How do you find all the zeros of #f(x)= 2x^3 + 3x^2+ 8x- 5#?

1 Answer
Aug 3, 2016

Answer:

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

Explanation:

#f(x) = 2x^3+3x^2+8x-5#

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 #-5# and #q# a divisor of the coefficient #2# of the leading term.

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

#+-1/2, +-1, +-5/2, +-5#

We find:

#f(1/2) = 2/8+3/4+8/2-5 = 1/4+3/4+4-5 = 0#

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

#2x^3+3x^2+8x-5#

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

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

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

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

Hence the other two zeros are:

#x = -1+-2i#