$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$ has all roots real Given the equation:
$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$
For what values of $m$ are all the roots real?
I've rewritten the equation as: $x^4-6x^3+11x^2-6x-m=0$
I'm quite sure this is done with Vieta's but didn't really figure out yet what should
I aim to get out of it in order to get $m \in [-1,\frac{9}{16}]$ which is the correct answer .
 A: Think about the graph of $y=x(x-1)(x-2)(x-3)$ – better yet, sketch it. Now you want to know, for what values of $m$ will the horizontal line $y=m$ cross the graph at four points. These are clearly the values of $m$ between the minimum value of $y$ and the local maximum of $y$ at $x=1.5$. It may take a little calculus to find the minimum. 
A: Let $u = x - \frac32.$ Then
\begin{align}
x(x−1)(x−2)(x−3) 
&= \left(u+\frac32\right)\left(u+\frac12\right)
   \left(u-\frac12\right)\left(u-\frac32\right) \\
&= \left(u^2 - \frac94\right)\left(u^2 - \frac14\right).
\end{align}
Let $v = u^2.$ Find the minimum value of the quadratic polynomial
$\left(v - \frac94\right)\left(v - \frac14\right)$ on the domain $v \geq 0.$
(It occurs when $v = \frac12\left(\frac94 + \frac14\right).$)
Working backwards, this is also the minimum value of $x(x−1)(x−2)(x−3).$
The fact that the minimum of $x(x−1)(x−2)(x−3)$ occurs at two different points means that if you translate the graph up just far enough so that it touches the $x$-axis at its two minima, you will have two double roots, which means all your roots are real (there are only four roots). If you translate the graph any higher then it doesn't meet the $x$ axis and you have no real roots at all.
As has already been pointed out, there are other translations of the graph that put its minima and a local maximum below the $x$-axis, leaving only two $x$-intercepts which give you two real roots (but not double roots, so the other two must be complex).
The symmetry of the polynomial 
$\left(u+\frac32\right)\left(u+\frac12\right)
\left(u-\frac12\right)\left(u-\frac32\right)$
says there should be a local minimum or maximum at $u=0$;
in fact, it's a local maximum.
A: Start from $x^4-6 x^3+11 x^2-6 x-m=0$ and substitute $x=z+\dfrac{3}{2}$
we get
$\left(z+\frac{3}{2}\right)^4-6 \left(z+\frac{3}{2}\right)^3+11 \left(z+\frac{3}{2}\right)^2-6 \left(z+\frac{3}{2}\right)-m=0$
Expand and reorder
$z^4-\frac{5 }{2}z^2+\frac{9}{16}-m=0$
substitute $z^2=w$
$w^2-\frac{5 }{2}w^2+\frac{9}{16}-m=0$
$w=\dfrac{1}{4} \left(5\pm 4 \sqrt{m+1}\right)$
to be real the solutions we need $w\ge 0$
$5\pm 4 \sqrt{m+1}\ge 0$
Begin with
$5+4 \sqrt{m+1}\ge 0\to 4 \sqrt{m+1}\ge -5$
verified for $m\ge -1$
Then we solve
$5- 4 \sqrt{m+1}\ge 0$
$4 \sqrt{m+1}\le 5$
$16(m+1)\le 25 \to 16m \le 9\to m \le \dfrac{9}{16}$
the equation has all real solutions if $\quad-1\le m \le \dfrac{9}{16}$
Hope this helps
