// Numbas version: exam_results_page_options {"duration": 0, "timing": {"timedwarning": {"action": "none", "message": ""}, "timeout": {"action": "none", "message": ""}, "allowPause": true}, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "Rachel's copy of Distributive law: expanding one set of brackets", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Rachel Staddon", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/901/"}], "preamble": {"js": "", "css": ""}, "variablesTest": {"condition": "", "maxRuns": 100}, "tags": [], "parts": [{"gaps": [{"correctAnswerStyle": "plain", "marks": 1, "scripts": {}, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "minValue": "{pmult*pxcoeff}", "type": "numberentry", "showFeedbackIcon": true, "maxValue": "{pmult*pxcoeff}", "correctAnswerFraction": false, "showCorrectAnswer": true}, {"correctAnswerStyle": "plain", "marks": 1, "scripts": {}, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "minValue": "{pmult*pconstant}", "type": "numberentry", "showFeedbackIcon": true, "maxValue": "{pmult*pconstant}", "correctAnswerFraction": false, "showCorrectAnswer": true}], "stepsPenalty": "1", "marks": 0, "variableReplacements": [], "prompt": "

Expand the expression $\\var{pmult}(\\var{pxcoeff}x+\\var{pconstant})$.

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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.

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

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[[0]] $a$ + [[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 $-(\\var{cx}x-\\var{-cy}y+\\var{cc})$.

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[[0]] $x$ + [[1]] $y$ + [[2]] 

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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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For all questions in this quiz, if you want to show indices (e.g., $x^2$) then input this as x^2 for $x^2$, y^3 for $y^3$, etc.

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Expand and simplify the following algebraic expression:

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$(x+\\var{const1})(x+\\var{const2})$

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You need to multiply everything in the first bracket by the second bracket. That is,

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$x (x + \\var{const2}) + \\var{const1} (x + \\var{const2})=x^2+\\var{const2}x+\\var{const1}x+(\\var{const1}\\times\\var{const2})$

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and then simplify your answer.

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You need to simplify your solution as far as possible.

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This question tests the method of expanding a pair of brackets.

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Using the method given by Show steps we have:

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\\[\\begin{eqnarray*}\\simplify[std]{ ({a}x+{b})({c}x+{d})}&=&\\simplify[std]{{a}x*({c}x+{d})+{b}({c}x+{d})}\\\\&=&\\simplify[std]{{a*c}x^2+{a*d}x+{b*c}x+{b*d}}\\\\&=&\\simplify[std]{{a*c}x^2+{(a*d+b*c)}x+{b*d}}\\end{eqnarray*}\\]

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$\\simplify[std]{({a}x+{b})({c}x+{d})}=\\;$[[0]].

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Your answer should be a quadratic in $x$ and should not include any brackets.

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You can click on Show steps for more information, but you will lose one mark if you do so.

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Do not include brackets in your answer. Input your answer as a quadratic in $x$, in the form $ax^2+bx+c$ for appropriate integers $a,\\;b,\\;c$.

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Input your answer as a quadratic in $x$, in the form $ax^2+bx+c$ for appropriate integers $a,\\;b,\\;c$.

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Input your answer as a quadratic in $x$, in the form $ax^2+bx+c$ for appropriate integers $a,\\;b,\\;c$.

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There are many ways to expand an expression such as $(ax+b)(cx+d)$.

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One way:

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\\[\\begin{eqnarray*} (ax+b)(cx+d)&=&ax(cx+d)+b(cx+d)\\\\&=&acx^2+adx+bcx+bd\\\\&=&acx^2+(ad+bc)x+bd\\end{eqnarray*}\\]

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Expand the following to give a quadratic in $x$.

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Expand $(ax+b)(cx+d)$.

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rebelmaths

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Questions about multiplying out brackets.

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