Question

Factor completely.

27m2+108m+96

(1 point)
Responses

3(3m−8)(3m−4)
3 open paren 3 m minus 8 close paren open paren 3 m minus 4 close paren

27(m+8)(m−4)
27 open paren m plus 8 close paren open paren m minus 4 close paren

not factorable
not factorable

3(3m+8)(3m+4)

Answer 1

To factor the expression \(27m^2 + 108m + 96\) completely, let's first look for common factors.

1. **Find the greatest common factor (GCF)**:
The coefficients \(27\), \(108\), and \(96\) have a GCF of \(9\):
\[
27m^2 + 108m + 96 = 9(3m^2 + 12m + \frac{96}{9}) = 9(3m^2 + 12m + \frac{32}{3})
\]
To avoid fractions, let's simplify it without converting to fractions for now.

Alternatively:
\[
27m^2 + 108m + 96 = 9(3m^2 + 12m + \frac{32}{3}) \text{ can be factored down more effectively.}
\]
Recheck to see if we can just factorize:

2. **Directly factor \(27m^2 + 108m + 96\)**:
Let's rewrite this expression:
\[
= 27m^2 + 108m + 96
\]
We can look for a quadratic factoring approach. Here, \(a=27\), \(b=108\), and \(c=96\).

To factor the quadratic, we can use the AC method. The product \(ac\) is:
\[
27 * 96 = 2592
\]
Now we need to find two numbers that multiply to \(2592\) and add up to \(108\).

The pair we find is \(72\) and \(36\) because:
\[
72 * 36 = 2592 \quad \text{and} \quad 72 + 36 = 108
\]

Now we can rewrite the middle term \(108m\):
\[
27m^2 + 72m + 36m + 96
\]

3. **Group the terms**:
\[
= (27m^2 + 72m) + (36m + 96)
\]

Factoring out common terms in each group:
\[
= 9m(3m + 8) + 12(3m + 8) = (3m + 8)(9m + 12)
\]
= \( (3m + 8)(3m+12) = (3m + 8)(3(m + 4)).
\]

The final factorization will be:
\[
= 3(3m + 8)(3m + 4) \text{ after factoring out the common parts.}
\]

So the correct answer is:
\[
\boxed{3(3m + 8)(3m + 4)}
\]