Question
Let $\alpha \neq \beta$ and consider the function
\begin{align*} p : \mathbb{R} & \rightarrow \mathbb{R} \\ x & \mapsto \det \begin{pmatrix} 1 & \alpha & \alpha ^2 \\ 1 & \beta & \beta^2 \\ 1 & x & x^2 \end{pmatrix} \end{align*}
Show that $p$ is a polynomial different from $0$ and of order $2$ and determine the roots of $p$.
My attempt
We start by calculating the determinant and have that
\begin{align*} \det A \begin{vmatrix} 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2 \\ 1 & x & x^2 \end{vmatrix} & = 1 \cdot \begin{vmatrix} \beta & \beta^2 \\ x & x^2 \end{vmatrix} - \alpha \cdot \begin{vmatrix} 1 & \beta^2 \\ 1 & x^2 \end{vmatrix} + \alpha^2 \begin{vmatrix} 1 & \beta \\ 1 & x \end{vmatrix} \\ & = \beta x^2 - x \beta^2 - \alpha x^2 + \alpha \beta^2 + \alpha^2 x - \alpha^2 \beta \\ & = (\beta - \alpha)x^2 + (\alpha^2 - \beta^2)x + (\alpha \beta^2 - \alpha^2 \beta) \end{align*}
which gives a polynomial of order $2$ and as $\alpha \neq \beta$ it is also different from the zero polynomial. To determine its roots I have used that p is a function defined as $p : \mathbb{R} \rightarrow \mathbb{R}$ which means that I can use the zero product property and by rewriting the determinant we have that $$ (\beta - \alpha)x^2 + (\alpha^2 - \beta^2)x + (\alpha \beta^2 - \alpha^2 \beta) = (\alpha - \beta)(\alpha - x)(x - \beta) $$ which must mean that $x = \alpha$ or $x = \beta$. I tried to calculate its roots with my CAS but it does not give this solution, rather, a very complicated one. Is this approach alright?
Thanks in advance.