What are the possible rational roots of #2x^4-x^3-6x+3=0# and then determine the rational roots?
1 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
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
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 = root(3)(3)#
#x = omega root(3)(3)#
#x = omega^2 root(3)(3)#
where