How do you find all the zeros of #P(x) = 3x^3 – 6x^2 – 3x – 18#?

1 Answer
Aug 7, 2016

Answer:

#P(x)# has zeros: #3# and #-1/2+-sqrt(7)/2i#

Explanation:

#P(x) = 3x^3-6x^2-3x-18 = 3(x^3-2x^2-x-6)#

By the rational root theorem, the only possible rational zeros of #x^3-2x^2-x-6# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constnat term #-6# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-6#

Substituting #x=3# we find:

#x^3-2x^2-x-6=(3)^3-2(3)^2-(3)-6=27-18-3-6 = 0#

So #x=3# is a zero and #(x-3)# a factor:

#x^3-2x^2-x-6#

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

#=(x-3)((x+1/2)^2+7/4)#

#=(x-3)((x+1/2)^2-(sqrt(7)/2i)^2)#

#=(x-3)(x+1/2-sqrt(7)/2i)(x+1/2+sqrt(7)/2i)#

So the other two zeros are:

#x = -1/2+-sqrt(7)/2i#