Question 360526
<pre>
{{{f(x) = 3x^4 + 2x^3 + 4x + 7}}}

By DesCarte's rule of signs:

There are no sign changes going left to right, 
so it has 0 positive zeros

Now we find f(-x)

{{{f(-x) = 3(-x)^4 + 2(-x)^3 + 4(-x) + 7}}}

{{{f(-x) = 3x^4 + 2(-x^3) - 4x + 7}}}

{{{f(-x) = 3x^4 - 2x^3 - 4x + 7}}}

There are 2 sign changes going left to right, 
so f(x) has either 2 negative zeros or 0 negative zeros.

A fourth degree equation has 4 zeros counting multiplicity,
and its imaginary zeros come in conjugate pairs. 

So the equation either has 2 negative zeros and one pair of
conjugate imaginary zeros or else it has 0 negative zeros
and two pairs of conjugate imaginary zeros, i.e., 4 imaginary
zeros.

Edwin</pre>