Dare to find my roots!

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 only one solution i.e $x=2$ . Now since $x=2$ setting $p(2)=0$ we get $\alpha =\dfrac{-64}{5}$ . Hence $\Large 64+5=\boxed{69}$ .

Hence $f(x)$ has only one root and independent of $\lambda$ $f(x)$ vanishes at $x=2$ when $\alpha=\dfrac{-64}{5}$ .