HMMT-Nov 2009 guts, Problem 14

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

Solution: Call the roots $-x_{1},-x_{2},-x_{3}$, and $-x_{4}$. Then $f(x)$ must factor as $\left(x+x_{1}\right)\left(x+x_{2}\right)(x+$ $\left.x_{3}\right)\left(x+x_{4}\right)$. If we evaluate $f$ at 1 , we get $\left(1+x_{1}\right)\left(1+x_{2}\right)\left(1+x_{3}\right)\left(1+x_{4}\right)=a+b+c+d+1=$ $2009+1=2010$. $2010=2 \cdot 3 \cdot 5 \cdot 67$. $d$ is the product of the four roots, so $d=(-1) \cdot(-2) \cdot(-4) \cdot(-66)$. $\fbox{528}$.

Answer: 528