How do you express 2/(x^3-x^2)2x3−x2 in partial fractions?
1 Answer
Mar 23, 2016
2/(x^3-x^2) = 2/(x-1)-2/x-2/x^22x3−x2=2x−1−2x−2x2
Explanation:
The denominator splits into factors as:
So we solve for a partial fraction decomposition of the form:
2/(x^3-x^2)2x3−x2
= A/(x-1)+B/x+C/x^2=Ax−1+Bx+Cx2
=(Ax^2+Bx(x-1)+C(x-1))/(x^3-x^2)=Ax2+Bx(x−1)+C(x−1)x3−x2
=((A+B)x^2+(C-B)x-C)/(x^3-x^2)=(A+B)x2+(C−B)x−Cx3−x2
Equating coefficients, we get:
{ (A+B = 0), (C-B = 0), (-C=2) :}
Hence we find:
{ (C = -2), (B = -2), (A = 2) :}
So:
2/(x^3-x^2) = 2/(x-1)-2/x-2/x^2
