Question

@jdoe0001 @zepdrix How do you decompose (5x^3-x^2+8x-55)/(x^4+5x^3+11x^2) into partial fractions?

Answer 1

@Luigi0210 please help

Answer 2

First thing to do is factorize the denominator:
\[x^4+5x^3+11x^2=x^2\left(x^2+5x+11\right)\]
The second factor is irreducible in the reals as far as I can tell.

Time for partial fractions:
\[\frac{5x^3-x^2+8x-55}{x^2\left(x^2+5x+11\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+5x+11}\]
Find a common denominator for the RHS and combine the fractions:
\[\frac{Ax\left(x^2+5x+11\right)+B\left(x^2+5x+11\right)+(Cx+D)x^2}{x^2\left(x^2+5x+11\right)}\]
Expand and pair up the coefficients by power of the \(x\) term:
\[\frac{Ax^3+5Ax^2+11Ax+Bx^2+5Bx+11B+Cx^3+Dx^2}{x^2\left(x^2+5x+11\right)}\]
\[\frac{(A+C)x^3+(5A+B+D)x^2+(11A+5B)x+11B}{x^2\left(x^2+5x+11\right)}\]
In the equation above, you need to have LHS = RHS, which means the coefficients of matching-power terms in the numerators must match:
\[\begin{cases}
A+C=5&x^3\text{ term}\\[1ex]
5A+B+D=-1&x^2\text{ term}\\[1ex]
11A+5B=8&x\text{ term}\\[1ex]
11B=-55&\text{constant term}
\end{cases}\]