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

Oct 17, 2016

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

#### Explanation:

$f \left(x\right) = 2 {x}^{4} + {x}^{2} - x + 6$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 6$

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 $f \left(x\right)$ has $0$ or $2$ positive Real zeros.

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

So the only possible remaining rational zeros are:

$\frac{1}{2} , 1 , \frac{3}{2} , 2 , 3 , 6$

We could try these, but actually there's a shortcut...

$2 {x}^{4} + {x}^{2} - x + 6 = 2 {x}^{4} + {\left(x - \frac{1}{2}\right)}^{2} + \frac{23}{4} \ge \frac{23}{4}$

for all Real values of $x$.

So all of the zeros of $f \left(x\right)$ are Complex.

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