Question

Use unique factorization to find the gcd in$\mathrm{ℂ}\left[x\right]$ of${\left(x-3\right)}^3{\left(x-4\right)}^4{\left(x-i\right)}^2$ and $\left(x-1\right)\left(x-3\right){\left(x-4\right)}^3$.

Short answer

The gcd is $\left(x-3\right){\left(x-4\right)}^3$.

Step-by-step solution

Given Polynomials

Consider the given gcd in$\mathrm{ℂ}\left[x\right]$ of ${\left(x-3\right)}^3{\left(x-4\right)}^4{\left(x-i\right)}^2$ and $\left(x-1\right){\left(x-3\right)}^4{\left(x-4\right)}^3$.

Obtain gcd in ℂx

By unique factorization, the gcd of${\left(x-3\right)}^3{\left(x-4\right)}^4{\left(x-i\right)}^2$ and $\left(x-1\right)\left(x-3\right){\left(x-4\right)}^3$ is a product irreducible.

The irreducible product appears in both factorization to the smallest power associate.

Therefore, the gcd is $\left(x-3\right){\left(x-4\right)}^3$.