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<title>Remainder Theorem and Factor Theorem</title>
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<h1 align="center">Remainder Theorem<br />
and Factor Theorem</h1>
<p align="center"><i>Or: how to avoid Polynomial Long Division when finding factors</i></p>
<p>Do you remember doing division in Arithmetic?</p>
<p align="center" class="larger"><img src="../numbers/images/remainder-7-2.svg" alt="7/2=3 remainder 1" /></p>
<p align="center" class="larger"><i> &quot;7 divided by 2 equals <b>3</b> with a <b>remainder of 1</b>&quot;</i></p>
<p>Each part of the division has names:</p>
<p align="center"><span class="larger"><img src="../numbers/images/division-7d2.svg" alt="dividend/divisor=quotient with remainder" /></span></p>
<p>Which can be <b>rewritten</b> as a sum like this:</p>
<p align="center"><img src="../numbers/images/division-multiply.svg" alt="7 = 2 times 3 + 1" /></p>
<h2>Polynomials</h2>
<p>Well, we can also <a href="polynomials-division-long.html">divide polynomials</a>.</p>
<p align="center"><span class="large">f(x) &divide; d(x) = q(x) with a remainder of r(x)</span></p>
<p>But it is better to write it as a sum like this: </p>
<p align="center" class="large"><img src="images/polynomial-division-names.svg" alt="f(x) = d(x) times q(x) + r(x)" /></p>
<p>Like in this example using <a href="polynomials-division-long.html">Polynomial Long Division</a>:</p>
<div class="example">
<h3>Example: 2x<sup>2</sup>&minus;5x&minus;1 divided by x&minus;3</h3>
<ul>
<li>f(x) is 2x<sup>2</sup>&minus;5x&minus;1</li>
<li>d(x) is x&minus;3</li>
</ul>
<p align="center"><img src="images/polynomial-long-division2.gif" alt="polynomial long division 2x^/2-5x-1 / x-3 = 2x+1 R 2" width="274" height="160" /></p>
<p>After dividing we get the answer <span class="large">2x+1</span>, but there is a remainder of <span class="large">2</span>.</p>
<ul>
<li>q(x) is 2x+1</li>
<li>r(x) is 2</li>
</ul>
<p>In the style <span class="large">f(x) = d(x)&middot;q(x) + r(x)</span> we can write:</p>
<p align="center" class="large">2x<sup>2</sup>&minus;5x&minus;1 = (x&minus;3)(2x+1) + 2</p>
</div>
<p>But you need to know one more thing:</p>
<div class="def">
<p>The <a href="degree-expression.html">degree</a> of r(x) is always less than d(x)</p>
</div>
<p>Say we divide by a polynomial of <b>degree 1</b> (such as &quot;x&minus;3&quot;) the remainder will have <b>degree 0</b> (in other words a constant, like &quot;4&quot;).</p>
<p>We will use that idea in the
&quot;Remainder Theorem&quot;:</p>
<h2>The Remainder Theorem</h2>
<p>When we divide <span class="large">f(x)</span> by the simple polynomial <span class="large">x&minus;c</span> we get:</p>
<p align="center"> <span class="large">f(x) = (x&minus;c)&middot;q(x) + r(x)</span></p>
<p><span class="large">x&minus;c</span> is <b>degree 1</b>, so <span class="large">r(x)</span> must have <b>degree 0</b>, so it is just some constant <span class="large">r</span> <i>:</i></p>
<p align="center"><span class="large">f(x) = (x&minus;c)&middot;q(x) + <span class="hi">r</span></span></p>
<p>Now see what happens when we have <span class="large">x equal to c</span>:</p>
<div class="tbl">
<div class="row"><span class="left">f(c) =</span><span class="right">(c&minus;c)&middot;q(c) + r</span></div>
<div class="row"><span class="left">f(c) =</span><span class="right">(0)&middot;q(c) + r</span></div>
