Using the method given by Show steps we first multiply out the first two brackets:
\n\\[\\begin{eqnarray*}\\simplify[std]{ ({a}y+{b})({c}y+{d})}&=&\\simplify[std]{{a}y*({c}y+{d})+{b}({c}y+{d})}\\\\&=&\\simplify[std]{{a*c}y^2+{a*d}y+{b*c}y+{b*d}}\\\\&=&\\simplify[std]{{a*c}y^2+{(a*d+b*c)}y+{b*d}}\\end{eqnarray*}\\]
\nAnd then multiply this by the third bracket:
\n\\[\\begin{eqnarray*}\\simplify[std]{({a}y+{b})({c}y+{d})({p}y+{q})}&=&\\simplify[std]{({a*c}y^2+{(a*d+b*c)}y+{b*d})({p}y+{q})}\\\\&=&\\simplify[std]{{a*c}y^2({p}y+{q})+{(a*d+b*c)}y*({p}y+{q})+{b*d}({p}y+{q})}\\\\&=&\\simplify[std]{{a*c*p}*y^3 +{a*c*q}*y^2+{p*(a*d+b*c)}y^2+{q*(a*d+b*c)}y+{b*d*p}y+{b*d*q}}\\\\&=&\\simplify[std]{{a*c*p}y^3+{a*c*q+a*d*p+p*b*c}y^2+{a*d*q+b*c*q+b*d*p}y+{b*d*q}}\\end{eqnarray*}\\]
\n\n ", "rulesets": {"std": ["all", "!noLeadingMinus", "!collectNumbers"]}, "parts": [{"stepspenalty": 1.0, "prompt": "
$\\simplify[std]{({a}y+{b})({c}y+{d})({p}y+{q})}=\\;$[[0]].
\nYour answer should be a cubic in $y$ and should not include any brackets.
\nYou can click on Show steps for more information, but you will lose one mark if you do so.
\nThere is a video in Show steps which expands three brackets, but uses the variable $x$ rather than $y$.
", "gaps": [{"notallowed": {"message": "Do not include brackets in your answer. Input your answer as a cubic in $y$, in the form $ay^3+by^2+cy+d$ for appropriate integers $a,\\;b,\\;c,\\;d$.
", "showstrings": false, "strings": ["(", "yy", "y*y"], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 2.0, "answer": "{a*c*p}y^3+{a*c*q+a*d*p+p*b*c}y^2+{a*d*q+b*c*q+b*d*p}y+{b*d*q}", "type": "jme"}], "steps": [{"prompt": "There are many ways to expand an expression such as $(ay+b)(cy+d)(py+q)$.
\nOne way is to expand the first two brackets, and then multiply the resulting quadratic in $y$ by $py+q$.
\nHence:
\n\\[\\begin{eqnarray*} (ay+b)(cy+d)&=&ay(cy+d)+b(cy+d)\\\\&=&acy^2+ady+bcy+bd\\\\&=&acy^2+(ad+bc)y+bd\\end{eqnarray*}\\]
\nand then work out $(acy^2+(ad+bc)y+bd)(py+q)$.
\nSee this video for more help:
\n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "extensions": [], "statement": "Expand the following to give a cubic in $y$.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..5)", "name": "a"}, "c": {"definition": "random(2..5)", "name": "c"}, "b": {"definition": "random(-9..9 except [0,a])", "name": "b"}, "d": {"definition": "random(-9..9 except [0,c])", "name": "d"}, "q": {"definition": "random(-3..3 except [0,b,d])", "name": "q"}, "p": {"definition": "random(1..3 except [a,c])", "name": "p"}}, "metadata": {"notes": "\n \t\t16/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\t", "description": "Expand $(ay+b)(cy+d)(py+q)$.
\nIncludes a video expanding three brackets, however uses the variable $x$ rather than $y$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}