Factorise $1+x^2$

Question

How do I factorise this expression?

$$1+x^2$$

An attempt: complete the square $(1-x)(1+x).$ teacher said no.

$x(1/x+x)$ again teacher said no.

She said is related to solving this $x^2+1=0$.

I got no idea, can anyone help me to solve it?

Answer 1

If you multiply these $(x-i)(x+i)$, where $i$ is the imaginary unit with the property that $i^2=-1$, you will get your expression.

Answer 2

In context:

$y=1+x^2.$

Assume there is a factorization

$(x^2+1)=(x-a)(x-b)$ where $a,b \in \mathbb{R}$, then

$a,b$ are the real roots , i.e.

$1+a^2=0$, and $1+ 1+ b^2=0$.

But: $y=1+x^2 >0$ (why?) for $x \in.\mathbb{R}.$

Hence no factorization in $ \mathbb{R}.$

But we know that a polynomial of degree $2$ has $2$ roots:

The roots are complex, refer to ProblemBook's answer.

Answer 3

Hint:

Why don't you listen to the teacher and solve the equation ?

$$x^2+1=0\iff x^2=-1\iff x=\pm i.$$

Then as the polynomial has these roots, it must be proportional to the binomials $(x-i)$ and $(x+i)$, which vanish at these roots.