# How do you divide (3x^3+4x^2-2x+5)/(x-1) ?

Mar 14, 2017

The remainder is $= 10$ and the quotient is $= 3 {x}^{2} + 7 x + 5$

#### Explanation:

We can perform a long division

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} + 4 {x}^{2} - 2 x + 5$$\textcolor{w h i t e}{a a a a}$$|$$x - 1$

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} - 3 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$$3 {x}^{2} + 7 x + 5$

$\textcolor{w h i t e}{a a a a a a}$$0 + 7 {x}^{2} - 2 x$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 7 {x}^{2} - 7 x$

$\textcolor{w h i t e}{a a a a a a a a a a}$$+ 0 + 5 x + 5$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$+ 5 x - 5$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$+ 0 + 10$

The remainder is $= 10$ and the quotient is $= 3 {x}^{2} + 7 x + 5$

$\frac{3 {x}^{3} + 4 {x}^{2} - 2 x + 5}{x - 1} = 3 {x}^{2} + 7 x + 5 + \frac{10}{x - 1}$