Find the max and min values of $\alpha$ for which $x^3-5x^2+8x+\alpha$ has 3 real roots. 
Find the max and min values of $\alpha$ for which $x^3-5x^2+8x+\alpha$ has 3 real roots.

I noticed that if we have 3 different roots then the set of solutions for $\alpha$ is open i.e no min/max.
I'm not sure how to approach the problem when also considering roots with degree $> 1$.
Maybe we could define $F(x,y,z,w,\alpha)= \alpha$
And using lagrange multipliers look for an ext under constraints:
$x^3-5x^2+8x+\alpha = 0$,
$(x-y)(x-z)(x-w)=0$
But this seems very unpleasant...
Would appreciate hints.
 A: If $f(x)=x^3-5x^2+8x+\alpha$, then $f'(x)=3x^2-10x+8$ and therefore $f'(x)>0$ if $x<\frac43$ or $x>2$, and $f(x)<0$ if $x\in\left(\frac43,2\right)$. That means that $f$ is strictly increasing on $\left(-\infty,\frac43\right]$ and on $[2,\infty)$ and it is strictly decreasing on$\left[\frac43,2\right]$. So, $f$ has $3$ distinct real roots if and only if $f\left(\frac43\right)>0$ and $f(2)<0$. This means that $\alpha\in\left(-4-\frac4{27},-4\right)$
Edit: If you're after those $\alpha$'s for which $x^3-5x^2+8x+\alpha$ has $3$ real roots when we count them with their multiplicities, then the inequalities that you should solve are $f\left(\frac43\right)\geqslant0$ and $f(2)\leqslant0$.
A: Imagine the graph of the polynomial $f(x)$. Here $f'(x)$ is equal to $0$ for two real points and let them be at $a$ and $b$. Observe that the $f(a).f(b)<0$ iff it has three distinct real roots as then $f(a)$ and $f(b)$ have opposite signs(visualize). Also $f(a).f(b)=0$  implies multiple real roots.  So the range of alpha can be derived from $f(a)f(b)\leqslant0$.
