Final answer:
The correct polynomial with zeros -5, 10i, and -10i is x³ + 5x² + 100x + 500. None of the provided options are correct since they have a leading term of 0x², which should not be the case.
Step-by-step explanation:
To find a polynomial with real coefficients that has the zeros -5, 10i, and -10i, we use the fact that if a polynomial has real coefficients, then its complex zeros must come in conjugate pairs. Hence, the polynomial will have (x + 5) as a factor for the zero -5, and (x - 10i)(x + 10i) for the zeros 10i and -10i.
First, let's multiply the factors for the imaginary zeros: (x - 10i)(x + 10i) = x² - (10i)² = x² + 100 since (10i)² = -100. Next, we multiply this by the factor for the real zero to get the polynomial: (x + 5)(x² + 100), which equals x³ + 5x² + 100x + 500.
The correct polynomial with the given zeros is thus x³ + 5x² + 100x + 500, which matches none of the given options because they all have a leading coefficient of 0 for the x² term, which is incorrect. The polynomial should not have a leading term of 0x², so none of the given options are correct.