Use a given factor of a polynomial to find the full factorisation of the polynomial through long division.
"}, "preamble": {"js": "", "css": ""}, "statement": "The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
", "name": "Marlon's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division", "variablesTest": {"condition": "", "maxRuns": "192"}, "variable_groups": [], "parts": [{"vsetrangepoints": 5, "showFeedbackIcon": true, "checkvariablenames": false, "checkingaccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "showpreview": true, "showCorrectAnswer": true, "variableReplacements": [], "scripts": {"mark": {"script": "if(this.credit>0) {\n // Parse the student's answer as a syntax tree\n var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n // Create the pattern to match against \n // we just want two sets of brackets, each containing two terms\n // or one of the brackets might not have a constant term\n // or for repeated roots, you might write (x+a)^2\n var rule = Numbas.jme.compile('m_all(m_any(x,x+m_pm(m_number),(x+m_pm(m_number))^m_number))*m_nothing');\n\n // Check the student's answer matches the pattern. \n var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n // If not, take away marks\n if(!m) {\n this.multCredit(0.5,'Your answer is not fully factorised.');\n }\n}", "order": "after"}}, "prompt": "Given that $(\\simplify{x+{z}})$ is a factor of \\[p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}*{u}+{z}*{u}+{y}*{z})*x+{y}*{u}*{z}}.\\]
\nFind the full factorisation of $p(x)$.
", "type": "jme", "answer": "(x+{y})(x+{u})(x+{z})", "marks": "2", "vsetrange": [0, 1], "expectedvariablenames": [], "checkingtype": "absdiff"}], "tags": ["factor theorem", "Factor Theorem", "Factorisation", "factorisation", "Factorise", "factorise", "Long division", "long division", "polynomials", "Polynomials", "taxonomy"], "advice": "For this question, we are given that $(\\simplify{x+{z}})$ is a factor of the polynomial
\n\\[p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},\\]
\nand we are then asked to find the full factorisation of $p(x)$.
\nWe know that $(\\simplify{x+{z}})$ is a factor of $p(x)$, so we can calculate the other factors of $p(x)$ through long division.
\n\\[
\\begin{align}
&\\simplify{x^2+({u}+{y})x+{u}{y}}\\\\
\\simplify{x+{z}} \\; &\\overline{)\\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\\\
&\\;\\,
\\simplify{x^3+{z}x^2}\\\\
&\\qquad\\quad
\\overline{\\simplify[all,noLeadingMinus]{({u}+{y})x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\\\
&\\qquad\\quad
\\simplify[all,noLeadingMinus]{({u}+{y})x^2+({u}{z}+{z}{y})x}\\\\
&\\qquad\\quad\\quad\\quad\\quad
\\overline{\\simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}}\\\\
&\\qquad\\quad\\quad\\quad\\quad
\\simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}\\\\
&\\qquad\\qquad\\quad\\quad\\quad
\\overline{0.}
\\end{align}
\\]
We can then factorise $\\simplify{x^2+({u}+{y})x+{u}{y}}$ into
\n\\[\\simplify{x^2+({u}+{y})x+{u}{y}} =(\\simplify{x+{y}})(\\simplify{x+{u}}).\\]
\nTherefore, the full factorisation of $p(x)$ is
\n\\[
\\begin{align}
p(x) &= \\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},\\\\
&= (\\simplify{x+{y}})(\\simplify{x+{z}})(\\simplify{x+{u}}).
\\end{align}
\\]
Factor 3.
", "templateType": "anything"}, "z": {"definition": "random(-3..3 except 0)", "group": "Ungrouped variables", "name": "z", "description": "Factor 1.
", "templateType": "anything"}, "u": {"definition": "random(-2..3 except 0 except -y)", "group": "Ungrouped variables", "name": "u", "description": "Factor 2.
", "templateType": "anything"}}, "ungrouped_variables": ["y", "u", "z"], "functions": {}, "type": "question", "contributors": [{"name": "Marlon Arcila", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/321/"}]}]}], "contributors": [{"name": "Marlon Arcila", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/321/"}]}