Quadratic Function Roots and Coefficients

$f(5) = -3f(2) \implies f(5) + 3f(2) = 0$. $(25a + 5b + c) + 3(4a + 2b + c) = 0 \implies 37a + 11b + 4c = 0 \quad \text{--- (i)}$

Also $3$ is a root, so $f(3) = 0 \implies 9a + 3b + c = 0 \quad \text{--- (ii)}$.

From (i) and (ii), we get $b = a$ and $c = -12a$.

Thus $f(x) = a(x^2 + x - 12) = a(x + 4)(x - 3)$.

The roots of $f(x) = 0$ are $3$ and $-4$. The other root is $-4$.

$f(x) = a(x^2 + x - 12)$. Here $b = a$ and $c = -12a$. So $a + b + c = a + a - 12a = -10a$.

Since $a$ is a non-zero constant whose exact value is not specified, $a + b + c$ cannot be uniquely determined.