How do you find all the zeros of #f(x)=2x^3-x^2-3x-1#?

1 Answer
Aug 2, 2016

Answer:

Zeros: #-1/2# and #1/2+-sqrt(5)/2#

Explanation:

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

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

That means that the only possible rational zeros of #f(x)# are:

#+-1/2, +-1#

We find:

#f(-1/2) = 2(-1/8)-(1/4)-3(-1/2)-1 = -1/4-1/4+3/2-1 = 0#

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

#2x^3-x^2-3x-1#

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

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

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

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

Hence the other two zeros are:

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