// Numbas version: finer_feedback_settings {"name": "Solving a monic quadratic by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Solving a monic quadratic by factorising", "tags": ["binomial", "Binomial", "factorisation", "Factorisation", "factorising", "factors", "Factors", "monic", "quadratic", "quadratics", "solving"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "
We aim to solve the quadratic $\\simplify{x^2+{linear}x+{const}}$ by factorisation.
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\n\nNote: In the gap above, enter the quadratic in factored form.
\n\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Since $(x+a)(x+b)=x^2+(a+b)x+ab$, when we are factorising a quadratic, such as $x^2+cx+d$, we must find the numbers $a$ and $b$ such that $c=a+b$ and $d=ab$.
\n\nIn the case of $\\simplify{x^2+{linear}x+{const}}$ we ask
\nwhat two numbers add to give $\\var{linear}$ and multiply to give $\\var{const}$?
\nTherefore the numbers must be $\\var{a}$ and $\\var{b}$, that is
\n$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).$
\nYou can check this by expanding the binomial product.
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"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The above factorisation and the null factor law implies $x=$ [[0]], [[1]].
\n\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Now, using the null factor law we have, either $\\simplify{x+{a}}=0$ or $\\simplify{x+{b}}=0$. In otherwords, either $x=\\var{-a}$ or $\\var{-b}$.
\n\n| \n | $\\simplify{x^2+{linear}x+{const}}$ | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n$ (\\simplify{x+{a}})(\\simplify{x+{b}})$ | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n$x$ | \n$=$ | \n$\\var{-a},\\var{-b}$ | \n