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Algebra Difficulty 4.7 AIME Find the answer

Let f(x)=x4+ax3+bx2+cx+df(x)=x^{4}+a x^{3}+b x^{2}+c x+d be a polynomial whose roots are all negative integers. If a+b+c+d=2009a+b+c+d=2009, find dd.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Call the roots −x1,−x2,−x3-x_{1},-x_{2},-x_{3}, and −x4-x_{4}. Then f(x)f(x) must factor as (x+x1)(x+x2)(x+x3)(x+x4)(x+x_{1})(x+x_{2})(x+x_{3})(x+x_{4}). If we evaluate ff at 1, we get (1+x1)(1+x2)(1+x3)(1+x4)=a+b+c+d+1=2010.2010=2⋅3⋅5⋅67(1+x_{1})(1+x_{2})(1+x_{3})(1+x_{4})=a+b+c+d+1=2010.2010=2 \cdot 3 \cdot 5 \cdot 67. dd is the product of the four roots, so d=(−1)⋅(−2)⋅(−4)⋅(−66)d=(-1) \cdot(-2) \cdot(-4) \cdot(-66).

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