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474 lines
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HTML
474 lines
21 KiB
HTML
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Partial Fractions</h1>
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<p class="center"><i>A way of "breaking apart" fractions with polynomials in them.</i></p>
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<h2>What are Partial Fractions?</h2>
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<p class="larger">We can do <i>this</i> directly:</p>
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<p class="center"><img src="images/partial-fractions-a.gif" alt="Partial Fractions" height="60" width="347"></p>
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<div class="example">
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<p>Like this:</p>
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<p class="center large"><span class="intbl"><em>2</em>
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<strong>x−2</strong></span> + <span class="intbl"><em>3</em>
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<strong>x+1</strong></span> = <span class="intbl">
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<em>2(x+1) + 3(x−2)</em>
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<strong>(x−2)(x + 1)</strong></span></p>
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<p>Which can be simplified using <a href="rational-expression-operations.html">Rational Expressions</a> to:</p>
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<p class="center large">= <span class="intbl">
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<em>2x+2 + 3x−6</em>
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<strong>x<sup>2</sup>+x−2x−2</strong></span></p>
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<p class="center large">= <span class="intbl">
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<em>5x−4</em>
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<strong>x<sup>2</sup>−x−2</strong></span></p>
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</div>
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<p> </p>
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<div class="center80">
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<p class="center large">... but how do we go in the opposite direction?</p>
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<p class="center large"><img src="images/partial-fractions-b.svg" alt="Partial Fractions"></p>
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</div>
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<p>That is what we are going to discover:</p>
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<p class="center larger">How to find the "parts" that make the single fraction <br>
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(the "<b>partial fractions</b>").</p>
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<h2>Why Do We Want Them?</h2>
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<p>First of all ... why do we want them?</p>
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<p class="center larger">Because the partial fractions are each <b>simpler</b>.</p>
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<p>This can help solve the more complicated fraction. For example it is very useful in <a href="../calculus/integration-introduction.html">Integral Calculus</a>.</p>
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<h2>Partial Fraction Decomposition</h2>
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<p>So let me show you how to do it.</p>
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<p class="center">The method is called <i>"Partial Fraction Decomposition"</i>, and goes like this:</p>
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<p><b>Step 1:</b> Factor the bottom</p>
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<p class="center"><img src="images/partial-fractions-1a.gif" alt="Partial Fractions" height="45" width="187"></p>
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<p><b>Step 2:</b> Write one partial fraction for each of those factors</p>
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<p class="center"><img src="images/partial-fractions-1b.gif" alt="Partial Fractions" height="47" width="242"></p>
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<p><b>Step 3:</b> Multiply through by the bottom so we no longer have fractions</p>
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<p class="center"><img src="images/partial-fractions-1c.gif" alt="Partial Fractions" height="29" width="244"></p>
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<p><b>Step 4:</b> Now find the constants A<sub>1</sub> and A<sub>2</sub></p>
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<p>Substituting the roots, or "zeros", of (x−2)(x+1) can help:</p>
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<p class="center"><img src="images/partial-fractions-1d.gif" alt="Partial Fractions" height="219" width="373"></p>
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<p><b>And we have our answer:</b></p>
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<p class="center"><img src="images/partial-fractions-1e.gif" alt="Partial Fractions" height="46" width="223"></p>
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<p> </p>
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<p> </p>
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<p>That was easy! <b><i>... almost too easy ... </i></b></p>
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<p class="center larger">... because it <b>can be a lot harder</b>!</p>
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<p>Now we go into detail on each step.</p>
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<h2>Proper Rational Expressions</h2>
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<p>Firstly, this only works for <b>Proper</b> Rational Expressions, where the degree of the top is <b>less than</b> the bottom.</p>
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<div class="def">
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<p class="center">The <b>degree</b> is the largest <b>exponent</b> the variable has.</p>
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</div>
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<ul class="larger">
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<li><span class="large">Proper</span>: the degree of the top is less than the degree of the bottom.
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr align="center">
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<td width="90">Proper:</td>
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<td width="90"><img src="images/rational-proper-2.gif" alt="x / (x^3 - 1)" height="46" width="66"></td>
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<td width="200">degree of top is 1<br>
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degree of bottom is 3</td>
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</tr>
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</tbody></table></li>
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<li><span class="large">Improper</span>: the degree of the top is greater than, or equal to, the degree of the bottom.
