According to the theorem, the possible rational roots are:
##+-1##, ##+-2##, ##+-3##, ##+-4##, ##+-6##, ##+-12##
##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.
