Question

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

Answer 1

This quartic equation has exactly `4` non-Real Complex roots.

Explanation:

`f(x) = 2x^4+x^2-x+6`

By the rational roots theorem, any rational zeros of `f(x)` 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 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 `f(x)` has `0` or `2` positive Real zeros.

The pattern of signs of coefficients of `f(-x)` is `+ + + +`. With no changes of sign, that means that `f(x)` has no negative Real zeros.

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 `x`.

So all of the zeros of `f(x)` are Complex.

By the Fundamental Theorem of Algebra, `f(x)` has exactly `4` zeros since it is of degree `4`.