// Numbas version: finer_feedback_settings {"name": "Partial Fractions: Proper Fractions 2b", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Partial Fractions: Proper Fractions 2b", "tags": [], "metadata": {"description": "
Rewrite the expression $\\frac{c}{kx^2+mx+n}$ as partial fractions in the form $\\frac{A}{kx+a}+\\frac{B}{x+b}$, where the quadratic $kx^2+mx+n=(kx+a)(x+b)$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{{c}/(({k}x^2+{a+k*b}x+{a*b}))} .\\]
", "advice": "To express\\[\\simplify{{c}/(({k}x^2+{a+k*b}x+{a*b}))} \\] as partial fractions, we must first factorise the denominator into linear factors:
\n\\[\\simplify{{c}/(({k}x^2+{a+k*b}x+{a*b}))}=\\simplify{{c}/(({k}x+{a})(x+{b}))}.\\]
\nNow the denominator is factorised, we want to set this equal to the sum of 2 fractions with denominators $\\simplify{{k}x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:
\n\\[ \\simplify{{c}/(({k}x+{a})(x+{b}))} = \\simplify{A/({k}x+{a}) + B/(x+{b})}.\\]
\nTo find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives
\n\\[ \\simplify{{c}=A(x+{b})+B({k}x+{a})}.\\]
\nThere are 2 methods of finding $A$ and $B$. The first is to choose suitable values for $x$ which will eliminate one of the terms, and the other is to compare the coefficients of each side of the equation. We will cover both methods here.
\nMethod 1:
\nTo find $A$, we can eliminate $B$ by setting $\\simplify[fractionNumbers]{x={-a/k}}$:
\n\\[ \\simplify[fractionNumbers]{{c}=A{b-a/k}} \\implies \\simplify[fractionNumbers]{A={c/(b-a/k)}}.\\]
\nSimilarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:
\n\\[ \\simplify{{c}=B{a-k*b}} \\implies \\simplify[fractionNumbers]{B={c/(a-k*b)}}.\\]
\nTherefore,
\n{check}
\nMethod 2:
\nBy comparing the coefficients of the $x$-terms and the constant terms we can form a pair of simultaneous equations to find $A$ and $B$.
\n\\[ \\begin{split} \\simplify{{c}}&\\,=\\simplify{A(x+{b})+B({k}x+{a})} \\\\ &\\,= \\simplify{A*x+{b}A+{k}B*x+{a}B} \\\\ &\\,= \\simplify{(A+{k}B)x+({b}A+{a}B)}\\end{split}\\]
\nComparing coefficients:
\n\\[ \\begin{split}&(x):\\quad 0 &\\,= \\simplify{A+{k}B} \\\\ &(c):\\quad \\var{c} &\\,= \\simplify{{b}A+{a}B} \\end{split} \\]
\nHence,
\n\\[A=\\simplify[fractionNumbers]{{Asol}},\\,B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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