What are the possible rational roots of #2x^4-x^3-6x+3=0# and then determine the rational roots?

1 Answer
Dec 30, 2016

Answer:

The "possible" rational roots are:

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

The only actual rational root is:

#1/2#

The other roots are the three cube roots of #3#.

Explanation:

Given:

#2x^4-x^3-6x+3 = 0#

By the rational roots theorem, any rational zeros of this quartic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #6# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational roots are:

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

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this quartic will factor by grouping:

#0 = 2x^4-x^3-6x+3#

#color(white)(0) = (2x^4-x^3)-(6x-3)#

#color(white)(0) = x^3(2x-1)-3(2x-1)#

#color(white)(0) = (x^3-3)(2x-1)#

Hence the only rational zero is #x=1/2# and the remaining three zeros are cube roots of #3#:

#x = root(3)(3)#

#x = omega root(3)(3)#

#x = omega^2 root(3)(3)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.