Let f(x) be a quadratic polynomial such that $\mathrm{f}(-2)+\mathrm{f}\left(3\right)=0$. If one of the roots of f(x) = 0 is -1, then the sum of the roots of f(x) = 0 is equal to :

$\begin{array}{l} \mathrm{Given} \mathrm{that} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{quadratic} \mathrm{polynomial} \mathrm{having} -1 \mathrm{is} \mathrm{a} \mathrm{root}.\\ \mathrm{Assume} \mathrm{that} \mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{ax}+\mathrm{b})\\ given that \\ \mathrm{f}(-2)+\mathrm{f}\left(3\right)=0\Rightarrow -1(-2\mathrm{a}+\mathrm{b})+4(3\mathrm{a}+\mathrm{b})=0\\ \\ 2\mathrm{a}-\mathrm{b}+12\mathrm{a}+4\mathrm{b}=0\\ 14\mathrm{a}+3\mathrm{b}=0\\ \frac{-\mathrm{b}}{\mathrm{a}}=\frac{14}{3}\end{array}$

Sum of roots $=\left(-1+\frac{-\mathrm{b}}{\mathrm{a}}\right)=-1+\frac{14}{3}=\frac{11}{3}$