Use the intermediate value theorem to show that the polynomial has a real zero between the given integers.
$f(x)=x^3-6x+2;$ between -6 and -2
Select the correct choice below and, if necessary, fill in the answer box(es) within your choice.
(Simplify your answers.)
A. Because $f(x)$ is a polynomial with $f(-6)=◻>0$ and $f(-2)=◻<0$, the function has a real zero between -6 and -2
B. Because $f(x)$ is a polynomial with $f(-6)=◻<0$ and $f(-2)=◻<0$, the function has a real zero between -6 and -2
C. Because $f(x)$ is a polynomial with $f(-6)=◻<0$ and $f(-2)=◻>0$, the function has a real zero between -6 and -2 .
D. Because $f(x)$ is a polynomial with $f(-6)=◻>0$ and $f(-2)=◻>0$, the function has a real zero between -6 and -2 .

Step 1 ：Given the polynomial function $f(x)=x^3-6x+2$, we are asked to determine if there is a real zero between -6 and -2.

Step 2 ：First, we calculate the values of $f(-6)$ and $f(-2)$.

Step 3 ：$f(-6)=(-6{)}^3-6\ast (-6)+2=-178$ and $f(-2)=(-2{)}^3-6\ast (-2)+2=6$.

Step 4 ：According to the Intermediate Value Theorem, if a function is continuous on a closed interval [a, b], and k is any number between f(a) and f(b), then there is at least one number c in the interval (a, b) such that f(c) = k.

Step 5 ：Since $f(-6)$ is negative and $f(-2)$ is positive, there must be a zero between -6 and -2 according to the Intermediate Value Theorem.

Step 6 ：$\boxed{{\displaystyle \text{Final Answer: A. Because }f(x)\text{ is a polynomial with }f(-6)=-178<0\text{ and }f(-2)=6>0,\text{ the function has a real zero between -6 and -2.}}}$