$(x^4-49)$ Is this polynomial completely factored over $\mathbb{C}$

Question

Is this polynomial completely factored over $\mathbb{C}$?

$(x^4-49)$ = $((x^2)^2 - 7^2)$ = $(x^2 + 7)(x^2 - 7)$

Thanks

Answer 1

No, because $\sqrt{7}$ and $i$ exists, so $(x^2-7)=(x-\sqrt{7})(x+\sqrt{7})$ and $(x^2+7)=(x-i\sqrt{7})(x+i\sqrt{7})$. More generally, in $\mathbb{C}$ you can factorize until you get linear terms (only "$x$" and no "$x^n$", with $n>1$).

Answer 2

If you want to factor the polynimal over $R$, you have: $$x^4-49=(x^2)^2-49=(x^2-7)(x^2+7)=(x-\sqrt{7})(x+\sqrt{7})(x^2+7)$$ Over $C$, instead: $$x^4-7=(x-i\sqrt{7})(x+i\sqrt{7})(x-7)(x+7)$$