<div class="row"><span class="left">f(c) =</span><span class="right">r</span></div>
</div>
<p>So we get this:</p>
<div class="def">
<p><b>The Remainder Theorem:</b></p>
<p align="center">When we divide a polynomial <span class="large">f(x)</span> by <span class="large">x&minus;c</span> the remainder is <span class="large">f(c)</span></p>
</div>
<p>So to find the remainder after dividing by <span class="large">x-c</span> we don't need to do any division: </p>
<p align="center" class="larger">Just calculate <span class="large">f(c)</span>.</p>
<p>Let us see that in practice:</p>
<div class="example">
<h3>Example: The remainder after 2x<sup>2</sup>&minus;5x&minus;1 is divided by x&minus;3</h3>
<p>(Our example from above)</p>
<p>We don't need to divide by <b>(x&minus;3)</b> ... just calculate <b>f(3)</b>:</p>
<p align="center" class="larger">2(3)<sup>2</sup>&minus;5(3)&minus;1 = 2x9&minus;5x3&minus;1 <br>
= 18&minus;15&minus;1 <br>
= <b>2</b></p>
<p>And that is the remainder we got from our calculations above.</p>
<p>We didn't need to do Long Division at all!</p>
</div>
<div class="example">
<h3>Example: The remainder after 2x<sup>2</sup>&minus;5x&minus;1 is divided by x&minus;5</h3>
<p>Same example as above but this time we divide by &quot;x&minus;5&quot;</p>
<p>&quot;c&quot; is 5, so let us check f(5):</p>
<p align="center" class="larger">2(5)<sup>2</sup>&minus;5(5)&minus;1 = 2x25&minus;5x5&minus;1 <br>
= 50&minus;25&minus;1 <br>
= <b>24</b></p>
<p>The remainder is <b>24</b></p>
<p>Once again ... We didn't need to do Long Division to find that. </p>
</div>
<h2>The Factor Theorem</h2>
<p>Now ...</p>
<div class="center80">
<p>What if we calculate <b>f(c)</b> and it is <b>0</b>?</p>
<p align="center">... that means the <b>remainder is 0</b>, and ...</p>
<p align="right">... <b>(x&minus;c) must be a factor</b> of the polynomial!</p>
</div>
<p>We see this when dividing whole numbers. For example 60 &divide; 20 = 3 with no remainder. So 20 must be a factor of 60.
</p>
<div class="example">
<h3>Example: x<sup>2</sup>&minus;3x&minus;4</h3>
<p align="center" class="larger">f(4) = (4)<sup>2</sup>&minus;3(4)&minus;4 = 16&minus;12&minus;4 = 0</p>
<p>so (x&minus;4) must be a factor of x<sup>2</sup>&minus;3x&minus;4</p>
</div>
<p>And so we have:</p>
<div class="def">
<p><b>The Factor Theorem:</b></p>
<p align="center">When <span class="large">f(c)=0</span> then <span class="large">x&minus;c</span> is a factor of <span class="large">f(x)</span></p>
<p><i>And the other way around, too:</i></p>
<p align="center">When <span class="large">x&minus;c</span> is a factor of <span class="large">f(x)</span> then <span class="large">f(c)=0</span></p>
</div>
<h2>Why Is This Useful?</h2>
<p>Knowing that <span class="large">x&minus;c</span> is a factor is the same as knowing that <span class="large">c </span>is a root (and vice versa).</p>
<div class="center80">
<p>The <b>factor &quot;x&minus;c&quot;</b> and the <b>root &quot;c&quot;</b> are the same thing</p>
<p>Know one and we know the other</p>
</div>
<p>For one thing, it means that we can quickly check if (x&minus;c) is a factor of the polynomial.</p>
<div class="example">
<h3>Example: Find the factors of 2x<sup>3</sup>&minus;x<sup>2</sup>&minus;7x+2</h3>
<p>The polynomial is degree 3, and could be difficult to solve. So let us plot it first:</p>
<p align="center"><img src="images/graph-2x3mx2m7xp2.gif" alt="graph of 2x^3-x^2-7x+2" width="248" height="134" /></p>
