# How do you divide ( -3x^3+ 12x^2-14x-6 )/(x + 1 )?

$\textcolor{b l u e}{\frac{- 3 {x}^{3} + 12 {x}^{2} - 14 x - 6}{x + 1} = - 3 {x}^{2} + 15 x - 29 + \frac{23}{x + 1}}$

#### Explanation:

From the given $\frac{- 3 {x}^{3} + 12 {x}^{2} - 14 x - 6}{x + 1}$

We have our
dividend $- 3 {x}^{3} + 12 {x}^{2} - 14 x - 6$
divisor $x + 1$

Let us begin the

$\textcolor{red}{\text{Long Division Method}}$

$\text{ " " " " } \underline{- 3 {x}^{2} + 15 x - 29} \leftarrow$the quotient
$x + 1 \lceiling - 3 {x}^{3} + 12 {x}^{2} - 14 x - 6$
" " " " " "underline(-3x^3-3x^2" " " " " " " " " " ")
$\text{ " " " " " " " " 0} + 15 {x}^{2} - 14 x - 6$
" " " " " " " " " " " "underline(15x^2+15x" " " " ")
$\text{ " " " " " " " " " " " " "0} - 29 x - 6$
$\text{ " " " " " " " " " " " " " " } \underline{- 29 x - 29}$
$\text{ " " " " " " " " " " " " " " " " " " 0} + 23 \leftarrow$the remainder

$\frac{- 3 {x}^{3} + 12 {x}^{2} - 14 x - 6}{x + 1} = - 3 {x}^{2} + 15 x - 29 + \frac{23}{x + 1}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{red}{\text{Another method is by Synthetic Division}}$

Our trial divisor comes from the divisor $x + 1$. Simply equate this to zero

$x + 1 = 0$ and $x = - 1$

Use the numerical coefficients of the dividend

$\text{ "x^3" " " "" "x^2" " " " " "x^1" " " " " } {x}^{0}$

$- 3 \text{ " " "+12" " " "-14" " " "-6" " } \lceiling - 1 \leftarrow$trial divisor
$\underline{\text{ " " " " " " "+3" " " "-15" " " "+29" }}$
$- 3 \text{ " " "+15" " " "-29" " " } + 23 \leftarrow$the remainder

The last set of numbers $- 3 , + 15 , - 29$ are the numerical coefficients of ${x}^{2} , {x}^{1} , {x}^{0}$ repectively so that our quotient is

$- 3 {x}^{2} + 15 x - 29$

since the remainder is $+ 23$, we write the last term $\frac{+ 23}{x + 1}$

$\frac{- 3 {x}^{3} + 12 {x}^{2} - 14 x - 6}{x + 1} = - 3 {x}^{2} + 15 x - 29 + \frac{23}{x + 1}$