Question

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

Answer 1

The table give general form of the partial fractions :

So , $\frac{1+3x+2x^2}{(1-2x)(1-x^2)}=\frac{1+3x+2x^2}{(1-2x)(1-x)(1+x)}$

$=>\frac{1+3x+2x^2}{(1-2x)(1-x)(1+x)}=\frac{A}{1-2x}+\frac{B}{1-x}+\frac{C}{1+x}$

$=>1+3x+2x^2=A(1-x)(1+x)+B(1-2x)(1+x)+C(1-2x)(1-x)$

For $x=1,=>1+3\left(1\right)+2(1{)}^2=A(0)+B(1-2\left)\right(1+1)+C(0)$

$=>6=-2B$

$=>B=-3$ $-\left(1\right)$

For $x=-1,=>1+3(-1)+2(-1{)}^2=A(0)+B\left(0\right)+C(1-2(-1)\left)\right(1-(-1))$

$=>0=6C$

$=>C=0$ $-\left(2\right)$

For $x=\frac{1}{2},=>1+3\left(\frac{1}{2}\right)+2{\left(\frac{1}{2}\right)}^2=A\left(\frac{1}{2}\right)(1+\frac{1}{2})+B\left(0\right)+C\left(0\right)$

$=>3=\frac{3}{4}A$

$=>A=4$ $-\left(3\right)$

• Therefore , from (1),(2) and (3) $=>\frac{1+3x+2x^2}{(1-2x)(1-x)(1+x)}=\frac{4}{1-2x}-\frac{3}{1-x}$