If x ^ 2 + X-2 and 2x-1 are factors of ax ^ 3 + BX ^ 2 + cx-5 and ax ^ 3 + BX ^ 2 + x-25 / 16 respectively, find a + B + C

If x ^ 2 + X-2 and 2x-1 are factors of ax ^ 3 + BX ^ 2 + cx-5 and ax ^ 3 + BX ^ 2 + x-25 / 16 respectively, find a + B + C

Let (x ^ 2 + X-2) * (MX + n) = ax ^ 3 + BX ^ 2 + cx-5, then there are two relations a + B + C = 2n, 2n = 5, so a + B + C = 5

Simplify ax ^ 2-3ax = 0

ax^2-3ax=0
ax（x-3）=0
When a = 0, the solution is all real numbers
When a ≠ 0, X1 = 0, X2 = 3

Simplification: 1 x (x + 1) + 1 (x + 1) (x + 2) + 1 (x + 2) (x + 3) + 1 (x + 3) (x + 4)

The original formula is: 1 X-1 x + 1 + 1 x + 1-1 x + 2 + 1 x + 2-1 x + 3 + 1 x + 3-1 x + 4 = 1 X-1 x + 4 = 4 x (x + 4)