Answer
The maximum number of real zeros is $5$.
The number of positive real zeros is $0$.
The number of negative real zeros is either 3 or 1.
Work Step by Step
We should remember that the maximum number of zeros of a polynomial function $f(x)$ cannot be greater than its degree.
We will consider Descartes' Rule of Signs for explaining this solution:
(1) The number of negative real zeros of a polynomial function $f(x)$ is either equal to the number of variations in the sign of the non-zero coefficients of $f(−x)$ or that number minus an even integer.
(2) The number of positive real zeros of a polynomial function $f(x)$ is either equal to the number of variations in the sign of the non-zero coefficients of $f(x)$ or that number minus an even integer.
We can notice from the given polynomial function that the highest degree is $5$, so the maximum number of real zeros is $5$.
$f(x)=x^5+x^4+x^2+x+1$ has 0 variations in the sign. So, the number of positive real zeros is $0$.
$f(−x)=-x^5+x^4+x^2-x+1$ has 3 variations in the sign. So, the number of negative real zeros is either 3 or 1.
