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

Sep 7, 2016

$4 {h}^{3} + 12 {h}^{2} + 17 h + 12$

#### Explanation:

Each term in the 2nd bracket must be multiplied by each term in the 1st bracket.

The following shows how this may be done.

$\left(\textcolor{red}{2 h + 3}\right) \left(2 {h}^{2} + 3 h + 4\right)$

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

now distribute each pair of brackets.

$= 4 {h}^{3} + 6 {h}^{2} + 8 h + 6 {h}^{2} + 9 h + 12$

and collecting like terms gives.

$= 4 {h}^{3} + \left(6 {h}^{2} + 6 {h}^{2}\right) + \left(8 h + 9 h\right) + 12$

$= 4 {h}^{3} + 12 {h}^{2} + 17 h + 12 \leftarrow \text{result of product}$

Sep 7, 2016

$\left(2 h + 3\right) \left(2 {h}^{2} + 3 h + 4\right) = \textcolor{\mathmr{and} a n \ge}{4 {h}^{3} + 12 {h}^{2} + 17 h + 12}$

#### Explanation:

color(red)(""(2h+3))(2h^2+3h+4)

using the distributive property
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{2 h} \left(2 {h}^{2} + 3 h + 4\right) \textcolor{red}{+ 3} \left(2 {h}^{2} + 3 h + 4\right)$

(no real change except color to see the separate terms more clearly)
color(white)("XXX")=color(blue)(2h(2h^2+3h+4))+color(green)(3(2h^2+3h+4)

expanding each term
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{b l u e}{4 {h}^{3} + 6 {h}^{2} + 8 h} + \textcolor{g r e e n}{6 {h}^{2} + 9 h + 12}$

grouping equal powers of the variable
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{b l u e}{4 {h}^{3}} + \left(\textcolor{b l u e}{6 {h}^{2}} + \textcolor{g r e e n}{6 {h}^{2}}\right) + \left(\textcolor{b l u e}{8 h} + \textcolor{g r e e n}{9 h}\right) + \left(\textcolor{g r e e n}{12}\right)$

Adding together equal powers of the variable
$\textcolor{w h i t e}{\text{XXX}} = 4 {h}^{3} + 12 {h}^{2} + 17 h + 12$