# How do you divide (x^4-2x^3-2x^2+9x+3)/(x^2-3) ?

May 2, 2016

$\frac{{x}^{4} - 2 {x}^{3} - 2 {x}^{2} + 9 x + 3}{{x}^{2} - 3} = \left({x}^{2} - 2 x + 1\right) + \frac{3 x + 6}{{x}^{2} - 3}$

#### Explanation:

I like to long divide the coefficients, not forgetting to include $0$'s for any missing powers of $x$. In our example, that means the missing $x$ term in the divisor, which is therefore repesented as $1 , 0 , - 3$.

The process is similar to long division of numbers.

We find a quotient $1 , - 2 , 1$, meaning ${x}^{2} - 2 x + 1$ and remainder $3 , 6$, meaning $3 x + 6$

So: $\frac{{x}^{4} - 2 {x}^{3} - 2 {x}^{2} + 9 x + 3}{{x}^{2} - 3} = \left({x}^{2} - 2 x + 1\right) + \frac{3 x + 6}{{x}^{2} - 3}$