Find the the least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real The least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real.
let $f(x) =x^4-4x^3 -8x^2 +a=0$
$  f'(x) = 4x^3 -12x^2 -16x $ 
Put $f'(x) = 0 $ we get x =0, -1, 4. How to proceed further, I am not getting any idea on this. Please guide thanks. 
 A: By your work, we have that $f(x) =x^4-4x^3 -8x^2 +a$ is a continuous function which is strictly monotone in each one of the following intervals: strictly decreasing in $(-\infty,-1]$, strictly increasing in $[-1,0]$, strictly decreasing in $[0,4]$, and strictly increasing in $[4,-\infty)$. Now apply the Intermediate Value Theorem. For example, for $x\in(-\infty,-1]$, $f(x)$ attains once and only once each value in $(\lim_{x\to-\infty}f(x),f(-1)]=(+\infty,-3+a]$. Hence we have one root in $(-\infty,-1)$  if and only if $-3+a<0$, i.e. $a< 3$.
Can you take it from here?
A: Let $f(x)=x^4-4x^3-8x^2+a$, $$f'(x)=4x^3-12x^2-16x=0 \Rightarrow x=4,-1,0$$
Next $f''(x)=12x(x-2)-16.$ so $f(x)$ has local minima at at $x=4,-1$ and loca max at $x=0.$ Hence $f_{min}=a-128, a-3$, $f_{max}=a$. For four real rootss: $f_{min}<0$ and
$f_{max}>0.$ Thus, for four real roots $$a \in( 0,3).$$
Both the integers $a=1,2$ are the solutions.
See the fig. for a=1, below
$f(x)$ for $a=1$">
See the fig. for $a=2$, below

A: Hint
I think that considering the second derivative could help.
$$f''(x)=12 x^2-24 x-16$$ shows that 


*

*$x=-1$ corresponds to a minimum since $f''(-1)=20$

*$x=0$ corresponds to a maximum since $f''(0)=-16$

*$x=4$ corresponds to a minimum since $f''(4)=80$
So, in order to have four real roots, you need ... ?
A: The easiest way to see what is going on is to sketch the curve - you have enough information for a sufficiently good sketch. Then changing the constant term shifts the curve vertically (moves it up or down).
How much do you need to shift it, and in what direction (up or down) to meet the condition?
I think a sketch would be a helpful guide to the mathematical analysis you need to do.
A: Hint: Consider $g(x) =x^4-4x^3 -8x^2$ and look for horizontal lines that cross the graph of $g$ four times (tangencies count twice). The values of $g$ at its critical points are the key in this argument.
