When x^4 + ax^3 + bx +c is divided by (x–1), (x+1), and (x+2), the remainders are 14, 0, and –18 respectively. Find the values of a, b, and c.
+128474 We have that
(1)^4 + a(1)^3 + b + c = 14
(-1)^4 + a(-1)^3 + b(-1) + c = 0
(-2)^4 + a(-2)^3 + b(-2) + c = -18 simplify
1 + a + b + c = 14
1 - a - b+ c = 0
16 - 8a -2b + c = -18
a + b + c = 13 (1)
-a - b + c = -1 (2)
-8a - 2b + c = -34 (3)
Adding (1) and (2) we get that
2c= 12
c = 6
Multiply (1) by 2 and we have
2a + 2b +2c = 26
2a + 2b + 2(6) = 26
2a + 2b = 14 (4)
And (3) becomes
-8a - 2b + 6 = -34
-8a - 2b = -40 (5)
Add (4) and (5)
-6a = -26
a = 26/6 = 13/3
And
a + b + c = 13
(13/3) + b + 6 =13
31/3 + b = 39/3
b = 8/3
{a, b, c } = { 13/3, 8/3 , 6 }
