$p(x) = x^3 + 2x^2 + kx − 2$ zeroes at $a$, $\frac{1}{a}$ and $b$ for what value of k? For what values of $k$ has $p(x) = x^3 + 2x^2 + kx − 2$ zeroes at $a$, $\frac{1}{a}$ and $b$? 
The options I'm given are $k=-7$, $k=1$, $ k=0$ and $k=2$.

The product of the roots is $2$ and also $a(1/a)b$. So $b = 2$. So $2$ is a root. What $k$ makes $p(2) = 0$?
