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      <text>2.1.5 (Division of polynomials)</text>
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      <text><![CDATA[<h1>Polynomial Division</h1>

<p>Factorising large numbers can make them simpler to understand, and the same thing is true with polynomials:</p><table style="text-align: center; margin-left:auto; margin-right:auto;" cellpadding="10">
<caption></caption>
<thead>
<tr>
<th scope="col">Expanded form</th>
<th scope="col">&nbsp;</th><th scope="col">Factorised form</th>
</tr>
</thead>
<tbody>
<tr>
<td>\(3417652\)</td>
<td>&nbsp;=</td><td>&nbsp; \(2^2\cdot7^3\cdot47\cdot53\)&nbsp;&nbsp;</td>
</tr>
<tr>
<td>&nbsp; \(x^3+3x^2-6x-8\)&nbsp;&nbsp;</td>
<td>&nbsp;=</td><td>\( (x+1)(x-2)(x+4) \)</td>
</tr>
</tbody>
</table>
<p>A key tool we use when factorising numbers is division (e.g. we spot that 2 is a factor, then divide by 2 to see what the other factor is), so we need to be able to do this with polynomials too.</p>
<br/>
<p>Dividing numbers is familiar, but recall that it is based on facts about multiplication, e.g. \[ 4\times3=12 \qquad \text{so} \qquad \dfrac{12}{3}=4\]</p>
<p>We can do the same sort of thing with polynomials, e.g.</p><p>\[ (x+4)(x-3)=x^2+x-12 \qquad \text{so} \qquad \dfrac{x^2+x-12}{x-3}=x+4. \]</p>

<p>Recall that if the numbers don't divide exactly we will get a "remainder", e.g.
\[ 
\dfrac{13}{3}=\dfrac{12+1}{3}=\dfrac{12}{3}+\dfrac{1}{3}=4+\dfrac13 \quad&nbsp; \text{i.e. }\unicode{x201C}\text{4 remainder 1''}
\]
We can use a similar process with polynomials.</p>
<p>A function which is the division of two polynomials is called a <b>rational function</b>, e.g. 
\[
\frac{x^2+x+1}{x-5}, \qquad \frac{x^3+x+1}{x-2} \qquad \frac{x^5+x^3=5x^2+1}{x^2+3}
\]
are all rational functions. Often dividing a polynomial \(P(x)\) by another is only useful if the degree of the denominator is less than or equal to the degree of the numerator. </p>
<br/>
<h3>Example</h3>

<p>Divide the polynomial \(2x^3 + 7x^2 + 10x +15\) by \(x+2\).</p>

<h3>Solution</h3>

<p>Note that \(2x^3 + 7x^2 + 10x +15 = (2x^2 +3x+4)(x+2)+7\). So, to divide by \(x+2\) we have
\[
\begin{align*}
\frac{2x^3 + 7x^2 + 10x +15}{x+2} &= \frac{(2x^2 +3x+4)(x+2)+7}{x+2}
\\
&= \frac{(2x^2 +3x+4)(x+2)}{x+2} + \frac{7}{x+2}
\\
&= 2x^2 +3x+4 + \frac{7}{x+2}.
\end{align*}
\]
We have now successfully performed the division, notice that the fraction \(\frac{7}{x+2}\) tells us that \(7\) is the remainder on performing this division. We can use this example to say something more generally about dividing by <i>linear</i> functions.</p>
<br/>
<p>If the polynomial \(P(x)\) is divided by the linear function \(ax+b\) until a constant remainder \(r\) is obtained then
\[
\frac{P(x)}{ax+b} = Q(x)+\frac{r}{ax+b}.
\]
Here, we call \(ax + b\) the <b>divisor</b> and it is often denoted \(D(x)\), \(Q(x)\) is called the <b>quotient</b> and \(r\) is the <b>remainder</b>. Notice that \(P(x) = Q(x) \cdot (ax+b) + r\).
</p>]]></text>
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