How do you find all the real and complex roots of #3x^4+8x^3+6x^2+3x-2#?

1 Answer
Jun 2, 2016

Answer:

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

Explanation:

#f(x) = 3x^4+8x^3+6x^2+3x-2#

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 hte constant term #-2# and #q# a divisor of the coefficient #3# of the leading term.

So the only possible rational zeros are:

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

We find:

#f(1/3) = 1/27+8/27+2/3+1-2 = 0#

#f(-2) = 48-64+24-6-2 = 0#

So #x = 1/3# and #x=-2# are zeros and #(3x-1)# and #(x+2)# are factors:

#3x^4+8x^3+6x^2+3x-2#

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

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

The zeros of #x^2+x+1# are the Complex cube roots of #1#, since #(x-1)(x^2+x+1) = x^3-1#

We can find them using the quadratic formula or whatever your preferred method is to get:

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