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

Feb 15, 2016

Partial fractions are $2 {x}^{2} + 7 + \frac{1}{4 x} - \frac{55}{8 \left(x - 2\right)} - \frac{57}{8 \left(x + 2\right)}$

#### Explanation:

To express (2x^5−x^3−1)/(x^3−4x) in partial fraction, first the degree of numerator should be less than that of denominator and then factorize denominator. As

(2x^5−x^3−1) = 2x^2(x^3-4x)+7(x^3−4x)+28x-1 and

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

(2x^5−x^3−1)/(x^3−4x) = 2x^2+7+(28x-1)/(x(x-2)(x+2).

Partial fractions of (28x-1)/(x(x-2)(x+2) are given by

$\frac{28 x - 1}{x \left(x - 2\right) \left(x + 2\right)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$

=[A(x-2)(x+2)+Bx(x+2)+Cx(x-2}}/(x(x-2)(x+2)

{A(x-2)(x+2)+Bx(x+2)+Cx(x-2)}/(x(x-2)(x+2

= {Ax^2-4A+Bx^2+2Bx+Cx^2-2Cx}/(x(x-2)(x+2

= {x^2(A+B+C)+x(2B-2C)-4A}/(x(x-2)(x+2

Hence $A + B + C = 0$, $2 B - 2 C = 28$ and $4 A = 1$

This gives $A = \frac{1}{4}$, while $B + C = - \frac{1}{4}$ and B-C=14$i . e .$B=55/8$\mathmr{and}$C=--57/8#

Hence partial fractions are $2 {x}^{2} + 7 + \frac{1}{4 x} - \frac{55}{8 \left(x - 2\right)} - \frac{57}{8 \left(x + 2\right)}$