Three roots of f(x)=x^4−2(x^3)+a(x^2)+bx+c are −5, −3 and 4. What is the value of a+b+c?

f(-5) = 625 + 250 + 25a - 5b + c = 0

25a - 5b + c = - 875--#1

f(-3) = 81 + 54 + 9a - 3b + c = 0
9a - 3b + c = -135 --#2

f(4) = 256 - 128 + 16a + 4b + c = 0
16a + 4b + c = -128 --#3

#1-#3 --> 9a - 9b = -747
a - b = -83 --- #4

#1-#2 --> 16a -2b = -740
8a - b = -370 ---#5

#5 - #4 ---> 7a = -287
a = -41

in #4:
-41 - b = -83
b = 42

in #3
-656 +168 + c = -128
c = 360

a+b+c = -41 +42 +360 = 361