PreCalc
posted by Anthony .
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.

PreCalc 
Reiny
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 = 4a
sub that into the first:
a  2(4a)  32 = 0
a + 8 + 2a  32 = 0
3a = 24
a = 8 , then b = 48 = 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 (x1) are factors,
so we must have
(2x+1)(x1)(.......) = 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 x1
then x = 1 or x = 1/2
confirmation:
http://www.wolframalpha.com/input/?i=solve+8x%5E3++12x%5E2+%2B+4+%3D+0
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