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

1 Answer
George C.
Oct 6, 2016

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.

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