Question

What is the rational expression as a sum of partial fractions?

Answer 1

$\frac{-x^2+2x-5}{x^3-x^2}$

Rewrite the expression as:

$\frac{-x^2+2x-5}{x^2(x-1)}$

The partial fraction expansion is of the form:

$\frac{-x^2+2x-5}{x^2(x-1)}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x^2}$

Multiply both sides by x²(x - 1):

$\begin{gathered} -x^2+2x-5=Ax^2+(x-1)(Bx+C) \\ -x^2+2x-5=-C+(A+B)x^2+(C-B)x \end{gathered}$

Equate the coefficients on both sides:

$\begin{gathered} -5=-C_{\text{ }}(1)_{} \\ 2=C-B_{\text{ }}(2) \\ -1=A+B_{\text{ }}(3) \end{gathered}$

So, from (1):

$C=5$

Replace C into (2):

$\begin{gathered} 2=5-B \\ B=3 \end{gathered}$

Replace B into (3):

$\begin{gathered} -1=A+3 \\ A=-4 \end{gathered}$

Therefore, the answer is:

$\frac{-x^2+2x-5}{x^2(x-1)}=\frac{-4}{x-1}+\frac{3}{x}+\frac{5}{x^2}$