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      <text>3.6.5.1 Improper fractions</text>
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      <text><![CDATA[<h4>Improper fractions</h4>
<br>
<p>When calculating the partial fractions of improper fractions an extra polynomial is added to any partial fractions that would normally arise. The added polynomial has degree <span class="math inline">\(n-d\)</span> where <span class="math inline">\(n\)</span>    is the degree of the numerator and <span class="math inline">\(d\)</span> is the degree of the denominator. Recall that</p>
<ul>
    <li> a polynomial of degree 0 is a constant, <span class="math inline">\(A\)</span> say, </li>
    <li>a polynomial of degree 1 has the form <span class="math inline">\(Ax+B\)</span>, </li>
    <li>a polynomial of degree 2 has the form <span class="math inline">\(Ax^2+Bx+C\)</span></li>
</ul>
<p>and so on. If, for example, the improper fraction is such that the numerator has degree 5 and the denominator has degree 3, then <span class="math inline">\(n-d=2\)</span>, and we need to add a polynomial of the form <span class="math inline">\(Ax^2+Bx+C\)</span>.</p>
<br>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>If a fraction is improper an additional term is included taking the form of a polynomial of degree \(n-d\), where \(n\) is the degree of the numerator and \(d\) is the degree of the denominator.</p>
</div>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Express as partial fractions</p>
<p>\[\dfrac{2x^2-x-2}{x+1}\]</p>
<h4 class="HELM_solution">Solution</h4>
<p>The fraction is improper because <span class="math inline">\(n=2\)</span>, <span class="math inline">\(d=1\)</span> and so <span class="math inline">\(d\le n\)</span>. Here <span class="math inline">\(n-d=1\)</span>, so we need to include as an extra
    term a polynomial of the form <span class="math inline">\(Bx+C\)</span>, in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction <span class="math inline">\(\dfrac{A}{x+1}\)</span>. So altogether
    we have </p>
<p>\[\dfrac{2x^2-x-2}{x+1}= \dfrac{A}{x+1}+ (Bx+C).\]</p>
<p> Multiplying both sides by <span class="math inline">\(x+1\)</span> we find</p>
<p>\[2x^2-x-2 = A+(Bx+C)(x+1) = Bx^2 + (C+B)x + (C+A).\]</p>
<p> Equating coefficients of <span class="math inline">\(x^2\)</span> gives</p>
<p>\[B=2.\]</p>
<p>Equating coefficients of <span class="math inline">\(x\)</span> gives</p>
<p>\[-1=C+B\]</p>
<p> and so </p>
<p>\[C=-1-B=-3.\]</p>
<p>Equating the constant terms gives</p>
<p>\[-2=C+A\]</p>
<p> and so</p>
<p>\[A=-2-C=-2-(-3)=1.\]</p>
<p>Finally, we have </p>
<p>\[\dfrac{2x^2-x-2}{x+1}= \dfrac{1}{x+1}+2x-3.\]</p>]]></text>
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