Question

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

Answer 1

Partial fractions are `2x^2+7+1/(4x)-55/(8(x-2))-57/(8(x+2))`

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^3-4x=x(x+2)(x-2)`

`(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

`(28x-1)/(x(x-2)(x+2))=A/x+B/(x-2)+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`, `2B-2C=28` and `4A=1`

This gives `A=1/4`, while `B+C=-1/4` and B-C=14`i.e.`B=55/8`and`C=--57/8#

Hence partial fractions are `2x^2+7+1/(4x)-55/(8(x-2))-57/(8(x+2))`