# How do you find the number of complex, real and rational roots of #2x^4+x^2-x+6=0#?

##### 1 Answer

This quartic equation has exactly

#### Explanation:

#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,