Question 11 • [INMT3] Integrated Math 3 Sem 1 (26-27)
For a polynomial with real coefficients, any nonreal complex root must occur with its conjugate. So if –5 + 2i is a root, –5 – 2i is also a root. These are two nonreal roots, and a third-degree polynomial has one root left; because nonreal roots come in conjugate pairs, that remaining root must be real. Thus the function has two imaginary (nonreal complex) roots and one real root: choice B.
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