$x\left(x-1\right)\left(x-2\right)\left(x-3\right)=m$ 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 .

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
