Answer
Maximum number of real zeros: $7$
The number of positive real zeros is either $3$ or $1$
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 $7$.
Since:
$$f\left( x\right) =-4x^{7}+x^{3}-x^{2}+2$$ has $3$ variations
The number of positive real zeros is either $3$ or $1$
Since
$$f\left( -x\right) =4x^{7}-x^{3}-x^{2}+2$$
Has $2$ variations
The number of negative real zeros is either $2$ or $0$.