Question

Resolve $\frac{x^4}{(x-1)(x-2)}$ into partial fractions.

Answer 1

$F=\frac{x^4}{(x-1)(x-2)}$
$=\frac{\left[\right(x-1)-(x-2\left)\right]x^4}{(x-1)(x-2)}$
$=\frac{x^4}{x-2}-\frac{x^4}{x-1}$
$=\frac{\left[\right(x-2)+2)]x^3}{x-2}-\frac{\left[\right(x-1)+1]x^3}{x-1}$
$=x^3+\frac{2x^3}{x-2}-x^3-\frac{x^3}{x-1}$
$=\frac{2x^3}{x-2}-\frac{x^3}{x-1}$
$=\frac{2[x-2+2]x^2}{x-2}-\frac{[x-1+1]x^2}{x-1}$
$=2x^2+\frac{4x^2}{x-2}-x^2-\frac{x^2}{x-1}$
$=x^2+\frac{4x^2}{x-2}-\frac{x^2}{x-1}$
$=x^2+\frac{4[x-2+2]x}{x-2}-\frac{[x-1+1]x}{x-1}$
$=x^2+4x+\frac{8x}{x-2}-x-\frac{x}{x-1}$
$=x^2+3x+\frac{8(x-2+2)}{x-2}-\frac{(x-1+1)}{x-1}$
$=x^2+3x+8+\frac{16}{x-2}-1-\frac{1}{x-1}$
$=x^2+3x+7+\frac{16}{x-2}-\frac{1}{x-1}$.
Hence
$\frac{x^4}{(x-1)(x-2)}=x^2+3x+7+\frac{16}{x-2}-\frac{1}{x-1}$