-1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Do you know of complex numbers? $\endgroup$ – Dave Apr 19 at 4:58
  • 1
    $\begingroup$ $(x-i)(x+i)$ where$ i=\sqrt{-1}$ $\endgroup$ – Tojrah Apr 19 at 5:01
  • 1
    $\begingroup$ 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 $1+x^2$. $\endgroup$ – amd Apr 19 at 5:26
  • 4
    $\begingroup$ @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. $\endgroup$ – David Peterson Apr 19 at 5:30
  • 2
    $\begingroup$ @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. $\endgroup$ – Ennar Apr 19 at 9:18
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is $\sqrt{-1}$? $\endgroup$ – amsmath Apr 19 at 5:04
  • $\begingroup$ It is not not a real number...so called complex number. $\endgroup$ – user421818 Apr 19 at 5:04
  • 1
    $\begingroup$ No, also in the complex numbers $\sqrt{-1}$ is undefined. What you mean is $i$, the imaginary unit. $\endgroup$ – amsmath Apr 19 at 5:07
  • 1
    $\begingroup$ @Antinous Nope. $\sqrt{-1}$ is not defined. $\endgroup$ – amsmath Apr 19 at 5:46
  • 1
    $\begingroup$ 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 $(0,1)\cdot(0,1)=(-1,0)$. $\endgroup$ – Antinous Apr 19 at 9:06
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.