To solve the problem, we will divide the first fraction by the second fraction by multiplying by the reciprocal of the second fraction.
Given:
\[ \frac{4x^2 - 49}{14x^2 + 53x + 14} \div \frac{4x^2 - 1}{14x^2 - 3x - 2} \]
This can be rewritten as:
\[ \frac{4x^2 - 49}{14x^2 + 53x + 14} \times \frac{14x^2 - 3x - 2}{4x^2 - 1} \]
Next, we will factor each polynomial where possible.
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Factoring \(4x^2 - 49\): This is a difference of squares. \[ 4x^2 - 49 = (2x - 7)(2x + 7) \]
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Factoring \(14x^2 + 53x + 14\): We'll use factoring techniques or the quadratic formula if necessary. This factors to: \[ 14x^2 + 53x + 14 = (7x + 2)(2x + 7) \]
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Factoring \(4x^2 - 1\): This is also a difference of squares. \[ 4x^2 - 1 = (2x - 1)(2x + 1) \]
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Factoring \(14x^2 - 3x - 2\): We can factor this quadratic as well. It factors to: \[ 14x^2 - 3x - 2 = (7x + 2)(2x - 1) \]
Now we substitute these factored forms into the expression:
\[ \frac{(2x - 7)(2x + 7)}{(7x + 2)(2x + 7)} \times \frac{(7x + 2)(2x - 1)}{(2x - 1)(2x + 1)} \]
Now we can cancel common factors:
- The \(2x + 7\) in the numerator and denominator cancels.
- The \(7x + 2\) in the numerator and denominator also cancels.
Thus, we are left with:
\[ \frac{(2x - 7)(1)}{(1)(2x + 1)} = \frac{2x - 7}{2x + 1} \]
This is our final answer.
So the quotient in simplest form is:
\[ \frac{2x - 7}{2x + 1} \]
Therefore, the correct response is:
Start Fraction 2 x minus 7 over 2 x plus 1 End Fraction
1 answer