Use a given factor of a polynomial to find the full factorisation of the polynomial through long division.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
", "variables": {"z": {"templateType": "anything", "description": "Factor 1.
", "definition": "random(-3..3 except 0)", "group": "Ungrouped variables", "name": "z"}, "y": {"templateType": "anything", "description": "Factor 3.
", "definition": "random(-2..3 except 0)", "group": "Ungrouped variables", "name": "y"}, "u": {"templateType": "anything", "description": "Factor 2.
", "definition": "random(-2..3 except 0 except -y)", "group": "Ungrouped variables", "name": "u"}}, "tags": [], "parts": [{"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": true, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "mustmatchpattern": {"pattern": "((`+-x^$n`? + `+- $n)^$n`?)`* * $z", "nameToCompare": "", "partialCredit": 0, "message": "Your answer is not fully factorised."}, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "(x+{y})(x+{u})(x+{z})", "valuegenerators": [{"value": "", "name": "x"}], "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 $p(x)$.
", "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "marks": "2", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}], "rulesets": {}, "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}
\\]