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr align="center">
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<td width="90">Improper:</td>
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<td width="90"><img src="images/rational-improper-1.gif" alt="(x^2 - 1) / (x + 1)" height="51" width="65"></td>
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<td width="200">degree of top is 2<br>
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degree of bottom is 1</td>
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</tr>
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</tbody></table></li>
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</ul>
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<p>If your expression is Improper, then do <a href="polynomials-division-long.html">polynomial long division</a> first.</p>
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<h2>Factoring the Bottom</h2>
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<p>It is up to you to factor the bottom polynomial. See <a href="factoring.html">Factoring in Algebra</a>.</p>
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<p>But don't factor them into <a href="../numbers/complex-numbers.html">complex numbers</a> ... you may need to stop some factors at quadratic (called irreducible quadratics because any further factoring leads to complex numbers):</p>
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<div class="example">
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<h3>Example: (x<sup>2</sup>−4)(x<sup>2</sup>+4)</h3>
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<ul>
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<div class="bigul">
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<li><b>x<sup>2</sup>−4</b> can be factored into (x−2)(x+2)</li>
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<li>But <b>x<sup>2</sup>+4</b> factors into complex numbers, so don't do it</li>
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</div>
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</ul>
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<p> </p>
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<p>So the best we can do is:</p>
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<p class="center large">(x−2)(x+2)(x<sup>2</sup>+4)</p>
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<p> </p>
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</div>
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<p>So the factors could be a combination of</p>
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<ul>
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<li>linear factors</li>
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<li>irreducible quadratic factors</li>
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</ul>
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<p><b>When you have a quadratic factor you need to include this partial fraction:</b></p>
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<p class="center large"><span class="intbl"><em>B<sub>1</sub>x + C<sub>1</sub></em><strong>(Your Quadratic)</strong></span></p>
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<h2>Factors with Exponents</h2>
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<p>Sometimes you may get a factor with an exponent, like <b>(x−2)<sup>3</sup></b> ...</p>
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<div class="center80">
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<p class="center larger">You need a partial fraction for each exponent from 1 up.</p>
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</div>
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<p>Like this:</p>
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<div class="example">
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<h3>Example:</h3>
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<p class="center large"><span class="intbl"><em>1</em><strong>(x−2)<sup>3</sup>
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</strong></span></p>
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<p>Has partial fractions</p>
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<p class="center large"><span class="intbl"><em>A<sub>1</sub></em><strong>x−2</strong></span> + <span class="intbl"><em>A<sub>2</sub></em><strong>(x−2)<sup>2</sup></strong></span> + <span class="intbl"><em>A<sub>3</sub></em>
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<strong>(x−2)<sup>3</sup></strong></span></p>
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</div>
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<p>The same thing can also happen to quadratics:</p>
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<div class="example">
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<h3>Example:</h3>
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<p class="center large"><span class="intbl"><em>1</em><strong>(x<sup>2</sup>+2x+3)<sup>2</sup></strong></span></p>
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<p>Has partial fractions:</p>
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<p class="center large"><span class="intbl"><em>B<sub>1</sub>x + C<sub>1</sub></em><strong>x<sup>2</sup>+2x+3</strong></span> + <span class="intbl"><em>B<sub>2</sub>x + C<sub>2</sub></em><strong>(x<sup>2</sup>+2x+3)<sup>2</sup></strong></span></p>
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</div>
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<p> </p>
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<h2>Sometimes Using Roots Does Not Solve It</h2>
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<p>Even after using the roots (zeros) of the bottom you can end up with unknown constants.</p>
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<p>So the next thing to do is:</p>
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<div class="center80">
