Question

Resolve $\frac{3x^2+x-2}{{(x-2)}^2(1-2x)}$ into partial fractions.

Answer 1

Assume $\frac{3x^2+x-2}{{(x-2)}^2(1-2x)}=\frac{A}{1-2x}+\frac{B}{x-2}+\frac{C}{{(x-2)}^2}$;
$\therefore 3x^2+x-2=A{(x-2)}^2+B(1-2x)(x-2)+C(1-2x)$
Let $1-2x=0$, then $A=-\frac{1}{3}$;
Let $x-2=0$, then $C=-4$
To find $B$, equate the coefficients of $x^2$, thus
$3=A-2B$l whence $B=-\frac{5}{3}$
$\therefore \frac{3x^2+x-2}{{(x-2)}^2(1-2x)}=-\frac{1}{3(1-2x)}-\frac{5}{3(x-1)}-\frac{4}{{(x-2)}^2}$