How do you express (x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) as partial fractions?

Oct 15, 2016

$\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2} = {x}^{2} + \frac{3}{x - 2} - \frac{x + 1}{{x}^{2} + 1}$

Explanation:

First note that:

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

Is the numerator divisible by either of these factors?

It is not divisible by $\left(x - 2\right)$ since $5$ is odd.

How about $\left({x}^{2} + 1\right)$ ?

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

So there are no common factors of the numerator and denominator.

Next note that:

$\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2} = \frac{{x}^{5} - 2 {x}^{4} + {x}^{3} - 2 {x}^{2} + 2 {x}^{2} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}$

$\textcolor{w h i t e}{\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}} = {x}^{2} + \frac{2 {x}^{2} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}$

$\textcolor{w h i t e}{\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}} = {x}^{2} + \frac{A}{x - 2} + \frac{B x + C}{{x}^{2} + 1}$

$\textcolor{w h i t e}{\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}} = {x}^{2} + \frac{A \left({x}^{2} + 1\right) + \left(B x + C\right) \left(x - 2\right)}{{x}^{2} - 2 {x}^{2} + x - 2}$

$\textcolor{w h i t e}{\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2}} = {x}^{2} + \frac{\left(A + B\right) {x}^{2} + \left(- 2 B + C\right) x + \left(A - 2 C\right)}{{x}^{2} - 2 {x}^{2} + x - 2}$

Equating coefficients we get the following system of linear equations:

$\left\{\begin{matrix}A + B = 2 \\ - 2 B + C = 1 \\ A - 2 C = 5\end{matrix}\right.$

Subtracting the third equation from the first, we get:

$B + 2 C = - 3$

Adding twice this to the second equation, we find:

$5 C = - 5$

Hence:

$C = - 1$

Substituting this value of $C$ into the second equation we find:

$- 2 B - 1 = 1$

Hence:

$B = - 1$

Then from the first equation we find:

$A = 2 - B = 3$

So:

$\frac{{x}^{5} - 2 {x}^{4} + {x}^{3} + x + 5}{{x}^{3} - 2 {x}^{2} + x - 2} = {x}^{2} + \frac{3}{x - 2} - \frac{x + 1}{{x}^{2} + 1}$