Factoring a sum of squares

[Nona No.1679179]

A person from another board(not 4chan) showed me this and I thought it was kinda cool. So, I'm sharing it with you guys.

Factor x^2+4 without using i.

I'll give the answer if I get too many "this is impossible FAGGOT!"

[Anonymous No.1679187]

(x+2)(x-2)

[Anonymous No.1679192]

It's not impossible, you're just a faggot.

[Nona No.1679202]

>>1679192
Okay.

[Anonymous No.1679199]

>>1679187
wrong

[Anonymous No.1679206]

>>1679187
lol no,

lrn2difference of squares

[Anonymous No.1679210]

>>1679187
Underage b&

[Anonymous No.1679222]

(x*x) + 4

[Nona No.1679250]

Bump?

[Anonymous No.1679254]

I think it's quite trivial to most people on this board to show that the polynomial x²+4 has the linear factors (x+2i) and (x-2i), and only those two factors. From there you could multiply or divide out constant terms, but any attempt to do so would ultimately leave you still with imaginary terms inside your linear terms.

However, one may use substitutions.

x²=u

Therefor the polynomial provided by the OP could clearly be factored down to u+4.

I'm not the OP

[Anonymous No.1679278]

i think OP means x^4 + 4.

[Nona No.1679283]

>>1679254
Not quite what I was looking for.

I only said factor it because I wanted them to go about this by the traditional meaning of factorization.

Normally it will result with an imaginary number but there's a method by factoring that eliminated i. This is why I thought it was cool.

[Anonymous No.1679284]

x^2+4 = x^6

(x)(x)(x)(x)(x)(x)

[Anonymous No.1679298]

>without using i.

It's impossible to factor <span class="math">x^2+4[/spoiler] without using <span class="math">\sqrt{-1}[/spoiler]. So we can just redefine <span class="math">\sqrt{-1}[/spoiler] to be a different letter, say "j".

<span class="math">x^2+4 = (x + 2j) (x - 2j)[/spoiler]

[Nona No.1679304]

>>1679278
Nope, x^2+4. Though, you can also factor x^4+4.

>>1679284
?

[Nona No.1679313]

>>1679298
Loophole'd but no. It is possible.

[Anonymous No.1679317]

(x+2+sqrt(4x)) (x+2-sqrt(4x))

[Anonymous No.1679319]

>>1679304

just tell us alrady

[Anonymous No.1679326]

>>1679304
>?

Poster was suggesting you were writing the equation incorrectly, without parenthesis. <span class="math">x^(2+4) = x^6[/spoiler]

[Nona No.1679331]

>>1679317
Winner.

[Nona No.1679334]

>>1679326
Oh, lol.

[Anonymous No.1679335]

well according to the fundamental theorem of algerbra the factors of a polynomial are also the roots and x^2+4 doesnt have any real roots, and thus it is not possible to factor it using real numbers.

so your "method" is probably just some trick which looks correct but is actually flawed.

[Anonymous No.1679336]

>>1679326

i meant just tell us the answer

[Nona No.1679355]

>>1679335
Yes, it can't be factored with real numbers but I just asked for it to be factored. Didn't give any bounds at all.

[Anonymous No.1679363]

>>1679331

Of course you realize that:

>>1679317

Is essentially

>>1679278

[Anonymous No.1679368]

>>1679254
Here

>>1679278
Regardless, even so x^4+4 would break down to (x²+2x+2)(x²-2x+2); (x-1-i)(x-1+i)(x+1-i)(x+1+i). Which again CAN NOT be factored without using i by simply multiplying or dividing each of those terms by constant terms. However similar substitutions may be made like my first one.

>>1679317
Do I even have to say it? FUCKING DOMAIN! That statement is NOT equivalent to OP's polynomial. They may be equal across the nonnegative numbers, but have different domains.

