What is the definition of a factor

Question

I know that $x-1$ is a factor of $x^2-1$, but is $x-i$ a factor of $x^2+1$. Do we consider $x-i$ to be a factor of $x^2+xi+2$?

Answer 1

In both cases the answer is YES if we are considering polynomials over the complex field and NO if we are considering polynomials over the real field.

Answer 2

Over the complex numbers, every polynomial $p(x)=a (x^n +b_{n-1}x^{n-1}+\ldots b_1 x + b_0)$ can be factored into $n$ (possibly identical) factors such that $$p(x)=a\prod_{m=1}^n (x-x_m)$$ with $x_m$ being the zeroes of the polynomial.

Thus, in particular $$x^2+1=(x-i)(x+i)$$ such that $x-i$ is a factor and $$x^2+ix+2=(x-i)(x+2i)$$ such that $x-i$ equally is a factor.

Answer 3

It does, unfortunately, depend on context. It depends on whether you're factorising over the integers (or rationals), the real numbers, or complex numbers. As you go much further in mathematics, there will be many other "things" you can factorise over, but we won't bother with that.

For example, over the integers, the polynomial $x^4 + 1$ cannot be factored. In other words, there are no two polynomials, with integer coefficients, whose product is $x^4 + 1$ (except trivially, $\pm 1$ multiplied to $\pm(x^4 + 1)$). But, if you factorise over the reals, $$x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - 2x^2 = (x^2 + 1 + \sqrt{2}x)(x^2 + 1 - \sqrt{2}x).$$ Over the reals, this cannot be factored any further. But, over the complex numbers, we get even more factors: $$(x^4 + 1) = \left(x - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)\left(x + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)\left(x + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)\left(x - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right).$$ Basically, the more possibilities we have for coefficients (integers to reals to complex numbers), the more possibilities we have for creating smaller factors, and you'll have to rely on context to know from where you may take your coefficients.