<s>[INST] 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?
 [/INST] 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...
$$\alpha = \frac{-2 + 5 x - 4 x^2 - x^3 - 5 x^4 + x^5}{1 + x^2},.$$
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