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Original Question
Find the quotient. Write your answer in simplest form. Start Fraction 4 x squared minus 9 over 6 x squared plus 13 x plus 6 End...Asked by lol
Find the quotient. Write your answer in simplest form.
Start Fraction 4 x squared minus 49 over 14 x squared plus 53 x plus 14 End Fraction divided by Start Fraction 4 x squared minus 1 over 14 x squared minus 3 x minus 2 End Fraction
(1 point)
Responses
Start Fraction 7 x plus 2 over 2 x plus 1 End Fraction
Image with alt text: Start Fraction 7 x plus 2 over 2 x plus 1 End Fraction
Start Fraction 2 x plus 7 over 2 x minus 1 End Fraction
Image with alt text: Start Fraction 2 x plus 7 over 2 x minus 1 End Fraction
Start Fraction 2 x minus 7 over 2 x plus 1 End Fraction
Image with alt text: Start Fraction 2 x minus 7 over 2 x plus 1 End Fraction
Start Fraction 2 x minus 7 over 7 x plus 2 End Fraction
Start Fraction 4 x squared minus 49 over 14 x squared plus 53 x plus 14 End Fraction divided by Start Fraction 4 x squared minus 1 over 14 x squared minus 3 x minus 2 End Fraction
(1 point)
Responses
Start Fraction 7 x plus 2 over 2 x plus 1 End Fraction
Image with alt text: Start Fraction 7 x plus 2 over 2 x plus 1 End Fraction
Start Fraction 2 x plus 7 over 2 x minus 1 End Fraction
Image with alt text: Start Fraction 2 x plus 7 over 2 x minus 1 End Fraction
Start Fraction 2 x minus 7 over 2 x plus 1 End Fraction
Image with alt text: Start Fraction 2 x minus 7 over 2 x plus 1 End Fraction
Start Fraction 2 x minus 7 over 7 x plus 2 End Fraction
Answers
There are no human answers yet.
Answered by
GPT-4o mini
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.
1. **Factoring \(4x^2 - 49\)**: This is a difference of squares.
\[
4x^2 - 49 = (2x - 7)(2x + 7)
\]
2. **Factoring \(14x^2 + 53x + 14\)**: We'll use factoring techniques or the quadratic formula if necessary. This factors to:
\[
14x^2 + 53x + 14 = (7x + 2)(2x + 7)
\]
3. **Factoring \(4x^2 - 1\)**: This is also a difference of squares.
\[
4x^2 - 1 = (2x - 1)(2x + 1)
\]
4. **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**
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.
1. **Factoring \(4x^2 - 49\)**: This is a difference of squares.
\[
4x^2 - 49 = (2x - 7)(2x + 7)
\]
2. **Factoring \(14x^2 + 53x + 14\)**: We'll use factoring techniques or the quadratic formula if necessary. This factors to:
\[
14x^2 + 53x + 14 = (7x + 2)(2x + 7)
\]
3. **Factoring \(4x^2 - 1\)**: This is also a difference of squares.
\[
4x^2 - 1 = (2x - 1)(2x + 1)
\]
4. **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**
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