// Numbas version: finer_feedback_settings {"name": "Marlon's copy of Factorising: Common factor", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [{"name": "part a", "variables": ["pmult", "pxcoeff", "pconstant", "primes"]}, {"name": "part b", "variables": ["nmult", "nxcoeff", "nconstant", "bp2", "bp1", "cf", "bx", "bc"]}, {"name": "part c", "variables": ["cprimes", "cx", "cy", "cc", "cmult", "ct1", "ct2", "ct3"]}], "advice": "", "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "showQuestionGroupNames": false, "question_groups": [{"pickingStrategy": "all-ordered", "name": "", "questions": [], "pickQuestions": 0}], "rulesets": {}, "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "notes": ""}, "statement": "", "type": "question", "tags": ["algebra", "common factor", "distributive law", "factor", "factorisation", "Factorisation", "factorising"], "parts": [{"showCorrectAnswer": true, "variableReplacements": [], "steps": [{"showCorrectAnswer": true, "variableReplacements": [], "prompt": "

We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

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The (largest) common factor of  $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is $\\var{pmult}$. Once we remove that factor from each term in $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ we are left with $\\var{pxcoeff}x+\\var{pconstant}$.

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That means $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}= \\var{pmult}(\\var{pxcoeff}x+\\var{pconstant})$.

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The expression $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is a sum and can be factorised (written as a product) by finding the largest common factor:

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 $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult} = $ [[0]] $\\large($ [[1]] $\\large)$

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We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

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The (largest) common factor of $\\simplify{{bp1}a+{bp2}}$ is $\\var{cf}$. Once we remove that factor from each term in $\\simplify{{bp1}a+{bp2}}$ we are left with $\\var{bx}a+\\var{bc}$.

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That means $\\simplify{{bp1}a+{bp2}}$ is $\\var{cf} = \\var{cf}(\\var{bx}a+\\var{bc})$.

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Factorise $\\simplify{{bp1}a+{bp2}}$

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 [[0]] $\\large($ [[1]] $\\large)$

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We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

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The (largest) common factor of $\\simplify{{ct1}x+{ct2}y+{ct3}}$ is $\\var{cmult}$. Once we remove that factor from each term in $\\simplify{{ct1}x+{ct2}y+{ct3}}$ we are left with $\\simplify{{cx}x+{cy}y+{cc}}$.

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That means $\\simplify{{ct1}x+{ct2}y+{ct3}} = \\simplify{{cmult}({ct1}x+{ct2}y+{ct3})}$.

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Factorise $\\simplify{{ct1}x+{ct2}y+{ct3}}$

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 [[0]] $\\large($ [[1]] $\\large)$

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