// Numbas version: exam_results_page_options {"name": "Partial Fractions: Quadratic Term in Denominator", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "name": "Partial Fractions: Quadratic Term in Denominator", "tags": [], "functions": {}, "statement": "", "advice": "

Suppose we want to express 

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$\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})/((x^2+{coeff1}x+{coeff0})(x-{a}))}$

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as the sum of its partial fractions.

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When the denominator contains a quadratic factor we have to consider the possibilty that the numebrator can contain a term in $x$. This is because if it did, the numerator would still be of lower degree than the denominator - this would be a proper fraction. So we write 

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$\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})/((x^2+{coeff1}x+{coeff0})(x-{a}))}=\\simplify{(Ax+B)/(x^2+{coeff1}x+{coeff0})+c/(x+{a})}$

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We multiply both sides by $\\simplify{((x^2+{coeff1}x+{coeff0})(x-{a}))}$ to give $\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})}=\\simplify{(Ax+B)(x+{a})+C(x^2+{coeff1}x+{coeff0})}$

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By evaluating both sides at $\\simplify{x=-{a}}$ we can find C$=\\var{a3}$

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Then by comparing coefficients we find A$=\\var{a1}$ and B$=\\var{a2}$

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Apply the same method for b) and c)

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Please split into partial fractions

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Express the following as a sum of partial fractions

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$\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})/((x^2+{coeff1}x+{coeff0})(x-{a}))}$

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[[0]]

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Express the following as a sum of partial fractions

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$\\simplify{({b1+b3}x^2+{b2-b*b1+b3*bcoeff1}x+{b3*bcoeff0-b2*b})/((x^2+{bcoeff1}x+{bcoeff0})(x-{b}))}$

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[[0]]

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Express the following as a sum of partial fractions

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$\\simplify{({c1+c3}x^2+{c2-c*c1+c3*ccoeff1}x+{c3*ccoeff0-c2*c})/((x^2+{ccoeff1}x+{ccoeff0})(x-{c}))}$

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[[0]]

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