How to factorise $x^2 + 1$ Keep in mind that I am not asking how to find the solutions to the equation $x^2 + 1 = 0$. I know that there will be imaginary answers In this case. I just want to know how to factor this using a logical method and obtaining 2 factors. Thanks for answering.
 A: Although you are not asking how to solve the equation $x^2+1=0$, usually the easiest way of factorising a polynomial is by solving the equation when the polynomial is equal to $0$. This is useful as a result of the factor theorem, which states:

Let $f(x)$ be a polynomial in $x$. If $f(a)=0$ then $(x-a)$ is a factor of $f(x)$ (and if $(x-a)$ is a factor of $f(x)$ then $f(a)=0$).

In our case then, solving the equation $x^2+1=0$ yields the solutions $x=\pm i$; hence the polynomial factorises as $(x+i)(x-i)$.

I hope that was helpful. If you have any questions please don't hesitate to ask :)
A: HINT:
Remember $a^2-b^2=(a+b)(a-b)$, $i^{2}=-1$.
So we can write, $a^2+b^2=a^2-(ib)^2=(a+ib)(a-ib)$
Putting $a=x$ and $b=1$ we get,
$x^2+1=(x+i)(x-i)$
A: Rewrite $x^2 + 1$ as $x^2 - (-1)$. This is a difference of two squares, which is easy to factorise.
$x^2 - (-1)\\$
$= x^2 - i^2\\$
$= (x + i)(x - i)$
A: I will suggest you $3$ different ways.
If you want to factorise $x^2+1$, remember that
$$x^2+1=(x-a)(x-b)$$
For general factorization we will find  $a,b\in\mathbb C$ such that $x^2+1=(x-a)(x-b)$.
We have,
$$x^2+1=x^2-x(a+b)+ab$$
$$ab=1, ~a+b=0$$
$$-a^2=1,~ -b^2=1$$
$$a^2=-1,~ b^2=-1$$
$$a=\pm \sqrt {-1}, ~b=-a=\mp \sqrt {-1}$$
$$a=\pm i, ~ b=\mp i$$
$$x^2+1=(x-i)(x+i)\\~~~~~~~~~~~~ =(x+i)(x-i).$$

But you can do faster,
$$\begin{align}x^2+1&=x^2-(-1)
\\&=x^2-i^2
\\&= (x-i)(x+i).\end{align}$$

Alternatively, the most common way for factorization you need to find the roots of polynomial. If, $r_1,r_2$ are the roots of $x^2+1$, then
$$x^2+1=(x-r_1)(x-r_2)$$
Thus, you need just to solve
$$x^2+1=0 \implies x_{1,2}=±i.$$
It is enough to take $r_1=i, r_2=-i.$
You get again the same result:
$$x^2+1=(x-r_1)(x-r_2)\\
~~~~~~~\thinspace=(x-i)(x+i).$$
