# How do you divide (2x^4 + 2x^3 - 6x^2 + 3x - 12)/( x^2 + 3x - 6)?

Dec 19, 2015

Long divide the coefficients to find:

$2 {x}^{4} + 2 {x}^{3} - 6 {x}^{2} + 3 x - 12$

$= \left(2 {x}^{2} - 4 x + 18\right) \left({x}^{2} + 3 x - 6\right) + \left(- 75 x + 96\right)$

That is:

$\frac{2 {x}^{4} + 2 {x}^{3} - 6 {x}^{2} + 3 x - 12}{{x}^{2} + 3 x - 6}$

$= 2 {x}^{2} - 4 x + 18 + \frac{- 75 x + 96}{{x}^{2} + 3 x - 6}$

#### Explanation:

I like to long divide the coefficients like this:

Write the dividend under the bar and the divisor to the left of the bar. In our example the dividend $\left(2 {x}^{4} + 2 {x}^{3} - 6 {x}^{2} + 3 x - 12\right)$ is represented by the sequence $2$, $2$, $- 6$, $3$, $- 12$ and the divisor $\left({x}^{2} + 3 x - 6\right)$ by the sequence $1$, $3$, $- 6$.

If we had any 'missing' powers of $x$ in either the dividend or the divisor then we would have to include a $0$ coefficient in the sequence representing it.

Then start writing the quotient above the bar, choosing the first term $\textcolor{b l u e}{2}$ so that when multiplied by the divisor it matches the leading term of the dividend.

Write the product of this term and the divisor $2$, $6$, $- 12$ under the dividend and subtract it to give a remainder.

Bring down the next term from the dividend alongside this remainder, then choose the next term $\textcolor{b l u e}{- 4}$ of the quotient to match the leading term of the remainder, when multiplied by the divisor.

Repeat until the 'degree' of the remainder is less than the divisor. In our case the remainder $- 75$, $96$ representing $\left(- 75 x + 96\right)$ is of degree $1$, less than the degree $2$ of the divisor.

This is our final remainder.

Being used to long dividing numbers, I find this an easy way to lay out and divide polynomials, but there are more compact ways to represent 'synthetic division'.