# How do you express (2x^4 + 9x^2 + x - 4)/(x^3 + 4x) in partial fractions?

Jan 27, 2018

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

#### Explanation:

Let us first divide $2 {x}^{4} + 9 {x}^{2} + x - 4$ by ${x}^{3} + 4 x$ to get the degree of numerator less than that of denominator and we get

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

Let us now get partial fractions of $\frac{{x}^{2} + x - 4}{x \left({x}^{2} + 4\right)}$, which will be of the form

$\frac{{x}^{2} + x - 4}{x \left({x}^{2} + 4\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 4}$

or ${x}^{2} + x - 4 = A \left({x}^{2} + 4\right) + x \left(B x + C\right) = \left(A + B\right) {x}^{2} + C x + 4 A$

andcomparing coefficients of like powers

$4 A = - 4$ or $A = - 1$, $C = 1$ and

$A + B = 1$ i.e. $B = 1 - A = 2$

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