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Consider the quadratic $ \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.

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a

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Quadratic factorisation that does not rely upon pattern matching.

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What are the the two linear factors of this quadratic?

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[[0]] and [[1]].

", "showCorrectAnswer": true, "sortAnswers": false, "gaps": [{"variableReplacementStrategy": "originalfirst", "vsetRange": [0, 1], "notallowed": {"partialCredit": 0, "showStrings": true, "message": "", "strings": ["/"]}, "checkingType": "absdiff", "type": "jme", "showFeedbackIcon": true, "useCustomName": false, "customMarkingAlgorithm": "", "valuegenerators": [{"value": "", "name": "x"}], "variableReplacements": [], "scripts": {"mark": {"order": "instead", "script": "\nvar variables = this.question.scope.variables;\nvar unwrap = Numbas.jme.unwrapValue;\n\nvar a = unwrap(variables.a);\nvar b = unwrap(variables.b);\nvar c = unwrap(variables.c);\n\ntry {\n // get the student's answers to the two gaps\n var gap0 = this.parentPart.gaps[0].studentAnswer;\n \n if (!gap0) {\n return; \n }\n \n // there are two roots: -b/a and -c. Make sure there is exactly one root in gap0\n \n var root0 = -1*b/a;\n var root1 = -1*c;\n var correctanswer = a + \"*x*x + (\" +(b+c*a) + \")*x\" + b*c;\n var gap0root0 = (\"0\" == unwrap(this.question.scope.evaluate(gap0,{\"x\": root0})));\n var gap0root1 = (\"0\" == unwrap(this.question.scope.evaluate(gap0,{\"x\": root1})));\n \n // check that gap0 is linear by verifying that it increases by\n // the same constant amount from 10 to 11 and 11 to 12.\n var e10 = unwrap(this.question.scope.evaluate(gap0,{\"x\": 10}));\n var e11 = unwrap(this.question.scope.evaluate(gap0,{\"x\": 11}));\n var e12 = unwrap(this.question.scope.evaluate(gap0,{\"x\": 12}));\n if ((e11 == e12) || (e11 - e10 !== e12 - e11)) {\n this.setCredit(0, \"$\" + gap0 + \"$ is not linear. 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You can use the quadratic formula to deduce that $\\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ has roots:

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$ x = \\frac{\\simplify{-({a}*{c}+{b})}\\pm\\sqrt{ (\\var{a*c+b})^2 - 4\\times(\\var{a})\\times(\\var{b*c}) }}{2\\times \\var{a}} = \\var{-1*c} \\text{ or } \\displaystyle \\simplify{-1*{b}/{a}}.$

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The roots determine the factors, but only upto a constant. In general, a quadratic with roots $ \\var{-1*c}$ and $\\simplify{-1*{b}/{a}}$ has the form:

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$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c})$

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for some constant term $C$. The only thing left to do is determine the value of the constant which makes:

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$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c}) = \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.

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Equating the coefficients of the $x^2$ terms in the left and right hand sides shows that $C=\\var{a}$. So

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$ (\\var{a}x + \\var{b}) \\times (x-\\var{-1*c}) = \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.

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