To express 2x5−x3−1x3−4x in partial fraction, first the degree of numerator should be less than that of denominator and then factorize denominator. As
(2x5−x3−1) = 2x2(x3−4x)+7(x3−4x)+28x−1 and
x3−4x=x(x+2)(x−2)
2x5−x3−1x3−4x = 2x2+7+28x−1x(x−2)(x+2).
Partial fractions of 28x−1x(x−2)(x+2) are given by
28x−1x(x−2)(x+2)=Ax+Bx−2+Cx+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
= Ax2−4A+Bx2+2Bx+Cx2−2Cxx(x−2)(x+2
= x2(A+B+C)+x(2B−2C)−4Ax(x−2)(x+2
Hence A+B+C=0, 2B−2C=28 and 4A=1
This gives A=14, while B+C=−14 and B-C=14i.e.B=55/8andC=--57/8#
Hence partial fractions are 2x2+7+14x−558(x−2)−578(x+2)
