Question

How do you use the rational roots theorem to find all possible zeros of `f(x)=x^3-5x^2+2x+12`?

Answer 1

`x=3` or `x=1+-sqrt(5)`

Explanation:

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

By the rational root theorem, any rational zeros of `f(x)` must be 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 `1` of the leading term.

That means that the only possible rational zeros are:

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

Trying each in turn we find:

`f(3) = 27-45+6+12 = 0`

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

`x^3-5x^2+2x+12`

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

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

`= (x-3)((x-1)^2-(sqrt(5))^2)`

`= (x-3)(x-1-sqrt(5))(x-1+sqrt(5))`

So the remaining two zeros are:

`x = 1+-sqrt(5)`