Question
Factor completely.
4f^2 –64f–68
4f^2 –64f–68
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Answered by
GPT-4o mini
To factor the expression \( 4f^2 - 64f - 68 \) completely, we can follow these steps:
1. **Factor out the greatest common factor (GCF)** from all the terms. In this case, the GCF is 4.
\[
4(f^2 - 16f - 17)
\]
2. **Now we need to factor the quadratic \( f^2 - 16f - 17 \)**. We are looking for two numbers that multiply to \(-17\) (the constant term) and add to \(-16\) (the coefficient of \(f\)).
The numbers that meet these criteria are \(-17\) and \(1\), since:
\[
-17 \times 1 = -17 \quad \text{and} \quad -17 + 1 = -16
\]
3. **Now we can express the quadratic as a product of binomials**:
\[
f^2 - 16f - 17 = (f - 17)(f + 1)
\]
4. **Putting it all together**, we have:
\[
4(f - 17)(f + 1)
\]
Thus, the completely factored form of \( 4f^2 - 64f - 68 \) is:
\[
\boxed{4(f - 17)(f + 1)}
\]
1. **Factor out the greatest common factor (GCF)** from all the terms. In this case, the GCF is 4.
\[
4(f^2 - 16f - 17)
\]
2. **Now we need to factor the quadratic \( f^2 - 16f - 17 \)**. We are looking for two numbers that multiply to \(-17\) (the constant term) and add to \(-16\) (the coefficient of \(f\)).
The numbers that meet these criteria are \(-17\) and \(1\), since:
\[
-17 \times 1 = -17 \quad \text{and} \quad -17 + 1 = -16
\]
3. **Now we can express the quadratic as a product of binomials**:
\[
f^2 - 16f - 17 = (f - 17)(f + 1)
\]
4. **Putting it all together**, we have:
\[
4(f - 17)(f + 1)
\]
Thus, the completely factored form of \( 4f^2 - 64f - 68 \) is:
\[
\boxed{4(f - 17)(f + 1)}
\]
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