# Detecting errors changing an odd number of bits using CRC

I was studying CRC from lecture notes of Schwarzkopf but I am stuck at this statement:

If $$G(x)$$ is a factor of $$E(x)$$, then $$G(1)$$ would also have to be $$1$$.

What does this statement means? It is the third point in the third last block of the article whose link is given. I know a similar question is asked but that is a specific question. I want to understand this particular statement.

• Suppose that $E(x) = G(x) H(x)$ and $E(x) = 1$. Then also $G(x) = 1$, since we're working modulo 2. Feb 17 '21 at 13:51

Recall that we are working modulo $$2$$. Thus $$E(x)$$ is a polynomial whose coefficients are $$0,1$$, and $$E(1) \in \{0,1\}$$.

By definition, $$G(x)$$ is a factor of $$E(x)$$ if there exists a polynomial $$H(x)$$ such that $$E(x) = G(x) H(x)$$.

In this case, we assume that $$E(1) = 1$$. Since $$E(1) = G(1) H(1)$$, this forces $$G(1) = 1$$.