Let P(x) be a polynomial with degree 2008 and leading coefficient 1 such that P(0)=2007,P(1)=2006,P(2)=2005,…,P(2007)=0. Determine the value of P(2008). You may use factorials in your answer.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Consider the polynomial Q(x)=P(x)+x−2007. The given conditions tell us that Q(x)=0 for x=0,1,2,…,2007, so these are the roots of Q(x). On the other hand, we know that Q(x) is also a polynomial with degree 2008 and leading coefficient 1 . It follows that Q(x)=x(x−1)(x−2)(x−3)⋯(x−2007). Thus P(x)=x(x−1)(x−2)(x−3)⋯(x−2007)−x+2007 Setting x=2008 gives the answer.
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