Answer
a.
$\displaystyle \frac{4+x}{(1+2x)(3-x)}=\frac{A}{1+2x}+\frac{B}{3-x}$
b.
$\displaystyle \frac{1-x}{x^{3}+x^{4}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{D}{x+1}$
Work Step by Step
a.
$\displaystyle \frac{4+x}{(1+2x)(3-x)}=\frac{A}{1+2x}+\frac{B}{3-x}$
b.
Factor the denominator.
$x^{3}+x^{4}=x^{3}(x+1)$
There are repeated factors, so
$\displaystyle \frac{1-x}{x^{3}+x^{4}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{D}{x+1}$
