let f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c =
"f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c = f ( 5 ) = 0 = 25 + 5 b + c - - - taking 25 to the other side - > 5 b + c = - 25 f ( - 3 ) = 0 = 9 - 3 b + c - - - taking - 3 b + c to the other side - > 3 b - c = 9 when we add these 2 equations , we get 8 b = - 16 - - - > b = - 2 and while substituting b = - 2 we get c = - 15 . b + c = - 17 - - - answer b"
negate(divide(subtract(power(3, 2), 5), add(3, 5)))