Question

27m^2+108m+96

Answer 1

To factor the quadratic expression \( 27m^2 + 108m + 96 \), we can start by looking for common factors and then apply factoring techniques.

First, we can factor out the greatest common factor (GCF) from the expression.

The GCF of \( 27m^2 \), \( 108m \), and \( 96 \) is 9. We can factor that out:

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

Now, simplify \( \frac{96}{9} \):

\[
\frac{96}{9} = \frac{32}{3}
\]

So, we rewrite the expression:

\[
27m^2 + 108m + 96 = 9(3m^2 + 12m + \frac{32}{3})
\]

However, this does not seem to lead us to neat integers. Let's revert to the original expression and attempt to factor it directly.

Now, we can use the quadratic formula or factor it directly without dividing:

To factor \( 27m^2 + 108m + 96 \), we look for factors of \( ac = 27 \times 96 = 2592 \) that add up to \( b = 108 \).

After listing factors, we find that \( 72 \) and \( 36 \) are suitable since:

\[
72 + 36 = 108
\]

Next, we can rewrite the quadratic:

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

Now, we can factor by grouping:

1. Group the first two terms and the last two terms:

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

2. Factor out \( 9m \) from the first group and \( 12 \) from the second group:

\[
9m(3m + 8) + 12(3m + 8)
\]

3. Now we can factor out the common binomial factor \( (3m + 8) \):

\[
(3m + 8)(9m + 12)
\]

4. The second term can be factored further since \( 9m + 12 \) has a common factor of 3:

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

5. Therefore, the complete factored form is:

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

So the final factorization of \( 27m^2 + 108m + 96 \) is:

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