How do you use the rational roots theorem to find all possible zeros of #P(x)=2x^3+9x^2+11x-8#?

1 Answer
Aug 12, 2016

Answer:

#P(x)# has zeros #1/2# and #5/2+-sqrt(7)/2i#

Explanation:

#P(x) = 2x^3+9x^2+11x-8#

By the rational roots theorem, any rational zeros of #P(x)# are 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 #2# of the leading term.

That means that the only possible rational zeros are:

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

We find:

#P(1/2) = 2(1/8)+9(1/4)+11(1/2)-8 = (1+9+22-32)/4 = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#2x^3+9x^2+11x-8#

#=(2x-1)(x^2+5x+8)#

#=(2x-1)((x-5/2)^2-25/4+8)#

#=(2x-1)((x-5/2)^2+7/4)#

#=(2x-1)((x-5/2)^2-(sqrt(7)/2i)^2)#

#=(2x-1)(x-5/2-sqrt(7)/2i)(x-5/2+sqrt(7)/2i)#

Hence the other two zeros are:

#x = 5/2+-sqrt(7)/2i#