Question

is X^2+7 in C[x] irreducible?

Answer 1

in C(x)? no

Answer 2

That was rather abrupt...

Answer 3

can you tell me why

Answer 4

(a-bi)^2 = a^2 + b^2

Answer 5

lol .... i knew it was something with abis

Answer 6

thank you

Answer 7

i^2 = -1

Answer 8

hey, @urbanderivative
A counter-question... a polynomial p(x) is not irreducible in F[x] if F contains some roots of p(x), right?

Answer 9

yes

Answer 10

Well, what are the roots of x^2 + 7
?
There are at most 2.

Answer 11

haha you removed the equation....

Answer 12

Oh Cr*p...
Sorry
\[\large x^2+7=x^2-(i\sqrt7)^2\]

Answer 13

Do apply difference of two squares again ;)

Answer 14

haha (x-sqr(7)i)(x+sqr(7)i)

Answer 15

That's right :)
So it can be factored into...
\[\huge (x+i\sqrt7)(x-i\sqrt7)\]So... it can be factored into polynomials... the question is, are these polynomials both in \[\huge \mathbb{C}[x]\]?

Answer 16

yes

Answer 17

Therefore, is x^2 + 7 irreducible in C[x]
?
I didn't think so ;)

Answer 18

haha yes, thank you

Answer 19

By the way, all polynomials with real coefficients have roots in C. So, the only polynomials with real coefficients that are irreducible in C are linear or constants.

Answer 20

^Follows from the fundamental theorem of Algebra :)
Have fun :D