<s>[INST] $x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer roots in the equation is? 
Let a,b,c,d be natural numbers.Now consider an equation 
$x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer
  roots in the equation is ?

Me: Plugged in $0,1,-1$.No use.At 0 the sign is negative and at infinity the sign is positive. At least one real root.Not sure if that will be integer though :/.Help please!
 [/INST] Let $P(x) = x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a$. 
For $x \le -1$, we have $-x \le x^2$, and thus, $P'(x) = 4x^3-3(a+b+c+d)x^2-2(a+b+c)x-(a+b)$ $\le 4x^3-(a+b+c+3d)x^2-(a+b) < 0$. 
Then, since $P(-1) = 1+b+d > 0$, we have that $P(x) > 0$ for all $x \le -1$, and thus, $P(x)$ has no negative integer roots. 
Trivially, $P(0) = -a < 0$, so $0$ is not a root of $P(x)$. 
By Descartes Rule of signs, $P(x)$ has at exactly one positive root. 
However, $P(a+b+c+d) = -(a+b+c)(a+b+c+d)^2-(a+b)(a+b+c+d)-a < 0$, and since $P(x) \ge x^4-(a+b+c+d)(x^3+x^2+x+1) = x^4-(a+b+c+d)\dfrac{x^4-1}{x-1}$ for $x \ge 0$, we have $P(a+b+c+d+1) \ge 1$. 
Thus, the one positive root of $P(x)$ is not an integer. 
Therefore, $P(x)$ has no integer roots.
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