Use the Intermediate Value Theorem to show that the polynomial $f(x)=x^{3}+x^{2}-2 x+4$ has a real zero between -4 and -2 Select the correct choice below and fill in the answer boxes to complete your choice. A. Because $f(x)$ is a polynomial with $f(-4)=\square>0$ and $f(-2)=\square>0$, the function has a real zero between -4 and -2 . B. Because $f(x)$ is a polynomial with $f(-4)=\square<0$ and $f(-2)=\square<0$, the function has a real zero between -4 and -2 . C. Because $f(x)$ is a polynomial with $f(-4)=\square<0$ and $f(-2)=\square>0$, the function has a real zero between -4 and -2 . D. Because $f(x)$ is a polynomial with $f(-4)=\square>0$ and $f(-2)=\square<0$, the function has a real zero between -4 and -2 .

Step 1 ：The Intermediate Value Theorem states that 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. In this case, we need to find the values of f(-4) and f(-2) and see if they have different signs. If they do, then by the Intermediate Value Theorem, there must be a zero between -4 and -2.

Step 2 ：Calculate the value of the function at -4 and -2: \(f(-4) = -36\) and \(f(-2) = 4\).

Step 3 ：Since \(f(-4) < 0\) and \(f(-2) > 0\), there must be a zero between -4 and -2 by the Intermediate Value Theorem.

Step 4 ：Final Answer: \(\boxed{\text{C. Because } f(x) \text{ is a polynomial with } f(-4)=-36<0 \text{ and } f(-2)=4>0, \text{ the function has a real zero between -4 and -2.}}\)