A cubic polynomial with zeroes α, β, γ satisfies α + β + γ = 5, αβ + βγ + γα = 2 and αβγ = −8.
Which cubic (leading coefficient 1) has these zeroes?

For $x^3+b{x}^2+cx+d:\alpha +\beta +\gamma =-b,\alpha \beta +\beta \gamma +\gamma \alpha =c$ and $\alpha \beta \gamma =-d$. So $b=-5,c=2$ and $d=8$, giving $x^3-5x^2+2x+8$. (Its zeroes are −1, 2 and 4: they add to 5, their pairwise products add to −2 + 8 − 4 = 2, and they multiply to −8.) The version ending in −8 copies the product straight into $d$ and misses the sign flip; $x^3+5x^2+2x-8$ reverses the signs of both $b$ and $d$; $x^3-5x^2-2x+8$ needlessly negates the pairwise sum.