Given that $\\displaystyle{(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0}$. Determine the set of possible values of $x$.
\n$x=$ [[0]]
\nNote: if your answer is $1$ and $2$ input set(1,2)
\n", "showFeedbackIcon": true, "sortAnswers": false, "variableReplacementStrategy": "originalfirst", "stepsPenalty": "", "type": "gapfill", "variableReplacements": [], "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "checkingType": "absdiff", "checkVariableNames": false, "extendBaseMarkingAlgorithm": true, "expectedVariableNames": [], "failureRate": 1, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "vsetRangePoints": 5, "type": "jme", "checkingAccuracy": 0.001, "variableReplacements": [], "scripts": {}, "showPreview": true, "marks": 1, "answer": "{tworoots}", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "answerSimplification": "fractionNumbers", "unitTests": []}], "customMarkingAlgorithm": "", "steps": [{"marks": 0, "variableReplacements": [], "scripts": {}, "extendBaseMarkingAlgorithm": true, "type": "information", "showFeedbackIcon": true, "showCorrectAnswer": true, "prompt": "Notice, this is a quadratic that has already been factorised. You don't need to expand it since we have the null factor law:
\n\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]
\n\nSince
\n\\[(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0,\\]
\nthis means $\\simplify{x+{a}}=0$ or $\\simplify{{b}x+{c}}=0$. Solving each of these equations gives $x=\\var{-a}$ or $x=\\simplify{-{c}/{b}}$.
\nFor this question, we input our answer as set$\\left(\\var{-a},\\simplify{-{c}/{b}}\\right)$.
", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "unitTests": []}], "unitTests": []}, {"extendBaseMarkingAlgorithm": true, "prompt": "Solve $\\displaystyle{\\simplify{{l}a}(\\simplify{a+{d}})(\\simplify{{f}a+{g}})\\left(\\simplify{{h}a+{j}/{k}}\\right)\\left(\\simplify{{m}a/{n}+{p}/{q}}\\right)=0}$ for $a$.
\n$a=$ [[0]]
\nNote: if your answer is $1$, $2$ and $3$ input set(1,2,3)
\n", "showFeedbackIcon": true, "sortAnswers": false, "variableReplacementStrategy": "originalfirst", "stepsPenalty": 0, "type": "gapfill", "variableReplacements": [], "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "checkingType": "absdiff", "checkVariableNames": false, "extendBaseMarkingAlgorithm": true, "expectedVariableNames": [], "failureRate": 1, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "vsetRangePoints": 5, "type": "jme", "checkingAccuracy": 0.001, "variableReplacements": [], "scripts": {}, "showPreview": true, "marks": 1, "answer": "{fiveroots}", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "answerSimplification": "fractionNumbers", "unitTests": []}], "customMarkingAlgorithm": "", "steps": [{"marks": 0, "variableReplacements": [], "scripts": {}, "extendBaseMarkingAlgorithm": true, "type": "information", "showFeedbackIcon": true, "showCorrectAnswer": true, "prompt": "Notice, this expression that has already been factorised. You don't need to expand it since we have the null factor law:
\n\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]
\nSince
\n\\[\\simplify{{l}a}(\\simplify{a+{d}})(\\simplify{{f}a+{g}})\\left(\\simplify{{h}a+{j}/{k}}\\right)\\left(\\simplify{{m}a/{n}+{p}/{q}}\\right)=0,\\]
\nthis means $\\simplify{{l}a}=0$, $\\simplify{a+{d}}=0$, $\\simplify{{f}a+{g}}=0$, $\\simplify{{h}a+{j}/{k}}=0$, or $\\simplify{{m}a/{n}+{p}/{q}}=0$. Solving each of these equations gives $x=0$, $x=\\var{-d}$, $x=\\var[fractionnumbers]{-g/f}$, $x=\\var[fractionnumbers]{-j/(k*h)}$, or $x=\\var[fractionnumbers]{(-p*n)/(q*m)}$.
\nFor this question, we input our answer as set$\\left(0,\\var{-d},\\var[fractionnumbers]{-g/f},\\var[fractionnumbers]{-j/(k*h)},\\var[fractionnumbers]{(-p*n)/(q*m)}\\right)$.
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