Answer
a) The degree of the given function $f(x)$ is $7$, so it will have a maximum number of real zeros of $7$.
b) The number of positive real zeros is either $3$ or $1$.
c) The number of negative real zeros is either $2$ or $0$.
Work Step by Step
Remember that the maximum number of zeros of a polynomial function $f(x)$ can not be greater than its degree.
Consider Descartes' Rule of Signs:
a) The number of positive real zeros of a function $f(x) $ will be 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.
b) The number of negative real zeros of $f(x)$ will be 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.
Since the degree of the given function $f(x)$ is $7$, it will have a maximum number of real zeros of $7$.
Now, $f( x) =-4x^{7}+x^{3}-x^{2}+2$ attains $3$ variations in the sign of the non-zero coefficients of $f(x)$. So, the number of positive real zeros is either $3$ or $1$
Next, $f( -x) =4x^{7}-x^{3}-x^{2}+2$ attains $3$ variations in the sign of the non-zero coefficients of $f(x)$. So, the number of negative real zeros is either $2$ or $0$.
