# How do you divide (3x ^ { 3} + 12x ^ { 2} - 21x - 23) \div ( x + 5)?

Jul 5, 2017

$3 {x}^{2} - 3 x - 6$ $r$ $7$

#### Explanation:

Firstly, you divide $3 {x}^{3}$ by $x$ to get $3 {x}^{2}$. Then you do $3 {x}^{2} \left(x + 5\right) = 3 {x}^{2} + 15 {x}^{2}$.

$3 {x}^{3} + 12 {x}^{2} - 3 {x}^{3} - 15 {x}^{2} = - 3 {x}^{2}$, the first value of the quptient is $3 {x}^{2}$.

Now you do $\frac{- 3 {x}^{2}}{x} = - 3 x$. Then you do $- 3 x \left(x + 5\right) = - 3 {x}^{2} - 15 x$.
$\left(3 {x}^{2} - 21 x\right) - \left(- 3 {x}^{2} - 15 x\right) = - 21 x + 15 x = - 6 x$. $- 3 x$ is the second part of the quotient.

Now you do $\frac{- 6 x}{x} = - 6$. Then you do $- 6 \left(x + 5\right) = - 6 x - 30$.
$\left(- 6 x - 23\right) - \left(- 6 x - 30\right) = - 23 + 30 = 7$. $- 6$ is the last part of the quotient. $7$ is the remainder.

Visual guide:

$\text{ } 3 {x}^{2} - 3 x - 6$ $r$ $7$
$\left(x + 5\right) \text{/} \left(3 {x}^{3} + 12 {x}^{2} - 21 x - 23\right)$
$\text{ } - \left(3 {x}^{3} + 15 {x}^{2}\right)$
$\text{ } \left(0 - 3 {x}^{2}\right)$
$\text{ } - \left(- 3 {x}^{2} - 15 x\right)$
$\text{ } \left(0 - 6 x\right)$
$\text{ } - \left(- 6 x - 30\right)$
$\text{ } \left(0 + 7\right)$