let f(x) be a quadratic polynomial such that f(-2) + f(3) = 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:

$f\left(-2\right)+f\left(3\right)=0.$ One root of f (x) = 0 is (-1)

Let's assume other root to be a

$\therefore f\left(x\right)=a\left(x+1\right)\left(x-\alpha \right)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a =    $\frac{14}{3}$

$\therefore$ sum of roots =   $\frac{14}{3}-1=\frac{11}{3}$