Answer
No positive real zero, 3 or 1 negative real zeros.
Work Step by Step
According to Descartes’s rule, for positive real zeros count the number of sign changes of the coefficients in the polynomial function (I).
$f\left( x \right)=\,{{x}^{3}}+2{{x}^{2}}+5x+4$
In the above polynomial function, the number of sign changes of the coefficients is equal to zero.
So there are 0 positive real zeros.
Similarly, for negative real zeros firstly replace $x$ from $-x$ in equation (I) and then count the sign changes in the coefficient of the polynomial.
$f\left( x \right)=\,{{x}^{3}}+2{{x}^{2}}+5x+4$
Replace x with -x,
$\begin{align}
& f\left( -x \right)=\,{{(-x)}^{3}}+2{{(-x)}^{2}}+5(-x)+4 \\
& f\left( -x \right)=-{{x}^{3}}+2{{x}^{2}}-5x+4 \\
\end{align}$
Here, the number of sign changes in the coefficient of the polynomial is equal to 3.
So, the negative real zeros are either equal to 3 or are less than this number by an even integer. This means that there are either 3 negative real zeros or 3-2=1 negative real zeros.
