$p(x) = x^3 + 2x^2 + kx − 2$ zeroes at $a$, $\frac{1}{a}$ and $b$ for what value of k?

Question

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 $k=2$.

Answer 1

Hint: Let the roots of the given cubic be $x_1,x_2,x_3$

Then, $$x_1\cdot x_2 \cdot x_3=2$$ and $$x_1+x_2+x_3=-2$$

Also, $$k=x_1\cdot x_2 +x_2\cdot x_3+ x_3\cdot x_1 $$

Now, you only need to put the values of $x_1,x_2,x_3$, solve for $a$ and $b$, and then find $k$. Can you take this from here now?

Answer 2

Hint: Instead of taking the equations for granted as in Gaurang's answer you can write $$p(x)=(x-a)(x-b) \left(x-\frac{1}{a} \right) $$ and expand this and compare the coefficients to the given coefficients of $p(x)$ to get the equations.

Answer 3

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