When we are finding a root to a polynomial say f(x), we put some constant

Question

When we are finding a root to a polynomial say $f(x)$, we put some constant $a=x$ and check whether it is zero or not.Suppose we have successfully found a constant say $a$, then we say that $x-a=0$ is a factor and it divides $f(x)$, but then how can $0$ divide a polynomial?

Answer 1

Well, if a polynomial $f(x)$ over the field $K$ has a zero $a\in K$, then $x-a$ divides $f(x).

Indeed, write with the division algorithm: $f(x) = q(x) \cdot (x-a) + r(x)$, where the remainder $r(x)=r$ is a polynomial of degree zero (less than the degree of $x-a$) or it is the zero polynomial (if you assign no degree for the zero).

Then $0=f(a) = q(a)(a-a) + r = r$ and so $r=0$.

Answer 2

There are two basic issues to clear here.

Let me clear the one that is more fundamental, namely your wondering how $0$ can divide a polynomial. Before polynomials, let us look at the integers. Note that in rings (which are just structures like the system of integers, or polynomials) the operation divides is not the same as the division operation in fields (like the system of rational numbers or rational functions). This is why in rings $0$ may divide $0$ (although $0$ cannot divide into a nonzero element of the ring), whereas in fields we cannot divide by $0.$ We usually say that the integer $B$ divides the integer $A$ provided there is an integer $q$ such that the equation $$A=Bq$$ is true. This is why $0$ may divide $0$ since we can write $0=0B.$

The same claim follows if instead of integers you regard the symbols as representing polynomials.

Now the second issue to clear out is that although $0$ can divide $0,$ hence we have $f(a)=Q(x-a)$ when $x=a,$ provided $f(a)=0,$ but the claim of the factor theorem is not that $x-a$ is the zero polynomial. So while it is true (by the explanation in the first paragraph) that $x-a$ divides into $f(x)$ at the particular point when $x=a$ and $f(x)=0,$ the claim of the factor theorem is that this factorisation is still valid even when $x\ne a.$

That is, we have that $f(x)=Q(x-a)$ for all $x,$ in particular for values of $x\ne a.$ This is simply because the equation

$$f(x)=Q(x-a)+r,$$ from the euclidean process, is actually an identity, so that it is valid for all values of $x.$ Hence the factor theorem is not as trivial as it might first seem. It is deep indeed.