$x-\frac{1}{2}$ નું શૂન્ય $\frac{1}{2}$ છે.             $\quad\left[\because x-\frac{1}{2}=0 \therefore x=\frac{1}{2}\right]$

તેથી જો $p(x)=x^{3}+3 x^{2}+3 x+1$ માં $x=\frac{1}{2}$ મૂકીએ તો

$ \therefore p\left(\frac{1}{2}\right) =\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1$

$=\frac{1}{8}+3\left(\frac{1}{4}\right)+\frac{3}{2}+1$

$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+\frac{1}{1}$

$=\frac{1+6+12+8}{8}$

$\therefore p\left(\frac{1}{2}\right) =\frac{27}{8}$

આમ, શેષ $=\frac{27}{8}$ મળે છે.