Dare to find my roots! 2

Let the expression be f(x). Write it as $g(x)+\lambda p(x)$ and since roots are independent of $\lambda$ we must have $g(x)=p(x)=0$ (at same value of $x$ ) so that $f(x)=0$ irrespective of $\lambda$ .

$f(x)=(\underbrace{x^5-x^3-4x^2-3x-2}_{\color{#D61F06}{g(x)}})+\lambda (\underbrace{5x^4+\alpha x^2-8x+\alpha}_{\color{#20A900}{p(x)}})=0$

$\color{#D61F06}{g(x)}=0\\\implies (x-2)(x^4+2x^3+3x^2+2x+1)=0\\\implies (x-2)(x^2+x+1)^2=0$

Since $f(x)$ has exactly two roots , $x^2+x+1=0\implies x=\omega,\omega^2$ . Now since $x=\omega$ ( $\omega^2$ will also work), setting $p(\omega)=0$ we get $\alpha(\omega^2+1)=3\omega\implies \alpha=\dfrac{3\omega}{\omega^2+1}=-3$ . Hence $\Large \alpha=\boxed{-3}$ .

$$ $f(x)$ has exactly two roots and independent of $\lambda$ and $f(x)$ vanishes at $x=\omega,\omega^2$ when $\alpha=-3$ .

NOTE:- $\omega,\omega^2$ are cube roots of unity.