In this question a polynomial means a polynomial with real coefficients. $$P(x)$$ is a polynomial of degree 2006 and $$P(x) \ge 0$$ for all real $$x$$. Then [ ]

If $$P(x)$$ had 2006 distinct real roots, every root would be simple, and a polynomial changes sign at a simple root, so $$P(x)$$ would be negative somewhere, a contradiction; hence (b) is true. The polynomial $$P(x) = x^{2006}$$ is non-negative, has no non-real root and is not a product of quadratics without linear factors, so (a) and (c) fail. The polynomial $$P(x) = (x-k)^{2006}$$ is non-negative and has the factor $$(x-k)^{2005}$$, so (d) fails. The answer is (b).