# How do you factor a^2b+6ab^2+9b^2?

Jan 18, 2017

See solution below

#### Explanation:

To factor this, because all terms are positive and there are two variables we know the factoring will be in the form:

$\left(\textcolor{red}{j} a + \textcolor{b l u e}{k} b\right) \left(\textcolor{g r e e n}{l} a + \textcolor{p u r p \le}{m} b\right)$

Where $\textcolor{red}{j}$, $\textcolor{b l u e}{k}$, $\textcolor{g r e e n}{l}$ and $\textcolor{p u r p \le}{m}$ are constants.

We also know because of the coefficients in the original problem:
$\textcolor{red}{j} \times \textcolor{g r e e n}{l} = 1$
$\textcolor{b l u e}{k} \times \textcolor{p u r p \le}{m} = 9$
$\left(\textcolor{red}{j} \times \textcolor{p u r p \le}{m}\right) + \left(\textcolor{b l u e}{k} \times \textcolor{g r e e n}{l}\right) = 6$

"Playing" with the factors for $9$ (1x9, 3x3, 9x1) Gives:

$\left(\textcolor{red}{1} a + \textcolor{b l u e}{3} b\right) \left(\textcolor{g r e e n}{1} a + \textcolor{p u r p \le}{3} b\right)$

$\left(a + 3 b\right) \left(a + 3 b\right) = {\left(a + 3 b\right)}^{2}$