Value of $\alpha$ for which $x^5+5\lambda x^4-x^3+(\lambda\alpha-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0$ has roots independent of $\lambda$

The question may be phrased incorrectly, as it is not possible to make the set of all roots independent of $\lambda$. The question I will answer is: For what value of $\alpha$ does the equation have some roots which are independent of $\lambda$?

As demonstrated in the question, $\alpha=-\frac{64}{5}$ is one possibility, which gives the root $x=2$, independent of $\lambda$.

But there is one other possibility that we can find by further factoring: $f(x) = (x-2)(x^2+x+1)^2$. Is there a value of $\alpha$ for which $g(x)$ shares a root with $x^2+x+1$? Setting $x=\omega$ with $\omega^3=1$ and $g(\omega)=0$ gives us $\alpha \omega^2 - 3\omega + \alpha = 0$. Reducing further using $\omega^2+\omega+1 = 0$ gives $-a\omega - 3\omega = 0$ or $a=-3$.

Indeed, we can verify that if $a=-3$, the original equation is divisible by $x^2 + x+1$ regardless of the value of $\lambda$.