1.

Pls sove this partial fraction

Answer»

2x4−2x3+x(2x−1)2(x−2)=Ax+B+C2x−1+D(2x−1)2+Ex−22x4−2x3+x(2x−1)2(x−2)=Ax+B+C2x−1+D(2x−1)2+Ex−2

More GENERALLY for

amxm+am−1xm−1+⋯+a1x+a0bnxn+bn−1xn−1+⋯+b1x+b0=amxm+am−1xm−1+⋯+a1x+a0∏(dix−ci)niamxm+am−1xm−1+⋯+a1x+a0bnxn+bn−1xn−1+⋯+b1x+b0=amxm+am−1xm−1+⋯+a1x+a0∏(dix−ci)NI

where m≥n,m≥n,

we can WRITE amxm+am−1xm−1+⋯+a1x+a0bnxn+bn−1xn−1+⋯+b1x+b0amxm+am−1xm−1+⋯+a1x+a0bnxn+bn−1xn−1+⋯+b1x+b0

=em−nxm−n+em−n−1xm−n−1+⋯+e1x+e0+∑(fi1dix−ci+fi2(dix−ci)2+⋯+fi(ni−1)(dix−ci)ni−1+fini(dix−ci)ni)


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