Denominators with linear factors, repeated factors and non-factorisable quadratics.
", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", ""], "variable_overrides": [[], [], [], []], "questions": [{"name": "Partial Fractions: Proper Fractions 4a", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "Rewrite the expression $\\frac{cx+d}{(kx+a)(x+b)}$ as partial fractions in the form $\\frac{A}{kx+a}+\\frac{B}{x+b}$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{({c}x+{d})/(({k}x+{a})(x+{b}))} .\\]
", "advice": "To express \\[ \\simplify{({c}x+{d})/(({k}x+{a})(x+{b}))} \\] as partial fractions, 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}x+{d})/(({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}x+{d}=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]{{d-c*a/k}=A{b-a/k}} \\implies \\simplify[fractionNumbers]{A={(d-c*a/k)/(b-a/k)}}.\\]
\nSimilarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:
\n\\[ \\simplify{{d-c*b}=B{a-k*b}} \\implies \\simplify[fractionNumbers]{B={(d-c*b)/(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}x+{d}} &\\,= \\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} \\]
\n\\[ \\begin{split}&(x):\\quad \\var{c} &\\,= \\simplify{A+{k}*B} \\\\ &(c):\\quad \\var{d} &\\,= \\simplify{{b}A+{a}B} .\\end{split} \\]
\nHence,
\n\\[A=\\simplify[fractionNumbers]{{Asol}},\\,B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{(k*d-c*a)}/(({k*b-a})({k}x+{a}))+{d-c*b}/(({a-k*b})(x+{b}))", "answerSimplification": "all", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": " `+-$n`?/($n`?*($n`?x+`+-$n`?))+`+-$n`?/($n`?*($n`?x+`+-$n`?)) `|`+-$n/`+-$n*(`+-$n`?/($n`?*($n`?x+`+-$n`?))+`+-$n`?/($n`?*($n`?x+`+-$n`?)))", "partialCredit": 0, "message": "", "nameToCompare": ""}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Partial Fractions: Proper Fractions 5b", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "Rewrite the expression $\\frac{n}{(x+a)(x+b)^2}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{B}{x+b}+\\frac{C}{(x+b)^2}$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{{n}/((x+{a})(x+{b})^2)}. \\]
", "advice": "To express \\[ \\simplify{{n}/((x+{a})(x+{b})^2)} \\] as partial fractions, we want to set this equal to the sum of three fractions with denominators $\\simplify{x+{a}}$, $\\simplify{x+{b}}$, and $\\simplify{(x+{b})^2}$. Since we have a distinct linear factor and a repeated linear factor, this tells us that the numerators will be constants, which we will call $A$, $B$, and $C$:
\n\\[ \\simplify{{n}/((x+{a})(x+{b})^2)} = \\simplify{A/(x+{a}) + B/(x+{b})+ C/(x+{b})^2}.\\]
\nTo 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
\n\\[ \\simplify{{n}=A(x+{b})^2+B(x+{a})(x+{b}) + C(x+{a})}.\\]
\n(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 only equating coefficients, but this is a more efficient process.)
\n\nTo find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:
\n\\[ \\simplify{{n}=A{(b-a)^2}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]
\nTo find $C$, we can eliminate $A$ and $B$ by setting $x=\\var{-b}$:
\n\\[ \\simplify{{n}=C{(a-b)}} \\implies C=\\simplify[fractionNumbers]{{Csol}}.\\]
\nFinally, by equating coefficients of the $x^2$-terms we can find $B$:
\n\\[ (x^2): \\quad 0 = \\simplify{A+B} \\implies B=-A. \\]
\nTherefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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\\\\[ \\\\simplify{{n}/((x+{a})(x+{b})^2)} = \\\\simplify{{Asol}/(x+{a})+{Bsol}/(x+{b})+{Csol}/(x+{b})^2}.\\\\]
\"", "description": "", "templateType": "long string", "can_override": false}, "sol2": {"name": "sol2", "group": "Ungrouped variables", "definition": "\"\\\\[ \\\\simplify{{n}/((x+{a})(x+{b})^2)} = \\\\simplify[all,fractionNumbers]{{n}/({(b-a)^2}(x+{a}))-{n}/({(b-a)^2}(x+{b}))+{n}/({(a-b)}(x+{b})^2)}.\\\\]
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))}. \\]
\n", "advice": "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
\n\\[ \\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})},\\]
\nwhere $A$, $B$, and $C$ are constants.
\nTo 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
\n\\[ \\simplify{{m}x^2+{n}x+{k}=A(x^2+{b}x+{c})+B*x(x+{a}) + C(x+{a})}.\\]
\n(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 only equating coefficients, but this is a more efficient process.)
\n\nTo find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:
\n\\[ \\simplify{{m*a^2-n*a+k}=A{(a^2-b*a+c)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]
\nTo find $C$, we can eliminate $B$ by setting $x=0$ and substituting in the result of $A$:
\n\\[ \\simplify{{k}={c}A+{a}C} \\implies C=\\simplify[all,fractionNumbers]{({k}-{c}A)/{a}}.\\]
\nHence,
\n\\[ C = \\simplify[fractionNumbers]{{Csol}}.\\]
\nFinally, by equating coefficients of the $x^2$-terms we can find $B$:
\n\\[ (x^2): \\quad \\var{m} = \\simplify{A+B} \\implies B=\\var{m}-A. \\]
\nTherefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify{{Asol}/(x+{a})+({Bsol}x+{Csol})/(x^2+{b}x+{c})}.\\\\]
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following quadratic expression in the form $(x+a)^2+b$ by using the method of completing the square:
\n\\[ \\simplify[unitFactor]{{a}x^2+{b}x+{c}} \\]
", "advice": "For a quadratic expression of the form \\[ x^2+bx+c,\\] when written in completed square form it becomes \\[ \\left(x+\\frac{b}{2}\\right)^2 - \\left(\\frac{b}{2}\\right)^2+c.\\]
\nSo, \\[ \\begin{split} \\simplify[unitFactor]{{a}x^2+{b}x+{c}} &\\,=\\simplify[unitFactor]{(x+{b/2})^2 - {(b/2)}^2 + {c}}, \\\\ \\\\&\\,=\\simplify[unitFactor]{(x+{b/2})^2 +{c-(b/2)^2}}. \\end{split} \\]
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