[hi No.1679393]

>>1679317
> (x+2+sqrt(4x)) (x+2-sqrt(4x))

http://mathworld.wolfram.com/PolynomialFactorization.html
http://mathworld.wolfram.com/Polynomial.html

This is not a polynomial.
You might as well 'factor' x^2+4 into
[x+4/x]*[x] or [(x^(3/2)+4x^(-1/2)]*[x^(1/2)]

[hi No.1679409]

>>1679368
>FUCKING DOMAIN! That statement is NOT equivalent to OP's polynomial

That 'factorization' is defined for all x.

[Nona No.1679416]

>>1679363
How?

>>1679368
Ahh, didn't notice its domain. I was just amazed how he "factored" it without using i. I'll show him this when he's on tomorrow then.

[Anonymous No.1679422]

CAN SOMEONE EXPLAIN WHAT ALL OF THIS MEANS

[hi No.1679430]

>>1679416
There is no problem with the domain.
You will need to take into account that x^(1/2) is multivalued though. The problem goes away if you define sqrt(x) to be either the principal or secondary value of x^(1/2)

[Anonymous No.1679437]

> There is no problem with the domain.

Yes, there is.

> You will need to take into account that x^(1/2) is multivalued though.

In which case it isn't a function anymore.

> The problem goes away if you define sqrt(x) to be either the principal or secondary value of x^(1/2)

In which case you have problems with the domain.

It's math. Definitions are important.

[hi No.1679445]

>>1679437
You haven't taken any classes in complex analysis, have you?

[Anonymous No.1679453]

>>1679445
so you are an advanced math person, and you think that is a legitimate factorization?

[Anonymous No.1679454]

>>1679445

What the fuck are you talking about? The OP insisted that he could do this in the real numbers.

If you change the fucking ring the polynomial's in, this is just fucking around.

[Anonymous No.1679468]

>>1679317

here you go bro:
http://www.mathsisfun.com/algebra/polynomials.html

not that neither division or square roots (fractional exponents) are allowed.

[Anonymous No.1679484]

>>1679393
this

[hi No.1679486]

>>1679453
No. See >>1679393

>>1679454
Also see >>1679393

OP's way of factoring this is clearly not factoring.
However, they are equivalent over the entire complex plane after you take into account that you are using either the principal root the secondary root for both terms. (The principal root is generally assumed when using a radical without the leading plus/minus, but in the realm of complex analysis, you rarely only consider one of the multiple solutions)

[Anonymous No.1679505]

>>1679416

Well, you're essentially just inventing a new variable y = sqrt(4x) and noting that y^4/16 + 4 = (y^2/4 + y + 2)(y^2/4 - y + 2).

(I'm considering the factoring of x^4 + 4 to be essentially the same as that of x^4 / 16 + 4).

[hi No.1679511]

>ZOMG NEGATIVE NUMBERS AREN'T ALLOWED UNDER SQUARE ROOT SIGNS GUISE... IT'S UNDEFINED

[hi No.1679535]

> ZOMG NEGATIVE NUMBERS AREN'T ALLOWED UNDER THE RADICAL SIGN GUISE... IT'S UNDEFINED

pic related, negative values of x clearly causing a problem

[Nona No.1679565]

>>1679535
So, this is a legit factorization after all?

[Anonymous No.1679581]

>>1679535
I think its because the imaginary parts are multplied by each other so they turn into non-imaginary numbers.

[Nona No.1679615]

>>1679581
Oh, well that would make sense then.

[hi No.1679619]

>>1679565
No. See my other post here >>1679393
It's equivalent, but not factored. The only factorization of the polynomial you posted has "i" in it because both roots are pure imaginary. You can't get around this.

Only under a really loose definition of "factoring" will it be right. But be careful with that because it's like saying that (10/3) and (3/2) are factors of 5 when they clearly are not.

[hi No.1679635]

>>1679581
Good job. You'll be doing complex analysis in no time.

[Nona No.1679642]

>>1679619
Alright, I understand now. Thanks.

[Anonymous No.1679796]

>>1679179
>mfw no one has mentioned you can express the answer to this in cosine form