Polynomial question

Let $f(x)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+...+{ a }_{ 1 }x+{ a }_{ 0 }$ .

If $x-1$ divides $f$ then there exists a polynomial $g$ so that $f(x)=g(x)(x-1)$ .

But $f(1)=g(1)(1-1)=0$ and $f(1)={ a }_{ n }+{ a }_{ n-1 }+...+{ a }_{ 1 }+{ a }_{ 0 }$ .

Which means that ${ a }_{ n }+{ a }_{ n-1 }+...+{ a }_{ 1 }+{ a }_{ 0 }=0$ .

Look at the simplest polynomial you can think of that is divisible by x–1, which would be $x^2 - x$ . The sum of the coefficient is 0.