<p>The curve crosses the x-axis at three points, and one of them <b>might be at 2</b>. We can check easily:</p>
<p align="center"><b>f(2)</b> = 2(2)<sup>3</sup>&minus;(2)<sup>2</sup>&minus;7(2)+2 <br>
= 16&minus;4&minus;14+2 <br>
= <b>0</b></p>
<p>Yes! <b>f(2)=0</b>, so we have found a root <b>and</b> a factor.</p>
<div class="center80">
<p align="center" class="large">So (x&minus;2) must be a factor of 2x<sup>3</sup>&minus;x<sup>2</sup>&minus;7x+2</p>
</div>
<p>&nbsp;</p>
<p>How about where it crosses near <b>&minus;1.8</b>?</p>
<p align="center"><b>f(&minus;1.8)</b> = 2(&minus;1.8)<sup>3</sup>&minus;(&minus;1.8)<sup>2</sup>&minus;7(&minus;1.8)+2 <br>
= &minus;11.664&minus;3.24+12.6+2 <br>
= <b>&minus;0.304</b></p>
<p>No, (x+1.8) is not a factor. We could try some other values near by and maybe get lucky. </p>
<p>But at least we know <span class="large">(x&minus;2)</span> is a factor, so let's use <a href="polynomials-division-long.html">Polynomial Long Division</a>:</p>
<div class="mono">
&nbsp;&nbsp;&nbsp;&nbsp;<span style="border-bottom: 1px solid black;">2x<sup>2</sup>+3x&minus;1&nbsp;&nbsp;&nbsp;&nbsp;</span><br>
x&minus;2)2x<sup>3</sup>&minus; x<sup>2</sup>&minus;7x+2<br>
&nbsp;&nbsp;&nbsp;&nbsp;<span style="border-bottom: 1px solid black;">2x<sup>3</sup>&minus;4x<sup>2</sup></span><br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3x<sup>2</sup>&minus;7x<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<span style="border-bottom: 1px solid black;">3x<sup>2</sup>&minus;6x</span><br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&minus;x+2<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<span style="border-bottom: 1px solid black;">&minus;x+2</span><br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0
</div>
<p>As expected the remainder is zero. </p>
<p>Better still, we are left with the <a href="quadratic-equation.html">quadratic equation</a> <b>2x<sup>2</sup>+3x&minus;1</b> which is easy to <a href="../quadratic-equation-solver.html">solve</a>.</p>
<p>It's roots are &minus;1.78... and 0.28..., so the final result is:</p>
<p class="center large">2x<sup>3</sup>&minus;x<sup>2</sup>&minus;7x+2 = (x&minus;2)(x+1.78...)(x&minus;0.28...)</p>
<p>We were able to solve a difficult polynomial.</p>
</div>
<h2>Summary</h2>
<div class="dotpoint">
<p><b>The Remainder Theorem:</b> </p>
<ul>
<li>When we divide a polynomial <span class="large">f(x)</span> by <span class="large">x&minus;c</span> the remainder is <span class="large">f(c)</span></li>
</ul>
</div>
<div class="dotpoint">
<p><b>The Factor Theorem:</b> </p>
<ul>
<li>When <span class="large">f(c)=0</span> then <span class="large">x&minus;c</span> is a factor of <span class="large">f(x)</span></li>
<li>When <span class="large">x&minus;c</span> is a factor of <span class="large">f(x)</span> then <span class="large">f(c)=0</span></li>
</ul>
</div><p>&nbsp;</p>
<div class="questions">
<script type="text/javascript">getQ(482, 483, 4014, 4015, 484, 485, 4016, 1124, 1125, 4017);</script>&nbsp;
<br />
Challenging Questions:
<a href="javascript:doQ(96)">1</a>
<a href="javascript:doQ(227)">2</a>
<a href="javascript:doQ(228)">3</a>
<a href="javascript:doQ(229)">4</a>
<a href="javascript:doQ(230)">5</a>
<a href="javascript:doQ(231)">6</a>
</div>
<div class="related"><a href="polynomials-division-long.html">Polynomial Long Division</a> <a href="index.html">Algebra Index</a></div>
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