// Numbas version: exam_results_page_options {"name": "1.3.3. Removing brackets from expressions of the form (a+b)(c+d)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "1.3.3. Removing brackets from expressions of the form (a+b)(c+d)", "tags": [], "metadata": {"description": "
Removing brackets from expressions of the form (a+b)(c+d). Part of HELM Book 1.3
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Sometimes it is necessary to consider two bracketed terms multiplied together. In the expression $(a+b)(c+d)$, by regarding the first bracket as a single term we can use the result in Key Point 14 to write it as $(a+b)c + (a+b)d$. Removing the brackets from each of these terms produces $ac + bc + ad + bd$. More concisely:
\n\\((a+b)(c+d)=(a+b)c+(a+b)d = ac + bc + ad + bd\\)
\nWe see that each term in the first bracketed expression multiplies each term in the second bracketed expression.
\n\nExample 38
\nRemove the brackets from $(3+x)(2+y)$
\nSolution
\nWe find $(3 + x)(2 + y) = (3 + x)(2) + (3 + x)y$
\n$= (3)(2) + (x)(2) + (3)(y) + (x)(y) = 6 + 2x + 3y + xy$
\nExample 39
\nRemove the brackets from $(3x + 4)(x + 2)$ and simplify your result.
\nSolution
\n$(3x + 4)(x + 2) = (3x + 4)(x) + (3x + 4)(2)$
\n$= 3x^2 + 4x + 6x + 8 = 3x^2 + 10x + 8$
\nExample 40
\nRemove the brackets from $(a+b)^2$ and simplify your result.
\nSolution
\nWhen a quantity is squared it is multiplied by itself. Thus
\n$(a + b)^2 = (a + b)(a + b) = (a + b)a + (a + b)b$
\n= a^2 + ba + ab + b^2 = a^2 + 2ab + b^2$
\n\n\\((a + b)^2 = a^2 + 2ab + b^2\\)
\n\\((a − b)^2 = a^2 − 2ab + b^2 \\)
\nRemove the brackets from $\\var{q1expr}$ and simplify the result.
", "answer": " {q1ans}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "notallowed": {"strings": ["(", ")"], "showStrings": false, "partialCredit": 0, "message": ""}, "valuegenerators": []}, {"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Example 41
\nExplain the distinction between $(x + 3)(x + 2)$ and $x + 3(x + 2)$.
\nSolution
\nIn the first expression removing the brackets we find
\n$(x + 3)(x + 2) = x^2 + 3x + 2x + 6$
\n$= x^2 + 5x + 6$
\nIn the second expression we have
\n$x + 3(x + 2) = x + 3x + 6 = 4x + 6$
\nNote that in the second expression the term (x + 2) is only multiplied by 3 and not by x.
\nExample 42
\nRemove the brackets from $(s^2 + 2s + 4)(s + 3)$.
\nSolution
\nEach term in the first bracket must multiply each term in the second. Working through all combinations systematically we have
\n$(s^2 + 2s + 4)(s + 3) = (s^2 + 2s + 4)(s) + (s^2 + 2s + 4)(3)$
\n$= s^3 + 2s^2 + 4s + 3s^2 + 6s + 12$
\n$= s^3 + 5s^2 + 10s + 12$
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