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 $\lambda$ 
Consider the equation $$x^5 + 5\lambda x^4 -x^3 + (\lambda \alpha -4)x^2 - (8\lambda +3)x + \lambda\alpha - 2 = 0$$ The value of $\alpha$ for which the roots of the equation are independent of $\lambda$ is _______

My approach: The equation can be rewritten as:
$$\underbrace{(x-2)(x^4 + 2x^3 + 3x^2 + 2x + 1)}_{f(x)} + \lambda\underbrace{(5x^4 + \alpha x^2 -8x + \alpha)}_{g(x)} = 0$$
For this equation to be valid independent of $\lambda$, $f(x) = g(x) = 0$. $f(x)$ has $2$ as one of it's roots. Solving $g(2) = 0$, the value of $\alpha$ comes out to be
$$\alpha = -\frac{64}{5}$$
which is unfortunately not the correct answer. Where is my approach breaking down?

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 $\lambda$.
