# How do you find the quotient of (4h^3+6h^2-3)div(2h+3) using long division?

Dec 31, 2017

$\left(4 {h}^{3} + 6 {h}^{2} - 3\right) \div \left(2 h + 3\right) = 2 {h}^{2} \text{ remainder } - 3$

#### Explanation:

Remember to leave a space for the $h$ term in the dividend.

Divide: $4 {h}^{3} \div 2 h = 2 {h}^{2}$

$\textcolor{w h i t e}{m m m m m m . m} 2 {h}^{2}$
2h+3 |bar(4h^3 +6h^2 +" " - 3

Multiply $2 {h}^{2}$ by both terms at the side (in the divisor)

$\textcolor{w h i t e}{m m m m m m . m} 2 {h}^{2}$
2h+3 |bar(4h^3 +6h^2 +" " - 3
$\textcolor{w h i t e}{\times \times \times} \underline{4 {h}^{3} + 6 {h}^{2}} \text{ "larr" }$ subtract
$\textcolor{w h i t e}{\times \times \times} 0 {h}^{3} + 0 {h}^{2} \textcolor{w h i t e}{m m m m} - 3 \text{ } \leftarrow$ remainder

$\left(4 {h}^{3} + 6 {h}^{2} - 3\right) \div \left(2 h + 3\right) = 2 {h}^{2} \text{ remainder } - 3$