How do you use the rational roots theorem to find all possible zeros of #f(x)=2x^3-7x^2-10x-3#?

1 Answer
Aug 8, 2016

Answer:

#f(x)# has zeros #-1/2# and #2+-sqrt(7)#

Explanation:

#f(x) = 2x^3-7x^2-10x-3#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-3# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

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

We find:

#f(-1/2) = 2(-1/2)^3-7(-1/2)^2-10(-1/2)-3#

#=-1/4-7/4+10/2-3=0#

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

#2x^3-7x^2-10x-3#

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

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

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

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

Hence the two other zeros are:

#x=2+-sqrt(7)#