# How do you divide ( 2x^4 - 5x^3 - 8x^2+17x+1 )/(x^2 - 2 )?

Jun 14, 2017

(2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1 )/(x^2 - 2)= (2 x^2 - 5 x - 4) + (7 x - 7 )/(x^2-2)

#### Explanation:

The first thing we should do when trying to divide two rational functions is to see if the denominator is a zero of the numerator.

Let $f \left(x\right) = 2 {x}^{4} - 5 {x}^{3} - 8 {x}^{2} + 17 x + 1$

$f \left(\sqrt{2}\right) = - 7 + 7 \sqrt{2}$

We know that any rational function can be written as: $f \left(x\right) = p \left(x\right) q \left(x\right) + r \left(x\right)$. Thus we can say:

$2 {x}^{4} - 5 {x}^{3} - 8 {x}^{2} + 17 x + 1 = \left(a {x}^{2} + b x + c\right) \left({x}^{2} - 2\right) + 7 x - 7$

This is because when $\left({x}^{2} - 2\right) = 0$, the remainder was $\left(7 x - 7\right)$. So $r \left(x\right) = 7 x - 7$. We already know $q \left(x\right)$ and we need to work out $p \left(x\right)$.

$2 {x}^{4} - 5 {x}^{3} - 8 {x}^{2} + 10 x + 8 = \left(a {x}^{2} + b x + c\right) \left({x}^{2} - 2\right)$

$a {x}^{4} = 2 {x}^{4} \Rightarrow a = 2$

$b {x}^{3} = - 5 {x}^{3} \Rightarrow b = - 5$

$- 2 c = 8 \Rightarrow c = - 4$

$2 {x}^{4} - 5 {x}^{3} - 8 {x}^{2} + 10 x + 8 = \left(2 {x}^{2} - 5 x - 4\right) \left({x}^{2} - 2\right)$

$2 {x}^{4} - 5 {x}^{3} - 8 {x}^{2} + 17 x + 1 = \left(2 {x}^{2} - 5 x - 4\right) \left({x}^{2} - 2\right) + 7 x - 7$

(2 x^4 - 5 x^3 - 8 x^2 + 17 x + 1 )/(x^2 - 2)= (2 x^2 - 5 x - 4) + (7 x - 7 )/(x^2-2)