Use the intermediate value theorem to show that the polynomial has a real zero between the given integers. \[ f(x)=3 x^{3}-5 x-2 \text {; between } 1 \text { and } 8 \] 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 \( f(1)=\square<0 \) and \( f(8)=\square>0 \), the function has a real zero between 1 and 8. B. Because \( f(x) \) is a polynomial with \( f(1)=\square<0 \) and \( f(8)=\square<0 \), the function has a real zero between 1 and 8. 8. Because \( f(x) \) is a polynomial with \( f(1)=\square>0 \) and \( f(8)=\square>0 \), the function has a real zero between 1 and C. 8. D. Because \( f(x) \) is a polynomial with \( f(1)=\square>0 \) and \( f(8)=\square<0 \), the function has a real zero between 1 and 8.

To apply the intermediate value theorem, we need to evaluate the polynomial at the given points. First, calculate \( f(1) \): \[ f(1) = 3(1)^3 - 5(1) - 2 = 3 - 5 - 2 = -4 \] Then, calculate \( f(8) \): \[ f(8) = 3(8)^3 - 5(8) - 2 = 3(512) - 40 - 2 = 1536 - 40 - 2 = 1494 \] Now we have \( f(1) = -4 < 0 \) and \( f(8) = 1494 > 0 \). Therefore, we can use the intermediate value theorem. A. Because \( f(x) \) is a polynomial with \( f(1)=-4<0 \) and \( f(8)=1494>0 \), the function has a real zero between 1 and 8. This option is correct! The intermediate value theorem states that if a continuous function takes on both negative and positive values, it must have a zero in between those points. Furthermore, what’s exciting about polynomials like this one is that they are continuous and differentiable everywhere, meaning we can always find some point where the function crosses the x-axis! So, use this property to your advantage every time you need to find roots of a polynomial!