When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 14, 0, and –16 respectively. Find the values of a, b, and c.
+130116
We have the following system of equations
a + b + c = 13 (1)
-a - b + c = -1 (2)
-8a -2b + c = -32 (3)
Adding (1) and (2) we have that 2c = 12 ⇒ c = 6
So (2) and (3) become
-a - b = -7 ⇒ -2a - 2b = - 14 ⇒ 2a + 2b = 14 (4)
-8a -2b = -38 (5)
Add (4) and (5) and we have that
-6a = -24 ⇒ a = 4
So
a + b + c = 13
4 + b + 6 = 13
b = 3
So a = 4, b = 3 and = 6
