How do you use the rational roots theorem to find all possible zeros of #f(x) = 3x^3 + 39x^2 + 39x + 27#?

1 Answer
Mar 25, 2016

Answer:

See explanation...

Explanation:

Before applying the rational roots theorem, note that all of the coefficients are divisible by #3#, so separate that out as a factor first:

#f(x) = 3x^3+39x^2+39x+27 = 3(x^3+13x^2+13x+9)#

Then applying the rational roots theorem to the remaining cubic factor, we can deduce that any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p#, #q# where #p# is a divisor of the constant term #9# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-3#, #+-9#

In addition note that all of the coefficients are positive, so there are no zeros for positive values of #x#. So the only possible rational zeros are:

#-1#, #-3#, #-9#

None of these is a zero, so #f(x)# has no rational zeros.

That is as much as we can learn from the rational roots theorem.

In fact #f(x)# has one Real irrational zero near #-12# and a pair of Complex zeros.