# How do you divide ( x^5 - x^3 - 3x^2 - 12x - 7 )/(x - 2 )?

$\frac{{x}^{5} - {x}^{3} - 3 {x}^{2} - 12 x - 7}{x - 2} =$

${x}^{4} + 2 {x}^{3} + 3 {x}^{2} + 3 x - 6 + \frac{- 19}{x - 2}$

#### Explanation:

First, arrange the dividend and divisor from highest to lowest degree of terms. The missing terms will have to be provided with a zero numerical coefficient.

the dividend:

${x}^{5} - {x}^{3} - 3 {x}^{2} - 12 x - 7 \text{ }$should be

${x}^{5} + 0 \cdot {x}^{4} - {x}^{3} - 3 {x}^{2} - 12 x - 7$

the divisor $x - 2$ is already arranged

Let us divide

" " " " " " " "underline(" "x^4+2x^3+3x^2+3x-6 " " " " " " " " " " ")
$x - 2 \text{ "|~" } {x}^{5} + 0 \cdot {x}^{4} - {x}^{3} - 3 {x}^{2} - 12 x - 7$
" " " " " " " "underline(" "x^5-2x^4"" " " " " " " " " " " " " " " " " " " " ")
$\text{ " " " " " " " " " " " " } 2 {x}^{4} - {x}^{3} - 3 {x}^{2} - 12 x - 7$
" " " " " " " " " " " " " "underline(2x^4-4x^3 " " " " " " " " " " " " " " " " ")
$\text{ " " " " " " " " " " " " " " " " } 3 {x}^{3} - 3 {x}^{2} - 12 x - 7$
" " " " " " " " " " " " " " " " " "underline(3x^3 -6x^2" " " " " " " " " " " " ")
$\text{ " " " " " " " " " " " " " " " " " " " " " } 3 {x}^{2} - 12 x - 7$
" " " " " " " " " " " " " " " " " " " " " " underline(3x^2-6x" " " " " " " " " ")
$\text{ " " " " " " " " " " " " " " " " " " " " " " " " " " } - 6 x - 7$
" " " " " " " " " " " " " " " " " " " " " " " " " " ""underline(-6x+12 " " " " ")
$\text{ " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " } - 19 \leftarrow$remainder

We write the answer in the form

$\left(\text{Dividend")/("divisor")="Quotient"+("Remainder")/("Divisor}\right)$

$\frac{{x}^{5} - {x}^{3} - 3 {x}^{2} - 12 x - 7}{x - 2} =$

${x}^{4} + 2 {x}^{3} + 3 {x}^{2} + 3 x - 6 + \frac{- 19}{x - 2}$

God bless....I hope the explanation is useful.