Question

factor completely 27m^2+108m+96

Answer 1

To factor the expression \(27m^2 + 108m + 96\) completely, we can start by looking for the greatest common factor (GCF) of the coefficients.

1. The GCF of \(27\), \(108\), and \(96\) is \(9\).

Now, we factor out \(9\):

\[
27m^2 + 108m + 96 = 9(3m^2 + 12m + \frac{96}{9})
\]

This simplifies to:

\[
9(3m^2 + 12m + \frac{32}{3})
\]

However, let's instead keep the entire polynomial and factor it directly from \(27m^2 + 108m + 96\):

Next, we look to factor the quadratic \(3m^2 + 12m + \frac{32}{3}\).

To simplify our effort, we can either complete the square or use the quadratic formula, but first, let's use the original quadratic:

### Quadratic Factoring:

1. Multiply \(a = 27\), \(b = 108\), and \(c = 96\) for the standard form \(ax^2 + bx + c\):

2. Find two numbers that multiply to \(a \cdot c = 27 \cdot 96 = 2592\) and add to \(b = 108\).

Let’s check the factor pairs of \(2592\):
- \(36 \cdot 72 = 2592\)
- \(36 + 72 = 108\)

With \(36\) and \(72\) found, we can rewrite the middle term:

\[
27m^2 + 36m + 72m + 96
\]

3. Now group the terms:

\[
(27m^2 + 36m) + (72m + 96)
\]

4. Factor by grouping:

\[
9m(3m + 4) + 24(3m + 4)
\]

5. Now factor out the common factor \((3m + 4)\):

\[
(3m + 4)(9m + 24)
\]

6. Notice that \(9m + 24\) can be factored further:

\[
9m + 24 = 3(3m + 8)
\]

So, substituting this back we have:

\[
(3m + 4)(3(3m + 8))
\]

Combining gives:

\[
3(3m + 4)(3m + 8)
\]

### Final Factored Expression:

Thus, the completely factored form of \(27m^2 + 108m + 96\) is:

\[
\boxed{3(3m + 4)(3m + 8)}
\]