How do you find the number of complex, real and rational roots of #2x^4+x^2-x+6=0#?
This quartic equation has exactly
#f(x) = 2x^4+x^2-x+6#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/2, +-1, +-3/2, +-2, +-3, +-6#
We could try these, but let's do some more analysis first.
Note that the pattern of signs of the coefficients is:
By Descartes' Rule of Signs, since there are two changes of sign, that means that
The pattern of signs of coefficients of
So the only possible remaining rational zeros are:
#1/2, 1, 3/2, 2, 3, 6#
We could try these, but actually there's a shortcut...
#2x^4+x^2-x+6 = 2x^4+(x-1/2)^2+23/4 >= 23/4#
for all Real values of
So all of the zeros of
By the Fundamental Theorem of Algebra,