When f(x)=2x^3+ax^2+bx+c is divided by (x^2-1) the remainder is x+1. If (x-2) is a factor of f(x), find the values of the constants a,b and c.

Question

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Answer 1

since x^2 - 1 = (x+1)(x-1)

for both f(1) and f(-1) we get a remainder of x+1

f(1) = 2+a+b+c = 1+1
a+b+c = 0 (#1)

f(-1) = -2+a-b+c = 0
a-b+c = 2 (#2)

also x-2 is a factor, so f(2) = 0
f(2) = 16+4a+2b+c = 0
4a + 2b+c = -16 (#3)
#1 - #2 ---> 2b=-2 , b = -1

new #3 : 4a -2 + c = -16 or 4a + c = -14

new #1 : a - 1 + c = 0 or a + c = 1

new #3 - new #1
3a = -15
a = -5
in #1
-5 + c = 1
c = 6

original #1
a+b+c=0
a-1+6=0
a=-5

a= -5
b= -1
c= 6

Answer 2

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