// Numbas version: finer_feedback_settings {"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": [{"functions": {}, "ungrouped_variables": ["coeff0", "coeff1", "a", "a1", "a2", "a3", "bcoeff0", "bcoeff1", "b", "b1", "b2", "b3", "ccoeff0", "ccoeff1", "c", "c1", "c2", "c3"], "name": "Partial Fractions - quadratic term in denominator", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "
Suppose we want to express
\n$\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})/((x^2+{coeff1}x+{coeff0})(x-{a}))}$
\nas the sum of its partial fractions.
\nWhen 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
\n$\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})/((x^2+{coeff1}x+{coeff0})(x-{a}))}=\\simplify{(Ax+B)/(x^2+{coeff1}x+{coeff0})+c/(x+{a})}$
\nWe multiply both sides by $\\simplify{((x^2+{coeff1}x+{coeff0})(x-{a}))}$ to give $\\simplify{({a1+a3}x^2+{a2-a*a1+a3*coeff1}x+{a3*coeff0-a2*a})}=\\simplify{(Ax+B)(x+{a})+C(x^2+{coeff1}x+{coeff0})}$
\nBy evaluating both sides at $\\simplify{x=-{a}}$ we can find C$=\\var{a3}$
\nThen by comparing coefficients we find A$=\\var{a1}$ and B$=\\var{a2}$
\n\nApply the same method for b) and c)
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