How do you express (x^5 + 1)/(x^6 - x^4)x5+1x6−x4 in partial fractions?

1 Answer
Bdub
Mar 8, 2016

(x^5+1)/(x^4(x+1)(x-1)) = -1/x^2 -1/x^4+1/(x-1)x5+1x4(x+1)(x−1)=−1x2−1x4+1x−1

Explanation:

(x^5+1)/(x^4(x^2-1)) =(x^5+1)/(x^4(x+1)(x-1))x5+1x4(x2−1)=x5+1x4(x+1)(x−1)
(x^5+1)/(x^4(x+1)(x-1))= A/x +B/x^2+C/x^3+D/x^4 +E/(x+1)+F/(x-1)x5+1x4(x+1)(x−1)=Ax+Bx2+Cx3+Dx4+Ex+1+Fx−1

x^5+1=A(x^3)(x^2-1)+B(x^2)(x^2-1)+C(x)(x^2-1)+D(x^2-1)+E(x^4)(x-1)+F(x^4)(x+1)x5+1=A(x3)(x2−1)+B(x2)(x2−1)+C(x)(x2−1)+D(x2−1)+E(x4)(x−1)+F(x4)(x+1)

x^5+1=Ax^5-Ax^3+Bx^4-Bx^2+Cx^3-Cx+Dx^2-D+Ex^5-Ex^4+Fx^5+Fx^4x5+1=Ax5−Ax3+Bx4−Bx2+Cx3−Cx+Dx2−D+Ex5−Ex4+Fx5+Fx4

1=A+E+F, 0=B-E+F, 0=-A+C,0=-B+D, 0=-C, 1=-D1=A+E+F,0=B−E+F,0=−A+C,0=−B+D,0=−C,1=−D

A=0,B=-1,C=0,D=-1,E=0,F=1A=0,B=−1,C=0,D=−1,E=0,F=1

(x^5+1)/(x^4(x+1)(x-1)) = -1/x^2 -1/x^4+1/(x-1)x5+1x4(x+1)(x−1)=−1x2−1x4+1x−1

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