Find the value of $a+b+c$

Let $P(x) = x^{4} + ax^{2} + bx + c$ . From the given information, we know that $P(2) = 0$ , $P(5) = 0$ and $P(-3) = 0$

Therefore we have the system of equations

$16 + 4a + 2b + c = 0$

$625 + 25a + 5b + c = 0$

$81 + 9a - 3b + c = 0$

Solving simultaneously yields $a = -27$ , $b = -14$ and $c = 120$ , and the value of $a + b + c = 79$ .

Let $f(x)=x^4+ax^2+bx+c$ . The sum of the roots is the coefficient of $x^3$ in this, so in this case it's zero; so the fourth root is $-4$ .

So $f(x)=(x-5)(x-2)(x+3)(x+4)$ .

To find $a+b+c$ , note that $f(1)=1+a+b+c=(1-5)(1-2)(1+3)(1+4)=80$ ; so $a+b+c=\boxed{79}$ .

The sum of the roots is $0$ , hence by Vieta $-0 = 2 + 5 + -3 + r, r = -4$ .

Thus using Vieta again, $a = 2 \cdot 5 + 2 \cdot -3 + 2 \cdot -4 + 5 \cdot -3 + 5 \cdot -4 + -3 \cdot -4 = -27$ , $-b = 2 \cdot 5 \cdot -3 + 2 \cdot 5 \cdot -4 + 2 \cdot -3 \cdot -4 + 5 \cdot -3 \cdot -4 \Rightarrow b = -14$ , and $c = 2 \cdot 5 \cdot -3 \cdot - 4 = 120$ . Hence $a+b+c = -27 + -14 + 120 = 79$ .