# How do you divide (x^4 - 8x^3 + 4x^2 + 12x)/(x^2-2x+2)?

May 1, 2016

$\frac{{x}^{4} - 8 {x}^{3} + 4 {x}^{2} + 12 x}{{x}^{2} - 2 x + 2} = \left({x}^{2} - 6 x - 10\right)$ with remainder $4 x + 20$.

#### Explanation:

I like to just long divide the coefficients, not forgetting to include $0$'s for any missing powers of $x$. In our example, that means the constant term of the dividend...

The process is similar to long division of numbers.

The result $1 , - 6 , - 10$ means ${x}^{2} - 6 x - 10$ and remainder $4 , 20$ means $4 x + 20$

So:

${x}^{4} - 8 {x}^{3} + 4 {x}^{2} + 12 x = \left({x}^{2} - 2 x + 2\right) \left({x}^{2} - 6 x - 10\right) + \left(4 x + 20\right)$

Or if you prefer:

$\frac{{x}^{4} - 8 {x}^{3} + 4 {x}^{2} + 12 x}{{x}^{2} - 2 x + 2} = \left({x}^{2} - 6 x - 10\right) + \frac{4 x + 20}{{x}^{2} - 2 x + 2}$