# How do you divide ( x^4-2x^3-x)/(x(x+4))?

$\frac{{x}^{4} - 2 {x}^{3} - x}{x \left(x + 4\right)} = \frac{{x}^{3} - 2 {x}^{2} - 1}{x + 4} = \textcolor{red}{{x}^{2} - 6 x + 24 - \frac{97}{x + 4}}$

#### Explanation:

From the given $\frac{{x}^{4} - 2 {x}^{3} - x}{x \left(x + 4\right)}$

We can reduce the degree of the dividend and the divisor

$\frac{{x}^{4} - 2 {x}^{3} - x}{x \left(x + 4\right)}$

factor the common monomial x

$\frac{x \left({x}^{3} - 2 {x}^{2} - 1\right)}{x \left(x + 4\right)}$

$\frac{\cancel{x} \left({x}^{3} - 2 {x}^{2} - 1\right)}{\cancel{x} \left(x + 4\right)}$

$\frac{{x}^{3} - 2 {x}^{2} - 1}{x + 4}$

Perform long division
" " " " ""underline(x^2-6x+24" " " " " ")
$x + 4 \lceiling {x}^{3} - 2 {x}^{2} + 0 \cdot x - 1$
"" " " " " underline(x^3+4x^2" " " " "" " " " "" " " " ")
$\text{ " " " " " " } - 6 {x}^{2} + 0 \cdot x - 1$
" " " " " " " "underline(-6x^2-24x" " " " "" " " " ")
$\text{ " " " " " " " " " " " " " } 24 x - 1$
$\text{ " " " " " " " " " " " " " } \underline{24 x + 96}$
$\text{ " " " " " " " " " " " " " " " " } - 97$$\leftarrow$remainder

We write our answer this way

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

$\frac{{x}^{4} - 2 {x}^{3} - x}{x \left(x + 4\right)} = {x}^{2} - 6 x + 24 - \frac{97}{x + 4}$

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