Question

f(x) = 1 / (x3 + x2 -2x)

Answer 1

What have we to find?

Answer 2

\[\frac{1}{x^3+x^2-2x} = \frac{1}{x (x^2-x-2)} = \frac{1}{x (x-2)(x+1)}\]
\[\rightarrow \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} = \frac{1}{x(x-2)(x+1)}\]
\[A(x-2)(x+1) + Bx(x+1) + Cx(x-2) = 1\]
\[A+B+C = 0\]
\[-A + B-2C = 0\]
\[-2A = 1\]
\[A = -\frac{1}{2}, B = \frac{1}{6}, C = \frac{1}{3}\]
Therefore:
\[\frac{1}{x^3 +x^2-2x} = -\frac{1}{2x} + \frac{1}{6(x-2)} + \frac{1}{3(x+1)}\]

Answer 3

Oh we were told to simplify!