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

1. anonymous

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: $\begin{cases} A+C=5&x^3\text{ term}\\[1ex] 5A+B+D=-1&x^2\text{ term}\\[1ex] 11A+5B=8&x\text{ term}\\[1ex] 11B=-55&\text{constant term} \end{cases}$