Using the method given by Show steps
\nFor part a)
\n\\[\\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*}\\]
\n\nFor part b)
\n\\[\\begin{eqnarray*}\\simplify[std]{ ({s}x+{t})({u}x+{v})}&=&\\simplify[std]{{s}x*({u}x+{v})+{t}({u}x+{d})}\\\\&=&\\simplify[std]{{s*u}x^2+{s*v}x+{t*u}x+{t*v}}\\\\&=&\\simplify[std]{{s*u}x^2+{(s*v+t*u)}x+{t*v}}\\end{eqnarray*}\\]
\n", "rulesets": {"std": ["all", "!noLeadingMinus", "!collectNumbers"]}, "parts": [{"stepsPenalty": "0", "prompt": "
$\\simplify[std]{({a}x+{b})({c}x+{d})}=\\;$[[0]].
\nYour answer should be a quadratic in $x$ 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.
", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "There are many ways to expand an expression such as $(ax+b)(cx+d)$.
\nOne way:
\n\\[\\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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", "partialCredit": 0}}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"stepsPenalty": 0, "prompt": "$\\simplify[std]{({s}x-{t})({u}x-{v})}=\\;$[[0]]
\nYour answer should be a quadratic in $x$ 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
", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "There are many ways to expand an expression such as $(sx-t)(ux-v)$.
\nOne way:
\n\\[\\begin{eqnarray*} (sx-t)(ux-v)&=&sx(ux+v)+t(ux+v)\\\\&=&sux^2-svx-tux+tv\\\\&=&sux^2-(sv+tu)x+tv\\end{eqnarray*}\\]
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