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

Aug 10, 2017

$2 x - 1$ and remainder of $\frac{- 3}{x - 1}$

#### Explanation:

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} \underline{2 x - 1}$
$\textcolor{w h i t e}{a a} x - 1$$|$$2 {x}^{2} - 3 x - 2$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} \underline{2 {x}^{2} - 2 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - x - 2$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{- x + 1}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - 3$

color(magenta)((2x^2-3x-2) / (x-1) = 2x-1 and remaindercolor(magenta)((-3)/(x-1)

Aug 10, 2017

$2 {x}^{2} - 3 x - 2 = \left(x - 1\right) \left(2 x - 1\right) - 3$

or

$\frac{2 {x}^{2} - 3 x - 2}{x - 1} = 2 x - 1 - \frac{3}{x - 1}$

#### Explanation:

In each section of the 'long division', the values on the quotient depend on what factor of the highest degree of the divisor can fit into the highest degree of the dividend. For example, $x$ fits $2 x$ times into $2 {x}^{2}$, $x$ fits -1 times into $- x$

{: (,,,2x,-1), (,,"---","---","---"), (x-1,")",2x^2, -3x, -2), (,,2x^2,-2x,), (,,"---","---",), (,,,-x,-2), (,,,-x,+1), (,,"---","---","---"), (,,,,-3) :}

Therfore,

$\frac{2 {x}^{2} - 3 x - 2}{x - 1} = 2 x - 1 - \frac{3}{x - 1}$

$2 {x}^{2} - 3 x - 2 = \left(x - 1\right) \left(2 x - 1\right) - 3$