Given a polynomial $p(x) = x^3 + kx - 2$, where $k \in \mathbb{R}$ is a constant and $p(x)$ has a double root at $x = \alpha$. Prove that $\alpha=-1$ and $k=-3$.
I plugged in $\alpha$ for $p(x)$ and $p'(x)$, but that didn't go anywhere, bc I would just be proving the values for $\alpha$ and $k$ by inspection. So I'm not sure how to tackle this problem.
I assume that, by saying "has a double root", you mean $\alpha$ is a root of multiplicity.
Then we can assume
$$f(x) = (x-\alpha)^2(x-\beta)$$
because $\deg(f) = 3$.
So we get
$$f(x) = (x-\alpha)^2(x-\beta)$$
$$= x^3 - (2\alpha+\beta)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2\beta$$
And this implies
$$x^3 - (2\alpha+\beta)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2\beta = x^3 + kx -2$$
$$\Rightarrow - (2\alpha+\beta) = 0, 2\alpha \beta + \alpha^2 =k, - \alpha^2\beta = -2$$
After solving this system of equations, we finally get
$$\alpha = -1, \beta = 2, k = -3$$