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      <text>9.3.8 (Proof of partial fractions for two linear factors)</text>
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      <text><![CDATA[<p>In order to be able to use partial fractions in this way, we need to know that such a decomposition always exists. Fortunately we can prove this as follows.</p>
<br/>
<p>Suppose we want to write the partial fractions decomposition for \(\dfrac{mx+c}{(x-a)(x-b)}\).</p>
<br/>
<p><b>Lemma:</b> We can always find \(A\) and \(B\) such that&nbsp;\(\dfrac{mx+c}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\).</p>
<br/>
<p><b>Proof:</b></p>
<p>Combining the right-hand side into a single term, we need:
\[
\dfrac{mx+c}{(x-a)(x-b)}=\dfrac{A(x-b)+B(x-a)}{(x-a)(x-b)}=\dfrac{(A+B)x-Ab-Ba}{(x-a)(x-b)}
\]
so by comparing the numerators we see that \(A\) and \(B\) must satisfy
\[
\begin{eqnarray*} 
A+B&amp;=&amp;m \\ 
-Ab-Ba&amp;=&amp;c 
\end{eqnarray*}
\]
Solving this system of equations, we find
\[ 
A= -\dfrac{am+c}{b-a},\quad B=\dfrac{bm+c}{b-a}
\]
So we can always find&nbsp;\(A\) and \(B\) to make this work, provided that \(a-b\neq0\).</p>
<br/>
<p>Note that this proof tells us nothing about the case where \(a=b\), that is, w e have only shown, so far that a partial fraction decomposition exists for a fraction with two linear, distinct factors in its denominator. In fact, we can form partial fractions for a wider range of rational functions, which we will see shortly.</p>]]></text>
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