TITLE: Let $f(x) = x^4 + ax^3 + bx^2 + cx + d$ be a polynomial whose roots are all negative integers. If $a + b + c + d = 2009,$ find $d.$
QUESTION [7 upvotes]: Let $f(x) = x^4 + ax^3 + bx^2 + cx + d$ be a polynomial whose roots are all negative integers. If $a + b + c + d = 2009,$ find $d.$

How should I approach this problem? I have tried to think of a way I could plug some value of $x$ in, but they haven't given us anything about that so that wouldn't work...how should I do it?

REPLY [7 votes]: Let the roots be $-n_1\geq-n_2\geq-n_3\geq-n_4$ where $0<n_1\leq n_2\leq n_3\leq n_4$ are positive integers. Then $f(x)=(x+n_1)(x+n_2)(x+n_3)(x+n_4)$
$$f(1)=1+a+b+c+d=2010\\\implies(1+n_1)(1+n_2)(1+n_3)(1+n_4)=2\times3\times5\times67\\\implies n_1=1,n_2=2,n_3=4,n_4=66\implies d=n_1n_2n_3n_4=528$$
The solution is unique as any other factorization of $2010$ (eq:$1\times5\times6\times 67$) implies $1+n_1=1$. However, $n_1>0$ and hence, this case does not arise.