# How do you express (4x^5 + 2x^4 + 3x^3 - x^2 - x + 1) /( x (x^2 + 1)) in partial fractions?

Oct 26, 2016

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

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

#### Explanation:

To express a rational algebraic expression in partial fractions, the degree of algebraic expression in numerator should be less than that of denominator. But here we have the degree of numerator higher so we first divide numerator by denominator, which is ${x}^{3} + x$.

Now $4 {x}^{5} + 2 {x}^{4} + 3 {x}^{3} - {x}^{2} - x + 1$

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

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

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

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

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

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

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

Now as the degree of numerator is less than that of denominator, let us workout partial fraction. Given denominator, these will be of type

$\frac{- 3 {x}^{2} + 1}{{x}^{2} \left(x + 1\right)} \Leftrightarrow \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 1}$

or (Ax(x+1)+B(x+1)+Cx^2)/(x^2(x+1)

or ((A+C)x^2+(A+B)x+B)/(x^2(x+1)

Hence $A + C = - 3$, $A + B = 0$ and $B = 1$

Hence we have $A = - 1$, $B = 1$ and $C = - 2$

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

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