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Let $f,g$ be two integer polynomials such that the sum of the coefficients of $f$ is negative, the sum of the coefficients of $g$ is positive. Prove that there exists a unique rational number $q$ such that $qf+g\in \mathbb{Q}[x]$ is divisible by $x-1$. As of my understanding $x=1$ is a root of $qf+g\in \mathbb{Q}[x]$. But it doesn't make it zero.

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What makes you think that $x=1$ doesn't make $qf+g$ zero? Do you have an example? –  Gerry Myerson Feb 9 '13 at 8:57
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The coefficient sum is simply the value at $x=1$. So we are given that $f(1)<0<g(1)$. Now $(x-1)|qf(x)+g(x)\iff qf(1)+g(1)=0\iff q=-\frac{g(1)}{f(1)}$. The existence and uniqueness of $q$ uses only that $f(1)\ne0$.

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Notice that if $qf(x)+g(x)$ is divisible by $x-1$ then $qf(1)+g(1)=0$ so $q=-g(1)/f(1)$.

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How do we no whether $x-1$ divides $q.f(x)+g(x)$ or not? That is one of my great problem. –  ftolessa Feb 9 '13 at 9:00
    
@user61455: $x-1$ divides $qf(x)+g(x)$ iff $1$ is a root of $qf(x)+q(x)$; but with such a value for $q$, it is clearly the case. –  Seirios Feb 9 '13 at 9:03
    
The importance of negativity and positivity of the sum of the coefficients of the polynomials $f(x)$ and $g(x)$ respectively is not clear to prove the result. –  ftolessa Feb 9 '13 at 9:10
    
@user61455: In fact, it is only important that $f(1) \neq 0$; however, you can show that $q$ is postive. –  Seirios Feb 9 '13 at 9:19
    
Perhaps what OP is missing is that if $p$ is any polynomial then the sum of the coefficients of $p$ is precisely $p(1)$. –  Gerry Myerson Feb 9 '13 at 11:53
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