// Numbas version: exam_results_page_options {"name": "Null factor law", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "c", "tworoots", "blist", "d", "f", "g", "h", "j", "k", "l", "plist", "m", "n", "p", "q", "fiveroots"], "name": "Null factor law", "tags": ["factorisation", "Factorisation", "factorised", "null factor law", "product", "solving", "zero divisor"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": "1", "prompt": "

Given that $\\displaystyle{(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0}$. Determine the set of possible values of $x$.

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$x=$ [[0]]

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Note: if your answer is $1$ and $2$ input set(1,2)

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Notice, this is a quadratic that has already been factorised. You don't need to expand it since we have the null factor law:

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\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]

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Since

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\\[(\\simplify{x+{a}})(\\simplify{{b}x+{c}})=0,\\]

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this means $\\simplify{x+{a}}=0$ or $\\simplify{{b}x+{c}}=0$. Solving each of these equations gives $x=\\var{-a}$ or $x=\\simplify{-{c}/{b}}$.

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For this question, we input our answer as set$\\left(\\var{-a},\\simplify{-{c}/{b}}\\right)$.

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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 $a$.

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$a=$ [[0]]

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Note: if your answer is $1$, $2$ and $3$ input set(1,2,3)

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Notice, this expression that has already been factorised. You don't need to expand it since we have the null factor law: 

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\\[\\text{If } ab=0, \\text{ then } a=0 \\text{ or } b=0.\\]

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Since

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\\[\\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,\\]

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this 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 $x=\\var[fractionnumbers]{(-p*n)/(q*m)}$.

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For 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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