Has the polynomial $z^4 + z^3 + 4z^2 + 2z + 3$ any real zero?

Question

I want to determine in which quadrants the zeros of the polynomial $z^4 + z^3 + 4z^2 + 2z + 3$ are. The argument principle comes to my mind, but to be able to apply that theorem, I must make sure that the polynomial doesn't have any zeros on the negative real axis. (Obviously it doesn't have any zeros on the positive real axis.) Is there an easy way of excluding the possibility of negative real zeros? The only way I can think of is cumbersome at best and flawed at worst.

Answer 1

Method-1:

You could use the second derivative test to prove that the function has a local minima at the point $(-0.276,2.737)$.

Since the local minima is greater than zero, this leads one to assume that the function does not cross the x-axis (in which case, its local minima would have ben less than or equal zero), hence, the function has no real roots.

Details: $$f(z)=z^4 + z^3 + 4z^2 + 2z + 3$$ $$f'(z)=4z^3+3z^2+8z+2$$ $$f''(z)=12z^2+6z+8$$

When $f'(z)=0$, leads to single real root at $z=-0.276$. This root may be found in any way including the Cardano's formula, or by other means. Now $f''(-0.276)=7.258112>0$, hence there is a local minima for $f(z)$ at $z=-0.276$.

Method-2:

You could use the Quartic Discriminant approach. When you do the calculations, you get the value of the Discriminant=$229$, which is positive. This indicates that $f(z)$ has either all roots as real numbers or all roots are complex number. Since the rational roots test for $f(z)$ fails, the roots of $f(z)$ are all complex numbers.

Finally, you can check Wolfram Alpha to obtain the exact 4 complex root values.

I hope this makes sense.