# Factorise $1+x^2$

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?

• Do you know of complex numbers? – Dave Apr 19 at 4:58
• $(x-i)(x+i)$ where$i=\sqrt{-1}$ – Tojrah Apr 19 at 5:01
• How exactly did you “complete the square?” You should check your own work: what do you get when you multiply out $(1+x)(1-x)$? It’s certainly not $1+x^2$. – amd Apr 19 at 5:26
• @amsmath I predict most would agree that $\sqrt{-1}$ is the notation for the principle square root of negative one. While you are free to disagree or to introduce ambiguity, it is standard. – David Peterson Apr 19 at 5:30
• @amsmath, you do realize that there is a way to define $z\mapsto \sqrt z\,\colon \mathbb C\to\mathbb C$? Choose, for example, principal branch, and let $\sqrt z = e^{1/2 \ln z}$. One can look at the limit $\lim_{\vartheta\to \pi^-}\sqrt {e^{i\vartheta}}$ to get $\sqrt{-1} = i$. The function won't be continuous on whole $\mathbb C$, but $\sqrt{-1} = i$ is very common and useful. Principal square root of a complex number. – Ennar Apr 19 at 9:18

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.

• What is $\sqrt{-1}$? – amsmath Apr 19 at 5:04
• It is not not a real number...so called complex number. – user421818 Apr 19 at 5:04
• No, also in the complex numbers $\sqrt{-1}$ is undefined. What you mean is $i$, the imaginary unit. – amsmath Apr 19 at 5:07
• @Antinous Nope. $\sqrt{-1}$ is not defined. – amsmath Apr 19 at 5:46
• Maybe you're thinking that we actually compute $\sqrt{-1}$? This is not so. Most people agree to define that symbol to be the ordered pair $(0,1)$, which is also commonly defined to be the imaginary unit denoted by $i$. These are symbols only. Symbols can be defined in any which way you like. You can define the complex numbers as ordered pairs $(a,b)$ with (one) property (being) $$(a,b)\cdot(c,d)=(ac-bd, ad+bc),$$ in which case I define the symbol (note not plural) $\sqrt{-1}$ to be the ordered pair $(0,1)$. Clearly then $(0,1)\cdot(0,1)=(-1,0)$. – Antinous Apr 19 at 9:06

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.

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.