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<p>Gather all powers of x together and then solve it as a <a href="systems-linear-equations.html">system of linear equations</a>.</p>
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</div>
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<p> </p>
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<p>Oh my gosh! That is a lot to handle! So, on with an example to help you understand:</p>
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<h2>A Big Example Bringing It All Together</h2>
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<p>Here is a nice big example for you!</p>
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<p class="center large"><span class="intbl"><em>x<sup>2</sup>+15</em><strong>(x+3)<sup>2 </sup>(x<sup>2</sup>+3)
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</strong></span></p>
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<ul>
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<div class="bigul">
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<li>Because <span class="large">(x+3)<sup>2</sup></span> has an exponent of 2, it needs two terms (<span class="large">A<sub>1</sub></span> and <span class="large">A<sub>2</sub></span>).</li>
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<li>And <span class="large">(x<sup>2</sup>+3)</span> is a quadratic, so it will need <span class="large">Bx + C</span>:</li>
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</div>
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</ul>
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<p class="center large"><span class="intbl"><em>x<sup>2</sup>+15</em><strong>(x+3)<sup>2</sup>(x<sup>2</sup>+3)</strong></span> = <span class="intbl"><em>A<sub>1</sub></em><strong>x+3</strong></span> + <span class="intbl"><em>A<sub>2</sub></em><strong>(x+3)<sup>2</sup></strong></span> + <span class="intbl"><em>Bx + C</em><strong>x<sup>2</sup>+3</strong></span></p>
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<p>Now multiply through by <b>(x+3)<sup>2</sup>(x<sup>2</sup>+3)</b>:</p>
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<p class="center"><span class="larger">x<sup>2</sup>+15 = (x+3)(x<sup>2</sup>+3)A<sub>1</sub> + (x<sup>2</sup>+3)A<sub>2</sub> + (x+3)<sup>2</sup>(Bx + C)</span></p>
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<p>There is a zero at <b>x = −3</b> (because x+3=0), so let us try that:</p>
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<p class="center"><span class="larger"><b>(−3)</b><sup>2</sup>+15 = 0 + (<b>(−3)</b><sup>2</sup>+3)A<sub>2</sub> + 0</span></p>
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<p>And simplify it to:</p>
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<p class="center"><span class="larger">24 = 12A<sub>2</sub></span></p>
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<p class="center">so<span class="large"> A<sub>2</sub>=2</span></p>
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<p>Let us replace <span class="large">A<sub>2</sub> </span>with<span class="large"> 2</span>:</p>
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<p class="center"><span class="larger">x<sup>2</sup>+15 = (x+3)(x<sup>2</sup>+3)A<sub>1</sub> + 2x<sup>2</sup>+6 + (x+3)<sup>2</sup>(Bx + C)</span></p>
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<p>Now expand the whole thing:</p>
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<p class="center"><span class="larger">x<sup>2</sup>+15 = (x<sup>3</sup>+3x+3x<sup>2</sup>+9)A<sub>1</sub> + 2x<sup>2</sup>+6 + (x<sup>3</sup>+6x<sup>2</sup>+9x)B + (x<sup>2</sup>+6x+9)C</span></p>
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<p>Gather powers of x together:</p>
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<p class="center"><span class="larger">x<sup>2</sup>+15 = x<sup>3</sup>(A<sub>1</sub>+B)+x<sup>2</sup>(3A<sub>1</sub>+6B+C+2)+x(3A<sub>1</sub>+9B+6C)+(9A<sub>1</sub>+6+9C)</span></p>
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<p>Separate the powers and write as a <a href="systems-linear-equations.html">Systems of Linear Equations</a>:</p>
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td align="right"><span class="large">x<sup>3</sup>:</span></td>
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<td align="right"> </td>
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<td align="right"><span class="larger"> 0 </span></td>
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<td align="center"><span class="larger"> = </span></td>
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<td><span class="larger">A<sub>1</sub>+B</span></td>
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</tr>
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<tr>
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<td align="right"><span class="large">x<sup>2</sup>:</span></td>
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<td align="right"> </td>
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<td align="right"><span class="larger"> 1 </span></td>
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<td align="center"><span class="larger"> = </span></td>
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<td><span class="larger">3A<sub>1</sub>+6B+C+2</span></td>
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</tr>
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<tr>
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<td align="right"><span class="large">x:</span></td>
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<td align="right"> </td>
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<td align="right"><span class="larger"> 0 </span></td>
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<td align="center"><span class="larger"> = </span></td>
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<td><span class="larger">3A<sub>1</sub>+9B+6C</span></td>
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</tr>
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<tr>
