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?
 A: 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 $\lambda$.
A: The problem with your answer is that it's true that $2$ would be a root independent on $\lambda$ but there may be other roots which depends on $\lambda$.
Indeed with $\alpha=-\frac{64}{5}$ your equation becomes
$$\frac{1}{5}(x-2)(5(1+x+x^2)^2+\lambda(5x+8)(4+x(2+5x))$$
which may have solutions different from $2$ and $\lambda$-dependent.
However I don't see how could you get all the solution independent of $\lambda$, unless you allow $\alpha$ to be $x$-dependent.
Indeed I would say that the only way to get all roots independent of $\lambda$ is to ask $g(x) = 0$ or $g(x) = f(x)$. Ando both require a dependence of $\alpha$ on $x$.
For $g(x) = 0$ you must have
$$\alpha = -\frac{5 x^4 - 8x}{1+x^2}\,.$$
For $g(x)=f(x)$ things look even worse...
$$\alpha = \frac{-2 + 5 x - 4 x^2 - x^3 - 5 x^4 + x^5}{1 + x^2},.$$
