How do you use the rational roots theorem to find all possible zeros of #x^3-3x-2#?

1 Answer
Jun 2, 2016

Answer:

#x^3-3x-2# has zeros #x=-1# with multiplicity #2# and #x=2#.

Explanation:

#f(x) = x^3-3x-2#

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 #-2# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#

We find:

#f(-1) = -1+3-2 = 0#

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

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

#-1# is also a zero of #x^2-x-2#:

If #x=-1# then #x^2-x-2 = 1+1-2 = 0#

Then we find:

#x^2-x-2 = (x+1)(x-2)#

So #x=2# is the last zero.