$\large x^5 +5 \lambda x^4-x^3 +(\lambda \alpha-4)x^2-(8\lambda +3)x+\lambda \alpha -2=0$

Consider the equation above, what is the value of $\large{\alpha}$ for which the equation has two roots independent of $\lambda$ ?

The answer is -3.

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Rewrite as $(x^5 - x^3 - 4x^2-3x-2) + \lambda ( 5x^4+\alpha x^2-8x+\alpha) = 0$ If this is supposed to have two roots independent of $\lambda$ , they should be roots of both of the polynomials in parentheses. The first polynomial factors as $(x-2)(x^2+x+1)^2$ . So at least one of the primitive third roots of unity $\omega$ is a root of the second polynomial.

So $0 = 5\omega^4 + \alpha \omega^2-8\omega + \alpha = - \omega ( \alpha + 3 )$ , so $\alpha = \fbox{-3}$ . It's not hard to check that both $\omega$ and $\omega^2$ are roots of the original equation if $\alpha = -3$ , so we are done.