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<td align="right"><span class="large">Constants:</span></td>
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<td align="right"> </td>
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<td align="right"><span class="larger"> 15 </span></td>
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<td align="center"><span class="larger"> = </span></td>
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<td><span class="larger">9A<sub>1</sub>+6+9C</span></td>
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</tr>
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</tbody></table>
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<p>Simplify, and arrange neatly:</p>
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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|
<td class="larger" align="right"> 0 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">B</td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> −1 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">3A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">6B</td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">C</td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"><span class="larger"> 0 </span></td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">3A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">9B</td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">6C</td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> 1 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">C</td>
|
|
</tr>
|
|
</tbody></table>
|
|
<p><b>Now solve. </b></p>
|
|
<p>You can choose your own way to solve this ... I decided to subtract the 4th equation from the 2nd to begin with:</p>
|
|
<table style="border: 0; margin:auto;">
|
|
<tbody>
|
|
<tr>
|
|
<td class="larger" align="right"> 0 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">B</td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"><b> −2 </b></td>
|
|
<td class="larger" align="center" width="30"><b> = </b></td>
|
|
<td class="larger" align="right"><b>2A<sub>1</sub></b></td>
|
|
<td class="larger" align="center" width="30"><b>+</b></td>
|
|
<td class="larger" align="right"><b>6B</b></td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> 0 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">3A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">9B</td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">6C</td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> 1 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">C</td>
|
|
</tr>
|
|
</tbody></table>
|
|
<p>Then subtract 2 times the 1st equation from the 2nd:</p>
|
|
<table style="border: 0; margin:auto;">
|
|
<tbody>
|
|
<tr>
|
|
<td class="larger" align="right"> 0 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">B</td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"><b> −2 </b></td>
|
|
<td class="larger" align="center" width="30"><b> = </b></td>
|
|
<td class="larger" align="right"> </td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"><b>4B</b></td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> 0 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">3A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">9B</td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">6C</td>
|
|
</tr>
|
|
<tr>
|
|
<td class="larger" align="right"> 1 </td>
|
|
<td class="larger" align="center" width="30"> = </td>
|
|
<td class="larger" align="right">A<sub>1</sub></td>
|
|
<td class="larger" align="center" width="30"> </td>
|
|
<td class="larger" align="right"> </td>
|
|
<td class="larger" align="center" width="30">+</td>
|
|
<td class="larger" align="right">C</td>
|
|
</tr>
|
|
</tbody></table>
|
|
<p>Now I know that <b>B = −(1/2)</b>. </p>
|
|
<p>We are getting somewhere!</p>
|
|
<p>And from the 1st equation I can figure that <b>A<sub>1</sub> = +(1/2)</b>.</p>
|
|
<p>And from the 4th equation I can figure that <b>C = +(1/2)</b>.</p>
|
|
|
|
<p>Final Result:</p>
|
|
<table style="border: 0; margin:auto;">
|
|
<tbody>
|
|
<tr>
|
|
<td class="large" align="center" width="120">A<sub>1</sub>=1/2</td>
|
|
<td class="large" align="center" width="120">A<sub>2</sub>=2</td>
|
|
<td class="large" align="center" width="120">B=−(1/2)</td>
|
|
<td class="large" align="center" width="120">C=1/2</td>
|
|
</tr>
|
|
</tbody></table>
|
|
<p> </p>
|
|
<p>And we can now write our partial fractions:</p>
|
|
<p class="center large"><span class="intbl">
|
|
<em>x<sup>2</sup>+15</em>
|
|
<strong>(x+3)<sup>2</sup>(x<sup>2</sup>+3)</strong>
|
|
</span> = <span class="intbl">
|
|
<em>1</em>
|
|
<strong>2(x+3)</strong>
|
|
</span> + <span class="intbl">
|
|
<em>2</em>
|
|
<strong>(x+3)<sup>2</sup></strong>
|
|
</span> + <span class="intbl">
|
|
<em>−x + 1</em>
|
|
<strong>2(x<sup>2</sup>+3)</strong>
|
|
</span></p>
|
|
|
|
<p><b>Phew!</b> Lots of work. But it can be done.</p>
|
|
<p align="right"><i>(Side note: It took me <b>nearly <br>
|
|
an hour</b> to do this, because
|
|
I had to <br>
|
|
fix
|
|
2 silly mistakes along the way!)</i></p>
|
|
<h2>Summary</h2>
|
|
|
|
<ul class="larger">
|
|
<li>Start with a <b>Proper</b> Rational Expressions (if not, do division first)</li>
|
|
<li>Factor the bottom into:
|
|
|
|
|
|
|
|
<ul>
|
|
<li>linear factors</li>
|
|
<li>or "irreducible" quadratic factors</li>
|
|
</ul></li>
|
|
<li>Write out a partial fraction for each factor (and every exponent of each)</li>
|
|
<li>Multiply the whole equation by the bottom</li>
|
|
<li>Solve for the coefficients by
|
|
|
|
|
|
|
|
<ul>
|
|
<li>substituting zeros of the bottom</li>
|
|
<li>making a system of linear equations (of each power) and solving</li>
|
|
</ul></li>
|
|
<li>Write out your answer!</li>
|
|
</ul>
|
|
<div class="questions">
|
|
|
|
<script>getQ(8073, 8074, 8075, 8076, 8077, 8078, 1259, 612, 613, 1260);</script>
|
|
|
|
</div>
|
|
|
|
<div class="related">
|
|
<a href="index.html">Algebra Index</a>
|
|
<a href="index-2.html">Algebra 2 Index</a>
|
|
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