# How do you divide (x^4 + 5x^2 + 12x + 3)/( 6x+8)?

Jun 9, 2016

$\implies {x}^{3} / 6 - \frac{2 {x}^{2}}{9} + \frac{61 x}{54} + \frac{40}{81} - \frac{77}{486 x + 648}$

#### Explanation:

$\text{ } {x}^{4} + 0 {x}^{3} + 5 {x}^{2} + 12 x + 3$
$\textcolor{m a \ge n t a}{\frac{1}{6} {x}^{3}} \left(6 x + 8\right) \to \text{ "color(white)(a)ul(x^4+4/3 x^3" ")" "larr "Subtract}$
$\text{ } 0 - \frac{4}{3} {x}^{3} + 5 {x}^{2} + 12 x + 3$
$\textcolor{m a \ge n t a}{- \frac{2}{9} {x}^{2}} \left(6 x + 8\right) \to \text{ "color(white)(a)ul(-4/3x^3-16/9x^2" ")" "larr "Subtract}$
$\text{ } 0 + \frac{61}{9} {x}^{2} + 12 x + 3$
$\textcolor{m a \ge n t a}{\frac{61}{54} x} \left(6 x + 8\right) \to \text{ "color(white)(a)ul(61/9x^2+244/27x)" "larr "Subtract}$
$\text{ } 0 + \frac{80}{27} x + 3$
$\textcolor{m a \ge n t a}{\frac{40}{81}} \left(6 x + 8\right) \implies \text{ "color(white)(a)ul(80/27x+320/81)" Subtract}$
$\text{ } 0 - \frac{77}{81}$

=> x^3/6-(2x^2)/9+(61x)/54+40/81-77/(81(6x+8)

$\implies {x}^{3} / 6 - \frac{2 {x}^{2}}{9} + \frac{61 x}{54} + \frac{40}{81} - \frac{77}{486 x + 648}$