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
The number of zeros of a polynomial can’t be greater than its degree.
$1)$ The number of positive real zeros of $f(x) $ either equals the number of variations in the sign of the nonzero coefficients of $f(x)$ or equals that number minus an even integer.
$2)$ The number of negative real zeros of $f(x)$ either equals the number of variations in the sign of the nonzero coefficients of $f(-x) $ or equals that number minus an even integer.
So, the maximum number of real zero here is $6$.
Since
$$f\left( x\right) =2x^{6}-3x^{2}-x+1$$ has $2$ variations.
The number of positive real zeros is either $2$ or $0$
Since
$$f\left( -x\right) =2x^{6}-3x^{2}+x+1 $$
has $2$ variations.
The number of negative real zeros is either $2$ or $0$.
