// Numbas version: exam_results_page_options {"name": "JD's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "metadata": {"description": "
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"}, "tags": [], "functions": {}, "variables": {"y": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(-2..3 except 0)", "name": "y", "description": "Factor 3.
"}, "u": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(-2..3 except 0 except -y)", "name": "u", "description": "Factor 2.
"}, "z": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(-3..3 except 0)", "name": "z", "description": "Factor 1.
"}}, "statement": "The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
", "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}
\\]
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 $p(x)$.
", "checkVariableNames": false}], "variablesTest": {"condition": "", "maxRuns": "192"}, "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}, {"name": "JD Ichwan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3389/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}, {"name": "JD Ichwan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3389/"}]}