# How do you find the product of (3r+2)(9r^2+6r+4)?

Aug 16, 2017

$27 {r}^{3} + 36 {r}^{2} + 24 r + 8$

#### Explanation:

$\text{each term in the second factor must be multiplied by }$
$\text{each term in the first factor}$

$\left(\textcolor{red}{3 r + 2}\right) \left(9 {r}^{2} + 6 r + 4\right)$

$= \textcolor{red}{3 r} \left(9 {r}^{2} + 6 r + 4\right) \textcolor{red}{+ 2} \left(9 {r}^{2} + 6 r + 4\right)$

$= \left(\textcolor{red}{3 r} \times 9 {r}^{2}\right) + \left(\textcolor{red}{3 r} \times 6 r\right) + \left(\textcolor{red}{3 r} \times 4\right)$
$\textcolor{w h i t e}{\times} + \left(\textcolor{red}{2} \times 9 {r}^{2}\right) + \left(\textcolor{red}{2} \times 6 r\right) + \left(\textcolor{red}{2} \times 4\right)$

$= 27 {r}^{3} + 18 {r}^{2} + 12 r + 18 {r}^{2} + 12 r + 8$

$= 27 {x}^{3} + 36 {r}^{2} + 24 r + 8$