Question

Find the values of a and b if f(x)=ax^3+bx^2+4 is exactly divisible by both (2x+1) and (x–1). With these values of a and b, solve the equation f(x)=0.

Answer 1

I assume you have studied the remainder theorem.
Then you must know that for your conditions,
f(-1/2) = (-1/8)a + (1/4)b + 4 = 0 ---> a - 2b - 32 = 0
f(1) = a + b + 4 = 0 ----> b = -4-a

sub that into the first:
a - 2(-4-a) - 32 = 0
a + 8 + 2a - 32 = 0
3a = 24
a = 8 , then b = -4-8 = -12

then your function is f(x) = 8x^3 - 12x^2 + 4

now solving this:
8x^3 - 12x^2 + 4 = 0
2x^3 - 3x^2 + 1 = 0
note , we already know that (2x+1) and (x-1) are factors,
so we must have
(2x+1)(x-1)(.......) = 2x^2 - 3x^2 + 1
looking at the front multiplications (2x)(x)(?) = 2x^3,
(?) = x
looking at the multiplication of the constants, (1)(-1)(?) = +1, (?) = -1
so our third factor is also x-1

then x = 1 or x = -1/2

confirmation:
http://www.wolframalpha.com/input/?i=solve+8x%5E3+-+12x%5E2+%2B+4+%3D+0