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

Nov 2, 2017

Taken you to a point where you can continue from. Method is demonstrated.

#### Explanation:

I am using no value place keepers. Example $0 {x}^{3}$

$\textcolor{w h i t e}{\text{ddddddddd.dddddddd.dd}} 6 {x}^{4} + 0 {x}^{3} + 0 {x}^{2} + 12 x + 4$
$\textcolor{m a \ge n t a}{+ 2 {x}^{2}} \left(3 {x}^{2} + 2 x + 5\right) \to \textcolor{w h i t e}{\text{d") ul(6x^4+4x^3+10x^2 larr" Subtract}}$
$\textcolor{w h i t e}{\text{dddddddddddddddddddddd}} 0 - 4 {x}^{3} - 10 {x}^{2} + 12 x + 4$
$\textcolor{m a \ge n t a}{- \frac{4}{3} x} \left(3 {x}^{2} + 2 x + 5\right) \to \textcolor{w h i t e}{\text{dd.dd")ul(-4x^3-8/3x^2-20/3 x larr" Sub.}}$
$\textcolor{w h i t e}{\text{dddddddddddddddddddddddddd}} 0 - \frac{22}{3} {x}^{2} + \frac{56}{3} x + 4$

This should be enough to demonstrate the method. I will let you take over from this point.

SO FAR WE HAVE AS AN ANSWER

$\frac{\textcolor{w h i t e}{\text{d")color(magenta)(2x^2-4/3x + ? + ("remainder}}}{3 {x}^{2} + 2 x + 5}$