Consider the polynomial $f\left(x\right)=1+2x+3x^2+4x^3$

Let $s$  be the sum of all distinct real roots of $f\left(x\right)$  and let $t=\left|s\right|$ , then the real number $s$  lies in the interval:

The given polynomial is $f\left(x\right)=1+2x+3x^2+4x^3$

$\Rightarrow f\left(x\right)=4x^3+3x^2+2x+1$

$\Rightarrow f\text{'}\left(x\right)=12x^2+6x+2$

Here $a=12,b=6,c=2$

$\Rightarrow$  discriminant $=6^2-\mathrm{4.12.2}\left(\because b^2-4ac\right)$

                         $=36-96$

                         $=-60<0$

$\therefore$  $f\text{'}\left(x\right)>0;\forall x\in R$

$\Rightarrow$ only one real root for $f\left(x\right)=0$

Also $f\left(0\right)=1,f\left(-1\right)=-2$

$\Rightarrow$ root must lie in $\left(-1,0\right)$

Taking average of 0 and $-1\Rightarrow f\left(\frac{-1}{2}\right)=\frac{1}{4}$

$\Rightarrow$  root must lie in $\left(-1,\frac{-1}{2}\right)$

similarly, $f\left(\frac{-3}{4}\right)=\frac{-1}{2}\Rightarrow$  root must lie in $\left(-\frac{3}{4},-\frac{1}{2}\right)$