# How do you divide (3x^3-6x^2+13x-4)/(x-3) ?

Jun 3, 2017

$\frac{3 {x}^{3} - 6 {x}^{2} + 13 x - 4}{x - 3} = \left(3 {x}^{2} + 3 x - 22\right) + \frac{62}{x - 3}$

#### Explanation:

We want to evaluate $\frac{3 {x}^{3} - 6 {x}^{2} + 13 x - 4}{x - 3}$

First let $f \left(x\right) = 3 {x}^{3} - 6 {x}^{2} + 13 x - 4$

Now, we'll take $f \left(3\right)$ to see if $\left(x - 3\right)$ is a factor of $f$ or not

$f \left(3\right) = 62$

So $\left(x - 3\right)$ isn't a factor.

But what can say is:

$3 {x}^{3} - 6 {x}^{2} + 13 x - 4 = \left(x - 3\right) \left(a {x}^{2} + b x + c\right) + 62$

$3 {x}^{3} - 6 {x}^{2} + 13 x - 66 = \left(x - 3\right) \left(a {x}^{2} + b x + c\right)$

$\therefore - 3 c = - 66 \Rightarrow c = 22$
and $a {x}^{3} = 3 {x}^{3} \Rightarrow a = 3$
and $- 9 {x}^{2} + b {x}^{2} = - 6 {x}^{2} \Rightarrow b = 3$

$\therefore 3 {x}^{3} - 6 {x}^{2} + 13 x - 4 = \left(x - 3\right) \left(3 {x}^{2} + 3 x - 22\right) + 62$

Dividing everything by $\left(x - 3\right)$ gives us

$\frac{3 {x}^{3} - 6 {x}^{2} + 13 x - 4}{x - 3} = \left(3 {x}^{2} + 3 x - 22\right) + \frac{62}{x - 3}$