Answer
The maximum number of real zeros is $3$
The number of positive real zeros is $1$
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 zeros here is $3$
Since
$$f\left( x\right) =-x^{3}-x^{2}+x+1 $$ has $1$ variation
The number of positive real zeros is $1$
Since
$$ f\left( -x\right) = x^{3}-x^{2}-x+1 $$
Has $2$ variations
Number of negative real zeros is either $2$ Or $0$
