## Rewriting the Reciprocal of a Quadratic Using the Binomial Expansion

We can use the binomial expansion to express
$f(x)= \frac{1}{2x^2+5x+2}$
as a sum of powers of
$x$
(a power series).
$\frac{1}{2x^2+5x+2}= \frac{A}{2+x}+ \frac{B}{1+2x} \rightarrow 1=A(1+2x)+B(2+x)$

If
$x= \frac{1}{2}$
then
$1=A(0)+B(\frac{3}{2}) \rightarrow B= \frac{2}{3}$
.
If
$x= -2$
then
$1=A(-3)+B(0) \rightarrow A= - \frac{1}{3}= \frac{2}{3}$
.
Then
$\frac{1}{2x^2+5x+2}= \frac{-1/3}{2+x}+ \frac{2/3}{1+2x}$
.
Using the general binomial expansion gives
\begin{aligned} \frac{-1/3}{2+x} &= \frac{-1}{6} \frac{1}{1+ \frac{x}{2}} \\ &= \frac{-1}{6} (1+ \frac{x}{2})^{-1} \\ &= \frac{-1}{6}(1+(-1) \frac{x}{2} + \frac{(-1)(-2)}{2!} (\frac{x}{2})^2+ \frac{(-1)(-2)(-3)}{3!} (\frac{x}{2})^3+... ) \\ &=-\frac{1}{6}+ \frac{x}{12}- \frac{x^2}{24}+ \frac{x^3}{48}+...\end{aligned}

\begin{aligned} \frac{2/3}{1+2x} &= \frac{2}{3} (1+2x)^{-1} \\ &= \frac{2}{3}(1+(-1)(2x) + \frac{(-1)(-2)}{2!} (2x)^2+ \frac{(-1)(-2)(-3)}{3!} (2x)^3+... ) \\ &=\frac{2}{3}- \frac{4x}{3}+ \frac{8x^2}{3}- \frac{16x^3}{3}+...\end{aligned}

$\frac{1}{2} - \frac{5x}{4} + \frac{21x^2}{8}- \frac{85x^3}{16}$