Question

Using the rational root theorem, what are the possible rational roots of `x^3-34x+12=0` ?

Answer 1

According to the theorem, the possible rational roots are:

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

Explanation:

`f(x) = x^3-34x+12`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integeres `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 eventually find that:

`f(color(blue)(-6)) = (color(blue)(-6))^3-34(color(blue)(-6))+12`

`color(white)(f(color(white)(-6))) = -216+204+12`

`color(white)(f(color(white)(-6))) = 0`

So `x=-6` is a rational root.

The other two roots are Real but irrational.