Answer
The maximum number of real zeros is $6$.
The number of positive real zeros is either 2 or 0.
The number of negative real zeros is either 2 or 0.
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 $6$, so the maximum number of real zeros is $6$.
$f(x)=2x^6-3x^2-x+1$ has 2 variations in the sign. So, the number of positive real zeros is either 2 or 0.
$f(−x)=2x^6-3x^2+x+1$ has 2 variations in the sign. So, the number of negative real zeros is either 2 or 0.
