# How do you factor 4m^3+9m^2-36m-81?

Nov 22, 2016

$4 {m}^{3} + 9 {m}^{2} - 36 m - 81 = \left(m - 3\right) \left(m + 3\right) \left(4 m + 9\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = m$ and $b = 3$, but first...

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms, so this cubic will factor by grouping:

$4 {m}^{3} + 9 {m}^{2} - 36 m - 81 = \left(4 {m}^{3} + 9 {m}^{2}\right) - \left(36 m + 81\right)$

$\textcolor{w h i t e}{4 {m}^{3} + 9 {m}^{2} - 36 m - 81} = {m}^{2} \left(4 m + 9\right) - 9 \left(4 m + 9\right)$

$\textcolor{w h i t e}{4 {m}^{3} + 9 {m}^{2} - 36 m - 81} = \left({m}^{2} - 9\right) \left(4 m + 9\right)$

$\textcolor{w h i t e}{4 {m}^{3} + 9 {m}^{2} - 36 m - 81} = \left({m}^{2} - {3}^{2}\right) \left(4 m + 9\right)$

$\textcolor{w h i t e}{4 {m}^{3} + 9 {m}^{2} - 36 m - 81} = \left(m - 3\right) \left(m + 3\right) \left(4 m + 9\right)$