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

Jun 12, 2018

$2 {x}^{2} + 7 x + 3$ with a remainder of $- 14 x - 5$.

#### Explanation:

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

We can use the polynomial long division method, begin by writing:

${x}^{2} - x + 2 | \overline{2 {x}^{4} + 5 {x}^{3} - 3 x + 1}$

The first question we have to ask is how many times does ${x}^{2}$ go into $2 {x}^{4}$. The answer is obviously $2 {x}^{2}$. We put this factor on top, so the next line of our working will look like:

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

We now have to multiply the divisor ${x}^{2} - x + 2$ by $2 {x}^{2}$ to get
$2 {x}^{4} - 2 {x}^{3} + 4 {x}^{2}$ and subtract this from the dividend, the next line of working will look like:

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} 2 {x}^{2}$
${x}^{2} - x + 2 | \overline{2 {x}^{4} + 5 {x}^{3} - 0 {x}^{2} - 3 x + 1}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} \underline{- \left(2 {x}^{4} - 2 {x}^{3} + 4 {x}^{2}\right)}$

Notice in the dividend their is no ${x}^{2}$ so I have added a $0 {x}^{2}$ term in place of where the ${x}^{2}$ term should be to make the working clearer. We do the subtraction to get:

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} 2 {x}^{2}$
${x}^{2} - x + 2 | \overline{2 {x}^{4} + 5 {x}^{3} - 0 {x}^{2} - 3 x + 1}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} \underline{- \left(2 {x}^{4} - 2 {x}^{3} + 4 {x}^{2} + 0 x + 0\right)}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 0 + 7 {x}^{3} - 4 {x}^{2} - 3 x + 1$

We now repeat the process, treating: $7 {x}^{3} - 4 {x}^{2} + 3 x + 1$ as the "new dividend." Keep repeating until the degree of the dividend is less than the degree of the divisor. The working, when written out in full will look like:

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots .} 2 {x}^{2} + 7 x + 3$
${x}^{2} - x + 2 | \overline{2 {x}^{4} + 5 {x}^{3} - 0 {x}^{2} - 3 x + 1}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} \underline{- \left(2 {x}^{4} - 2 {x}^{3} + 4 {x}^{2} + 0 x + 0\right)}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots .} 0 + 7 {x}^{3} - 4 {x}^{2} - 3 x + 1$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \underline{- \left(7 {x}^{3} - 7 {x}^{2} + 14 x + 0\right)}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} 0 + 3 {x}^{2} - 17 x + 1$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \underline{- \left(3 {x}^{2} - 3 x + 6\right)}$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} - 14 x - 5$

So the result is:

$2 {x}^{2} + 7 x + 3$ with a remainder of $- 14 x - 5$. We can also use this to rewrite the original fraction as:

$\frac{2 {x}^{4} + 5 {x}^{3} - 3 x + 1}{{x}^{2} - x + 2} = 2 {x}^{2} + 7 x + 3 + \frac{- 14 x - 5}{{x}^{2} - x + 2}$