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      <text>2.1.16 (Factor theorem)</text>
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      <text><![CDATA[<h1>Factor Theorem</h1>

<p>Building on the division process from the previous page, we have seen that we can take any polynomial \(P(x)\) and number \(a\) and write
\[ 
P(x) = (x-a)Q(x) +&nbsp; r, 
\]
where \(Q(x)\) is a polynomial with degree less than the degree of \(P(x)\), and \(r\) is a number. For example, 
\[ 
\frac{x^2+x-10}{x-3} = x+4 + \frac{2}{x-3} 
\]
so multiplying throughout by \(x-3\), we get
\[ 
x^2+x-10 = (x-3)(x+4) + 2 
\]
i.e. we have found \(Q(x)=x+4\) and \(r=2\).</p>

<p>Using this observation, we can establish two theorems which are useful when working with polynomials - the factor and remainder theorems.</p>
<br/>
<h3>Theorem</h3>

<p>\(x-a\) is a factor of \(P(x)\) if and only if \(P(a)=0\).</p>

<p><b>Proof</b></p><p><span>Because this is an "if and only if" statement, we need to prove both directions, (we will talk more about this in week 3).</span></p>
<ul>
<li>If \(P(a)=0\) then 
\[
\begin{eqnarray} 
P(x) &amp;=&amp; (x-a)Q(x)+P(a)
\\
&amp;=&amp; (x-a)Q(x). 
\end{eqnarray}
\] 
So we have written \(P(x)\) as a product of \(x-a\) and another polynomial, meaning \(x-a\) is a factor of \(P(x)\).</li>
<li>If \(x-a\) is a factor of&nbsp;\(P(x)\), this means we can write \(P(x)=(x-a)Q(x)\) for some polynomial \(Q(x)\).
Using this expression, \(P(a)=(a-a)Q(a)=0\times Q(a)=0\).</li>
</ul>
<p>Since we have proved both directions, this completes the proof.</p>
<br/>
<p>For the example above, the fact that \(P(-4)=2\) tells us that \(x+4\) is not a factor of \(P(x)\).</p>]]></text>
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