// Numbas version: exam_results_page_options {"name": "Distributive law: expanding one set of brackets", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickQuestions": 0, "questions": [], "pickingStrategy": "all-ordered"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "", "notes": ""}, "name": "Distributive law: expanding one set of brackets", "tags": ["algebra", "distributive law", "expanding", "expanding brackets"], "ungrouped_variables": [], "rulesets": {}, "functions": {}, "parts": [{"steps": [{"showCorrectAnswer": true, "variableReplacements": [], "marks": 0, "variableReplacementStrategy": "originalfirst", "type": "information", "prompt": "

The number in front of the bracket is multiplying the bracketed term, that is, each term in the brackets.

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For example, $3(5x+6)$ means $3\\times (5x+6)$ which means $3\\times 5x+3\\times 6$, and so expanding $3(5x+6)$ gives $15x+18$.

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The expression $\\var{pmult}(\\var{pxcoeff}x+\\var{pconstant})$ is factorised (written as a product), we can expand the expression (so it is written as a sum) to get 

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[[0]] $x$ + [[1]]

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The number in front of the bracket is multiplying the bracketed term, that is, each term in the brackets. Further, recall that a negative multiplied by a negative is a positive.

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For example, $-3(5a-6)$ means $-3\\times (5a-6)$ which means $(-3)\\times 5a+(-3)\\times (-6)$, and so expanding $3(5a+6)$ gives $-15a+18$.

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Expand $\\var{nmult}(\\var{nxcoeff}a-\\var{-nconstant})$.

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[[0]] $a$ + [[1]]

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A negative sign in front of a bracket is a common way to signify $-1$ times the bracketed term. The result is that it changes the sign of everything in the brackets.

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For example, $-(5x-y+6)$ means $-1\\times (5x-y+6)$ which means $(-1)\\times 5x+(-1)\\times (-y)+(-1)\\times 6$, and so expanding $-(5x-y+6)$ gives $-5x+y-6$.

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Expand $-(\\var{cx}x-\\var{-cy}y+\\var{cc})$.

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[[0]] $x$ + [[1]] $y$ + [[2]] 

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