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

May 18, 2017

Given: $\frac{{x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}{{x}^{2} - 3}$

Write the divisor with a 0 coefficient for missing term

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{\textcolor{w h i t e}{\left({x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3\right)}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$

To determine the first term in the quotient divide the first term in the dividend by the first term in the divisor ${x}^{4} / {x}^{2} = {x}^{2}$ write ${x}^{2}$ in the quotient:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} \textcolor{w h i t e}{2 {x}^{3} + 2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$

Multiply the term in the quotient by the divisor:

${x}^{2} \left({x}^{2} + 0 x - 3\right) = {x}^{4} + 0 {x}^{3} - 3 {x}^{2}$

Make it negative:

$- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}$

Write it below the dividend:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} \textcolor{w h i t e}{2 {x}^{3} + 2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$

Perform the addition and bring down the x term:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} \textcolor{w h i t e}{2 {x}^{3} + 2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$

To find the next term in the quotient, divide the first term in the sum by the first term of the divisor:

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

Write it as the next term of the quotient:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x \textcolor{w h i t e}{2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$

Multiply the term in the quotient by the divisor:

$- 2 x \left({x}^{2} + 0 x - 3\right) = - 2 {x}^{3} + 0 {x}^{2} + 6 x$

Make it negative:

$2 {x}^{3} + 0 {x}^{2} - 6 x$

Write it below the sum:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x \textcolor{w h i t e}{2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{2 {x}^{3} + 0 {x}^{2} - 6 x}$

Perform the additions and bring down the constant term:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x \textcolor{w h i t e}{2 {x}^{2} - 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{2 {x}^{3} + 0 {x}^{2} - 6 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 5 {x}^{2} - 9 x + 3$

To find the next term in the quotient, divide the first term in the sum by the first term of the divisor:

$\frac{5 {x}^{2}}{x} ^ 2 = 5$

Write it as the next term of the quotient:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x + 5 \textcolor{w h i t e}{- 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{2 {x}^{3} + 0 {x}^{2} - 6 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 5 {x}^{2} - 9 x + 3$

Multiply the term in the quotient by the divisor:

$5 \left({x}^{2} + 0 x - 3\right) = 5 {x}^{2} + 0 x - 15$

Make it negative:

$- 5 {x}^{2} + 0 x + 15$

Write it below the sum:

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x + 5 \textcolor{w h i t e}{- 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{2 {x}^{3} + 0 {x}^{2} - 6 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 5 {x}^{2} - 9 x + 3$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{- 5 {x}^{2} + 0 x + 15}$

$\textcolor{w h i t e}{\frac{{x}^{2} + 0 x - 3}{\textcolor{b l a c k}{{x}^{2} + 0 x - 3}}} \frac{{x}^{2} - 2 x + 5 \textcolor{w h i t e}{- 3 x + 3}}{| \textcolor{w h i t e}{x} {x}^{4} - 2 {x}^{3} + 2 {x}^{2} - 3 x + 3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- {x}^{4} - 0 {x}^{3} + 3 {x}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 2 {x}^{3} + 5 {x}^{2} - 3 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \underline{2 {x}^{3} + 0 {x}^{2} - 6 x}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} 5 {x}^{2} - 9 x + 3$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{- 5 {x}^{2} + 0 x + 15}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} - 9 x + 18$
Because $- 9 x + 18$ is of order 1 and the divisor is of order 2, we stop and declare $- 9 x + 18$ to be the remainder.