Question

What are all the possible rational zeros for `f(x)=5x^3-9x^2-12x-2` and how do you find all zeros?

Answer 1

The zeros of `f(x)` are:

`-1/5" "` and `" "1+-sqrt(3)`

Explanation:

Given:

`f(x) = 5x^3-9x^2-12x-2`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-2` and `q` a divisor of the coefficient `5` of the leading term.

That means that the only possible rational zeros are:

`+-1/5, +-2/5, +-1, +-2`

First check `f(+-1)`...

`f(1) = 5-9-12-2 = -18`

`f(-1) = -5-9+12-2 = -4`

We can deduce that not all of the zeros are rational, since there are not enough factors to go around.

Trying the other possibilities, we eventually find:

`f(-1/5) = -5(1/125)+9(1/25)+12(1/5)-2`

`color(white)(f(-1/5)) = (-1-9+60-50)/25 = 0`

So `x=-1/5` is a zero and `(5x+1)` a factor:

`5x^3-9x^2-12x-2 = (5x+1)(x^2-2x-2)`

`color(white)(5x^3-9x^2-12x-2) = (5x+1)(x^2-2x+1-3)`

`color(white)(5x^3-9x^2-12x-2) = (5x+1)((x-1)^2-(sqrt(3))^2)`

`color(white)(5x^3-9x^2-12x-2) = (5x+1)(x-1-sqrt(3))(x-1+sqrt(3))`

So the zeros are:

`-1/5" "` and `" "1+-sqrt(3)`