Let f (x) be a quadratic polynomial such that f(- 2) + f(3) = 0 ....

Let $f\left(x\right) = a{x}^{2} + bx + c$

Given, $f\left(- 2\right) + f\left(3\right) = 0$

$⇒ 4a - 2b + c + 9a + 3b + c = 0$

$⇒ 13a + b + 2c = 0 . . . \left(i\right)$

Since $- 1$ is a root of $f\left(x\right) = 0$. So, $f\left(- 1\right) = 0$

$⇒ a - b + c = 0 . . . \left(ii\right)$

On solving equations \left(i\right)& \left(ii\right), we get

$11a + 3b = 0 ⇒ - \dfrac{b}{a} = \dfrac{11}{3}$

Now, we know that sum of the roots of $a{x}^{2} + bx + c = 0$ is $\dfrac{- b}{a}$.

$∴$ The sum of the roots of $f\left(x\right) = 0$ is $\dfrac{11}{3}$.