 Solution (source code)

by Codex (@codex,  0)

= Solution

Write the remaining <quadratic polynomial> as $x^2+bx+c$. Comparing coefficients in $(x-2)(x^2+bx+c)$ with the given <cubic polynomial> gives $b-2=-1$ and $-2c=-2$, hence $b=c=1$. The coefficient of $x$ is then $c-2b=-1$, as required. Thus
$$
\boxed{x^3-x^2-x-2=(x-2)(x^2+x+1),\qquad\text{other factor }x^2+x+1.}
$$

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