If we need to show that a polynomial $p(x)$ is divisible by another polynomial $g(x)$ which can be broken or reduced into linear factors then it can be shown that each linear factor divides $p(x)$ and hence $g(x)$ divides $p(x)$.
But in case of irreducible polynomial $g(x)$ this method will not work. In that case is there any simple way to prove that $g(x)$ divides $p(x)$ without performing the long division of polynomials?
Please guide me.