// Numbas version: exam_results_page_options {"name": "Luis'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, "preventleave": false, "showfrontpage": false}, "question_groups": [{"questions": [{"functions": {}, "preamble": {"js": "", "css": ""}, "type": "question", "metadata": {"description": "

Use a given factor of a polynomial to find the full factorisation of the polynomial through long division.

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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": "Luis's copy of Finding the full factorisation of a polynomial, using the Factor Theorem and long division", "extensions": [], "tags": ["factor theorem", "Factor Theorem", "Factorisation", "factorisation", "Factorise", "factorise", "Long division", "long division", "polynomials", "Polynomials", "taxonomy"], "ungrouped_variables": ["y", "u", "z"], "advice": "

For this question, we are given that $(\\simplify{x+{z}})$ is a factor of the polynomial

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\$p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},\$

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and we are then asked to find the full factorisation of $p(x)$.

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We know that $(\\simplify{x+{z}})$ is a factor of $p(x)$, so we can calculate the other factors of $p(x)$ through long division.

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\\\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} \

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We can then factorise $\\simplify{x^2+({u}+{y})x+{u}{y}}$ into

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\$\\simplify{x^2+({u}+{y})x+{u}{y}} =(\\simplify{x+{y}})(\\simplify{x+{u}}).\$

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Therefore, the full factorisation of $p(x)$ is

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\\\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} \

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Factor 3.

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Factor 1.

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Factor 2.

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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}}.\$

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Find the full factorisation of $p(x)$.

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