// Numbas version: finer_feedback_settings
{"name": "Partial Fractions: Proper Fractions 6a", "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 6a", "tags": [], "metadata": {"description": "<p>Rewrite the expression $\\frac{nx+k}{(x+a)(x+b)(x+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{B}{x+b}+\\frac{C}{x+c}$.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p>Rewrite the following expression as partial fractions:</p>\n<p>\\[ \\simplify{({n}x+{k})/((x+{a})(x+{b})(x+{c}))}. \\]</p>", "advice": "<p>To express \\[ \\simplify{({n}x+{k})/((x+{a})(x+{b})(x+{c}))} \\] as partial fractions, we want to set this equal to the sum of three fractions with denominators $\\simplify{x+{a}}$, $\\simplify{x+{b}}$, and $\\simplify{x+{c}}$. Since these are all distinct linear factors, this tells us that the numerators will be constants, which we will call $A$, $B$, and $C$:</p>\n<p>\\[&nbsp;\\simplify{({n}x+{k})/((x+{a})(x+{b})(x+{c}))} = \\simplify{A/(x+{a}) + B/(x+{b})+ C/(x+{c})}.\\]</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{{n}x+{k}=A(x+{b})(x+{c})+B(x+{a})(x+{c}) + C(x+{a})(x+{b})}.\\]</p>\n<p>There are two methods of finding $A$, $B$, and $C$. The first is to choose suitable values for $x$ which will eliminate two of the terms, and the other is to compare the coefficients of each side of the equation. We will cover both methods here.</p>\n<p><strong>Method 1:</strong></p>\n<p>To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:</p>\n<p>\\[ \\simplify{{k}-{n*a}=A{(b-a)(c-a)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]</p>\n<p>Finding $B$ by setting $x=\\var{-b}$:</p>\n<p>\\[ \\simplify{{k}-{n*b}=B{(a-b)(c-b)}} \\implies B=\\simplify[fractionNumbers]{{Bsol}}.\\]</p>\n<p>Finally, setting $x=\\var{-c}$ we can find C:</p>\n<p>\\[ \\simplify{{k}-{n*c}=C{(a-c)(b-c)}} \\implies C=\\simplify[fractionNumbers]{{Csol}}.\\]</p>\n<p>Therefore,</p>\n<p>{check}</p>\n<p></p>\n<p><strong>Method 2:</strong></p>\n<p>By comparing the coefficients of the $x^2$-terms, $x$-terms and the constant terms we can form a set of simultaneous equations to find $A$, $B$ and $C$.</p>\n<p>\\[ \\begin{split} \\simplify{{n}x+{k}} &amp;\\,= \\simplify{A(x+{b})(x+{c})+B(x+{a})(x+{c}) + C(x+{a})(x+{b})} \\\\ &amp;\\,= \\simplify{A(x^2+{b+c}x+{b*c})+B(x^2+{a+c}x+{a*c})+C(x^2+{a+b}x+{a*b})} \\\\ &amp;\\,=\\simplify{(A+B+C)x^2+({b+c}A+{a+c}B+{a+b}C)x+{b*c}A+{a*c}B+{a*b}C} . \\end{split} \\]</p>\n<p>\\[ \\begin{split}(x^2)&amp;: \\quad 0 &amp;\\,= \\simplify{A+B+C} \\\\ (x)&amp;: \\quad \\var{n} &amp;\\,= \\simplify{{b+c}A+{a+c}B+{a+b}C} \\\\ (c)&amp;:\\quad \\var{k} &amp;\\,= \\simplify{{b*c}A+{a*c}B+{a*b}C} .\\end{split} \\]</p>\n<p>Hence,</p>\n<p>\\[A=\\simplify[fractionNumbers]{{Asol}},\\,B=\\simplify[fractionNumbers]{{Bsol}}, \\, C=\\simplify[fractionNumbers]{{Csol}}, \\]</p>\n<p>and</p>\n<p>{check}</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(-6..6)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(-6..6 except a)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": 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