Find the real values of the parameter m such that all the roots of the equation $x(x-1)(x-2)(x-3)=m$ are real. If K is the largest value of m then find the value of 16K is______.

Multiplying x with $x-3 and x-1$ with x-2, we can preserve the symmetry of the equation, yielding:
$(x^2-3x)(x^2-3x+2)=m\text{.}$

With the substitution $x^2-3x=y$ we obtain the quadratic equation
$y^2+2y-m=0$
This has real roots if its discriminant $\mathrm{\Delta }$ is nonnegative, that is
${\mathrm{\Delta }}_1=2^2-4(-m)=4(1+m)\ge 0\text{,}$

which is equivalent to $m\ge -1.$ Returning to the substitution, for the quadratic equation  $x^2-3x-y=0$ to have real roots, its discriminant ${\Delta }_2$ must be nonnegative. In particular,
${\mathrm{\Delta }}_2=9+4y\ge 0.$
This implies that both roots of  $y^2+2y-m=0$ should be at least $-\frac{9}{4}$ This is true if the smaller root is at least $-\frac{9}{4}$From the quadratic formula we obtain the inequality $-1-\sqrt{1+m}\ge -\frac{9}{4}-1-\sqrt{1+m}\ge -\frac{9}{4}$which is equivalent to $m\le \frac{9}{16}$Putting this together with $m\ge -1$ from before, we obtain$-1\le m\le \frac{9}{16}\text{.}$