# How do you divide (x^4-9x^2-2) / (x^2+3x-1) ?

Jun 29, 2018

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

#### Explanation:

One method involves separating out multiples of the denominator from the numerator, starting with the highest degree term. This is equivalent to polynomial long division.

We find:

$\frac{{x}^{4} - 9 {x}^{2} - 2}{{x}^{2} + 3 x - 1}$

$= \frac{\left({x}^{4} + 3 {x}^{3} - {x}^{2}\right) - 3 {x}^{3} - 8 {x}^{2} - 2}{{x}^{2} + 3 x - 1}$

$= \frac{{x}^{2} \left({x}^{2} + 3 x - 1\right) - \left(3 {x}^{3} + 8 {x}^{2} + 2\right)}{{x}^{2} + 3 x - 1}$

$= {x}^{2} - \frac{3 {x}^{3} + 8 {x}^{2} + 2}{{x}^{2} + 3 x - 1}$

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

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

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

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

$= {x}^{2} - 3 x + 1 - \frac{6 x + 1}{{x}^{2} + 3 x - 1}$