The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
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\nFind the full factorisation of $p(x)$.
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"}, "preamble": {"css": "", "js": ""}, "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}
\\]