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.
Best Answer
Polynomial division has low computational cost compared to calculating the root of a polynomial, which is nearly impossible for high degree (in fact, general solution by radicals is impossible via Galois theory). But if you’re really not willing to use the easiest method, extend $\mathbb R$ to $\mathbb C$ so every polynomial can be splitted into linear factors.
Your polynomial will be divisible by the irreducible one if all the roots $\alpha_1, \ldots, \alpha_k$ are also roots and since it’s irreducible over $\mathbb R$, it has no multiple roots. So not only you’ll have to compute the roots of a polynomial which is extremely costly but also apply them in a higher-degree polynomial.
So, in synthesis: use polynomial long division.