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Are we talking about not-necessarily-different real roots? If so, part (ii) is impossible, since an odd-degree polynomial will have always an odd number of real roots (some of them possibly equal). Are we talking about not-necessarily-real different roots? If so, part (i) is impossible, since the polynomial should be of the form $ (x + r)^5$, yielding $ \lambda = r$ and $ - 1 = 10r^2$, obviously impossible for arbitrary $ \lambda$. Finally, if we write the polynomial as $ \left(x^5 - x^3 - 4x^2 - 3x - 2\right) + \lambda(5x^4 + \alpha x^2 - 8x + \alpha) = p(x) + \lambda q(x)$, obviously any root that is independent of $ \lambda$ must satisfy $ p(x) = q(x) = 0$, but frankly I do not know what to look for now because as I stated above I do not understand what is meant by exactly one root and exactly two roots... Edit: maybe there is an erratum in the problem, since "suspiciously enough", $ \alpha=-3$ makes $ q(x)=p'(x)$, so any $ \lambda$-independent (possibly multiple) root would be a double root of $ p(x)$, but $ p(x)$ does not have a double real root... Edit 2: OK, silly me! The one or two roots are obviously $ \lambda$-independent roots, regardless of how may real or complex roots the polynomial has!!!

Continuing with the idea above that any $ \lambda$-independent root must satisfy $ p(x)=q(x)=0$, note that $ p(x)=(x-2)(x^2+x+1)^2$ has the only real root $ x=2$. Now, any $ \lambda$-independent root will also be a root of $ r(x)=5p(x)-xq(x)=(\alpha+5)x^3+12x^2+(\alpha+15)x+10$. Then... (i) in order to have exactly one root, it must be real, since complex roots come in conjugate pairs both for $ p(x)$ and $ q(x)$ because their coefficients are real (unless stated otherwise I am assuming $ \alpha$ and $ \lambda$ to be real), and it must be the only real root of $ p(x)$, so $ 0=8(\alpha+5)+48+2(\alpha+15)+10$ yielding $ \alpha=-\frac{64}{5}$. (ii) in order to have exactly two roots, they must be complex, and roots of both $ (x^2+x+1)^2$ (ie, of $ x^2+x+1$) and of $ q(x)$. Then, since $ q(x)=(5x^2-5x+\alpha)(x^2+x+1)-3x-\alpha x$, we must have $ \alpha=-3$, in which case we find $ q(x)=p'(x)$, and the complex double roots of $ p(x)$ are also roots of $ q(x)$, or $ \lambda$-independent roots of the given polynomial.