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      <text>9.3.7 (Binomial theorem with partial fractions)</text>
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      <text><![CDATA[<h1>Partial fractions</h1>

<p>We saw in the previous section, that we can use a version of the Binomial Theorem with negative powers. This means that we can consider expansions of the form \(\frac{1}{(a+b)^n}\). One way that such expressions naturally arise is when performing <i>partial fraction decompositions</i>. In this section we will consider partial fractions and how we can perform such decompositions.</p>
<br/>
<p>We can sum rational functions by first expressing them with a common denominator, just as we do with numbers e.g. :
\[
\frac{2}{x-3} + \frac{3}{x-1} = \frac{2(x-1) + 3(x-1)}{(x-3)(x-1)} = \frac{2x -2 + 3x -9}{(x-3)(x-1)} = \frac{5x-11}{(x-3)(x-1)}
\]
The idea, of partial fractions is to reverse this process. Essentially by undoing the steps above, we can write \(\dfrac{5x-11}{(x-3)(x-1)}\) as a sum of terms of the form \(\dfrac{A}{x-3}\) and \(\dfrac{B}{x-1}\). This is known as the <b>partial fraction decomposition</b> of the rational function. We will see how to do this for other rational functions of a similar type, as shown in the following example.</p>
<br/>
<h3>Example</h3>

<p>Find the partial fraction decomposition of \(\dfrac{4x+8}{(x+5)(x+1)}\).</p>

<h3>Solution</h3>

<p>We wish to write the given fraction in the form
\[
\dfrac{4x+8}{(x+5)(x+1)} = \frac{A}{x+5} + \frac{B}{x+1}.
\]
The idea is to manipulate the right hand side of this equation in order to find the values of \(A\) and \(B\). Firstly, we can cross multiply the right hand side to write it as one fraction to obtain
\[
\dfrac{4x+8}{(x+5)(x+1)} = \frac{A}{x+5} + \frac{B}{x+1} =\frac{A(x+1) + B(x+5)}{(x+5)(x+1)}.
\]
Now the left hand side and the right hand side have the same denominator, so in order for them to be equal, the numerators must be the same, i.e. \(4x +8 = A(x+1)+B(x+5)\). In order to find \(A\) and \(B\) we can either equate the coefficients of \(x\) as well as the constant terms, or we can notice that substituting \(x=-1\) and \(x=-5\) into the rhs gets rid of the \(A\) and \(B\) terms respectively
\[
\begin{align*}
&x = -1: \:\:\:\:\:\: 4B = 4 \implies B = 1
\\
&x = -5: \:\:\:\:\:\: -4A = -12 \implies A = 3.
\end{align*}
\]
Hence the solution is 
\[
\dfrac{4x+8}{(x+5)(x+1)} =  \frac{3}{x+5} + \frac{1}{x+1}.
\]</p>]]></text>
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