Use the intermediate value theorem to show that the polynomial function has a real zero between the numbers given. \( x^{4}-6 x^{3}-30 x^{2}+48 x+82 ;-2 \) and -1 \( f(-2)=\square \) (Simplify your answer. Type an integer or a fraction.)

To apply the intermediate value theorem, we first need to calculate the values of the polynomial function \( f(x) = x^{4} - 6x^{3} - 30x^{2} + 48x + 82 \) at the endpoints of the interval, which are \( x = -2 \) and \( x = -1 \). Calculating \( f(-2) \): \[ f(-2) = (-2)^{4} - 6(-2)^{3} - 30(-2)^{2} + 48(-2) + 82 = 16 + 48 - 120 - 96 + 82 = 16 + 48 - 120 - 96 + 82 = -70 \] Thus, \( f(-2) = -70 \). Next, let's compute \( f(-1) \): \[ f(-1) = (-1)^{4} - 6(-1)^{3} - 30(-1)^{2} + 48(-1) + 82 = 1 + 6 - 30 - 48 + 82 = 1 + 6 - 30 - 48 + 82 = 11 \] So, \( f(-1) = 11 \). Now we have \( f(-2) = -70 \) and \( f(-1) = 11 \). Since \( f(-2) < 0 \) and \( f(-1) > 0 \), by the intermediate value theorem, there must be at least one real zero of the polynomial function in the interval \( (-2, -1) \).