x + 4x + 6 = 3x(2x + 3) + 2(2x + 3) = (3x + 2)(2x + 3). \]
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Factoring \(6x^2 + x - 2\):
- We look for two numbers that multiply to \(6 \cdot -2 = -12\) and add to \(1\). The numbers are \(4\) and \(-3\): \[ 6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2). \]
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Factoring \(4x^2 - 1\):
- This is also a difference of squares: \[ 4x^2 - 1 = (2x - 1)(2x + 1). \]
Now substituting these factored forms back into our expression gives us:
\[ \frac{(2x - 3)(2x + 3)}{(3x + 2)(2x + 3)} \times \frac{(2x - 1)(3x + 2)}{(2x - 1)(2x + 1)}. \]
Next, we can cancel the common factors:
- The \( (3x + 2) \) from the numerator of the second fraction and the denominator of the first fraction.
- The \( (2x - 1) \) from the numerator of the second fraction and the denominator of the second fraction.
- The \( (2x + 3) \) from the numerator of the first fraction and the denominator of the first fraction.
This simplifies to:
\[ \frac{(2x - 3)}{(2x + 1)}. \]
Thus, the final answer in simplest form is:
\[ \frac{2x - 3}{2x + 1}. \]
So the correct response is:
\[ \text{Start Fraction } 2x - 3 \text{ over } 2x + 1 \text{ End Fraction}. \]
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