Use the intermediate value theorem to show that the polynomial has a real zero between the given integers.
$f(x)=x^{3}-6 x+2 ;$ 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 $f(-6)=\square> 0$ and $f(-2)=\square< 0$, the function has a real zero between -6 and -2
B. Because $f(x)$ is a polynomial with $f(-6)=\square< 0$ and $f(-2)=\square< 0$, the function has a real zero between -6 and -2
C. Because $f(x)$ is a polynomial with $f(-6)=\square< 0$ and $f(-2)=\square> 0$, the function has a real zero between -6 and -2 .
D. Because $f(x)$ is a polynomial with $f(-6)=\square> 0$ and $f(-2)=\square> 0$, the function has a real zero between -6 and -2 .

Step 1 ：Given the polynomial function \(f(x)=x^{3}-6 x+2\), 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 ：\(f(-6) = (-6)^{3}-6*(-6)+2 = -178\) and \(f(-2) = (-2)^{3}-6*(-2)+2 = 6\).

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