Question

How to solve below given partial fraction ... can u kindly explain... ( 8-4v+11v^2-2v^3)÷{(1-0.25v)(1-v+0.5v^2)}

Answer 1

We check the 2nd degree polynomial if we can factorize it.  It is not factorizable and it has only imaginary roots.  so we keep it as it is.

$\frac{(8-4v+11v^2-2v^3)}{(1-0.25v)(1-v+0.5v^2)} \\\\=\frac{8(8-4v+11v^2-2v^3)}{(4-v)(2-2v+v^2)} \\\\=+16+\frac{A}{4-v}+\frac{Bx+c}{2-2v+v^2}$

We write +16 on RHS because the coefficient of (highest degree term) v^3 in numerator is -16 and in denominator - 1 * +1 = -1, and hence  -16 v^3/- v^3.

Since 4-v is first degree polynomial, write A in the numerator.  B x +C in the second fraction because the other factor is a second degree polynomial.

$8(8-4v+11v^2-2v^3)\\=16(4-v)(2-2v+v^2)+A(2-2v+v^2)+(Bv+c)(4-v)\\\\-16v^3=-16v^3\\+88v^2=32v^2+64v^2+Av^2-Bv^2\\-32v=-32v-128v-2Av+4Bv-Cv\\64=128+2A+4C$

Solve the above for A, B and C.
You get answer