How do you use the rational roots theorem to find all possible zeros of #f(x)=3x^3-17x^2+18x+8#?

1 Answer
Jun 5, 2016

Answer:

#x = -1/3#, #x = 2#, #x = 4#

Explanation:

#f(x) = 3x^3-17x^2+18x+8#

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 #8# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3#, #+-2/3#, #+-1#, #+-4/3#, #+-2#, #+-8/3#, #+-4#, #+-8#

We find:

#f(-1/3) = -1/9-17/9-6+8 = 0#

So #x=-1/3# is a zero and #(3x+1)# a factor of #f(x)#

#3x^3-17x^2+18x+8#

#= (3x+1)(x^2-6x+8)#

#= (3x+1)(x-2)(x-4)#

Hence the other two zeros are #x=2# and #x=4#