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{"name": "AE6 Partial Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "AE6 Partial Fractions", "tags": [], "metadata": {"description": "<p>Rewrite the expression $\\frac{mx^2+nx+k}{(x+a)(x^2+bx+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{Bx+C}{x^2+bx+c}$.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p>Rewrite the following expression as partial fractions:</p>\n<p>\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))}. \\]</p>\n<p></p>", "advice": "<p>To express \\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} \\] as partial fractions, we want to set this equal to the sum of two fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x^2+{b}x+{c}}$. Since we have a linear factor and a quadratic factor, this tells us that the form of the partial fractions will be</p>\n<p>\\[&nbsp;\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\simplify{A/(x+{a}) + (B*x+C)/(x^2+{b}x+{c})},\\]</p>\n<p>where $A$, $B$, and $C$ are constants.</p>\n<p>To find the values of $A$, $B$, and $C$, we want to first multiply this equation by the denominator of the left-hand side. This gives</p>\n<p>\\[ \\simplify{{m}x^2+{n}x+{k}=A(x^2+{b}x+{c})+B*x(x+{a}) + C(x+{a})}.\\]</p>\n<p><strong>(Note: To find $A$, $B$, and $C$, we will use a combination of choosing suitable values of $x$ to eliminate terms, and equating coefficients. It can be solved by <em>only</em> equating coefficients, but this is a more efficient process.)</strong></p>\n<p></p>\n<p>To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:</p>\n<p>\\[ \\simplify{{m*a^2-n*a+k}=A{(a^2-b*a+c)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]</p>\n<p>To find $C$, we can eliminate $B$ by setting $x=0$ and substituting in the result of $A$:</p>\n<p>\\[ \\simplify{{k}={c}A+{a}C} \\implies C=\\simplify[all,fractionNumbers]{({k}-{c}A)/{a}}.\\]</p>\n<p>Hence,</p>\n<p>\\[ C = \\simplify[fractionNumbers]{{Csol}}.\\]</p>\n<p>Finally, by equating coefficients of the $x^2$-terms we can find $B$:</p>\n<p>\\[ (x^2): \\quad \\var{m} = \\simplify{A+B} \\implies B=\\var{m}-A. \\]</p>\n<p>Therefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]</p>\n<p>and</p>\n<p>{check}</p>\n<p><strong><a href=\"https://www.sheffield.ac.uk/mash/maths-resources/directory#P\" target=\"_blank\" rel=\"noopener\">Use this link to find some resources which will help you 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