Question

What are all the possible rational zeros for `f(x)=8x^3-2x+24` and how do you find all zeros?

Answer 1

`f(x)` has zeros:

`x_1 = -3/2`

`x_(2,3) = 3/4+-sqrt(23)/4i`

Explanation:

`f(x) = 8x^3-2x+24 = 2(4x^3-x+12)`

By the rational roots theorem, any rational zeros of `4x^3-x+12` are expressible in the form `p/q` for integers `p,q` with `p` a divisor of the constant term `12` and `q` a divisor of the coefficient `4` of the leading term.

So the only possible rational zeros of `4x^3-x+12` and thus `f(x)` are:

`+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-4, +-6, +-12`

By Descartes' rule of signs we can tell that `f(x)` has `0` or `2` positive zeros and `1` negative zero.

So let's try the negative possibilities first:

`f(-1/4) = 2(4(-1/64)-(-1/4)+12) = 2(-1/16+1/4+12) != 0`

`f(-1/2) = 2(4(-1/8)-(-1/2)+12) = 2(-1/2+1/2+12) = 24 != 0`

`f(-3/4) = 2(4(-27/64)-(-3/4)+12) = 2(-27/16+3/4+12) != 0`

`f(-1) = 2(-4+1+12) = 18 != 0`

`f(-3/2) = 2(4(-27/8)-(-3/2)+12) = 2(-27/2+3/2+24/2) = 0`

So `x=-3/2` is a zero and `(2x+3)` a factor:

`4x^3-x+12 = (2x+3)(2x^2-3x+4)`

The remaining quadratic factor has zeros given by the quadratic formula:

`x = (3+-sqrt((-3)^2-4(2)(4)))/(2*2)`

`=(3+-sqrt(9-32))/4`

`=(3+-sqrt(-23))/4`

`=(3+-sqrt(23)i)/4`

`=3/4+-sqrt(23)/4i`