Binomial expansion with partial fractions

In a nutshell

Partial fractions can be used to break apart more complicated expressions into sums of simpler expressions, and hence help you find their expansions.

Finding the expansion of more complicated expressions

Given an expression of the form $$\dfrac{P(x)}{Q(x)}$$​ where the degree of $$P(x)$$​ is strictly less than that of $$Q(x)$$​, you can often use partial fractions to rewrite this expression as a sum of expressions of the form $$\dfrac{A}{(ax + b)}$$​ and $$\dfrac{A}{(ax + b)^2}$$​. The binomial expansion formula allows you to find expansions for all such terms (valid when $$|x| < \left|\dfrac{a}{b}\right|$$​ for values of $$n$$ other than are non-negative integers). The sum of these expansions gives the expansion for the original expression.

Example 1

Find the first three terms in the expansion of $$\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)}$$ and find the range of values of $$x$$ for which this expansion is valid.

Use partial fractions:

$$\begin{aligned}\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} &= \dfrac{A}{2-x} + \dfrac{B}{(2-x)^2} + \dfrac{C}{3 + 2x}\\\\7x^2 - 17x - 29&=(3+2x)(2-x)A + (3+2x)B + (2-x)^2C\\\\7x^2 - 17x - 29 &= (C - 2A)x^2 + (A + 2B -4C)x + (6A + 3B + 4C)\end{aligned}$$​​​

Compare coefficients to obtain the system of linear equations:

$$\begin{aligned}7 &= C - 2A\\-17 &= A + 2B - 4C\\-29 &= 6A + 3B + 4C\end{aligned}$$​​

Solve these to get the values:

$$\begin{aligned}A &= -3\\B &= -5\\C &= 1\end{aligned}$$​​

So, you have that:

​$$\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} = \dfrac{-3}{2-x} + \dfrac{-5}{(2-x)^2} + \dfrac{1}{3 + 2x}$$​​

The expansion of $$\dfrac{-3}{2-x} = -\dfrac32\left(1 - \dfrac{x}{2}\right)^{-1}$$ is:

​$$\begin{aligned}-\dfrac32\left(1 - \dfrac{x}{2}\right)^{-1} &= -\dfrac32\left[1 - \left(-\dfrac{x}{2}\right) + \dfrac{(-1)(-2)}{2!}\left(-\dfrac{x}{2}\right)^2 + \dots\right]\\&= -\dfrac32 - \dfrac{3x}{4} - \dfrac{3x^2}{8} + \dots\end{aligned}$$​​

And this is valid while $$|x| < 2$$.

​

The expansion of $$\dfrac{-5}{(2-x)^2} = -\dfrac54\left(1-\dfrac{x}{2}\right)^{-2}$$ is:

$$\begin{aligned}-\dfrac54\left(1 - \dfrac{x}{2}\right)^{-2} &= -\dfrac54\left[1 -2\left(-\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(-\dfrac{x}{2}\right)^2 + \dots\right]\\&= -\dfrac54 - \dfrac{5x}{4} - \dfrac{15x^2}{16} + \dots\end{aligned}$$​​

And this is valid while $$|x| < 2$$.

The expansion of $$\dfrac{1}{3 + 2x} = \dfrac13\left(1 + \dfrac{2x}{3}\right)^{-1}$$ is:

$$\begin{aligned}\dfrac13\left(1 + \dfrac{2x}{3}\right)^{-1} &= \dfrac13\left[1 - \left(\dfrac{2x}{3}\right) + \dfrac{(-1)(-2)}{2!}\left(\dfrac{2x}{3}\right)^2 + \dots\right]\\&= \dfrac13 - \dfrac{2x}{9} + \dfrac{4x^2}{27} + \dots\end{aligned}$$​​

And this is valid while $$|x| < \dfrac32$$.

Adding these together, you have that the first three terms in the expansion of $$\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)}$$​ are:

​$$-\dfrac{29}{12} - \dfrac{20x}{9} - \dfrac{503x^2}{432} + \dots$$​​

And this is only valid when all three other series have valid expansions, i.e. while $$|x| < \dfrac32$$.

Therefore, the first three terms in the expansion of $$\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)}$$ are $$\underline{-\dfrac{29}{12} - \dfrac{20x}{9} - \dfrac{503x^2}{432} + \dots}$$ and this is valid while $$\underline{|x| < \dfrac32}$$.