# Prove that a polynomial `p(x)` is divisible by `(x-1)` if the sum of coefficients of `p(x)` is zero.

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Let the polynomial be

`p(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0.`

We see that

`p(1)=a_n(1)^n+...+a_1(1)+a_0=a_n+...+a_1+a_0.`

According to the assumption that the coefficients add to zero, we have `p(1)=0.`

If `p(1)=0,` then by the Factor Theorem, `(x-1)` divides `p(x).`

**Sources:**

first if `(x-1) |p(x)` is that: `p(x)=(x-1)q(x)` ,so that:

`p(1)=0`

and `p(x)=sum_(k=0)^n a_k x^k` (1)

gets:

`p(1)=sum_(k=0)^na_k=0` (2)

On the other side, if `p(x) ` verifies (2) it means `p(1)=0`

So we know from polynomial rules:

`p(x)-p(1)=q(x)(x-1)` (3)

Being: `p(1)=0` :

`p(x)=q(x)(x-1)` that means `(x-1)|p(x)`

so we can affirm:

A polynom `p(x)=sum_(k=0)^n a_kx^k`

verifies: `sum_(k=0)^na_k=0` if only if, `(x-1)|p(x)`