Question

Prove that x^2+4 is an irreducible polynomial in R[x]
Please help

Answer 1

Do you have the Eisenstein criterion yet?

Answer 2

Not yet.

Answer 3

hmm its easy with that, \(2|4\) but \(2^2\cancel{|}1\)

Answer 4

What all do you have?

Answer 5

I stuck at this:
Suppose x^2 +4 is reducible, then it can be factored as
(ax+b)(cx+d) , it gives me x = -b/a and x =-d/c are roots of the x^2+4
but how to argue that they are not in real?

Answer 6

Can you use the fact that this can have at most two roots?

Answer 7

another way to approach is open the bracket and get
\[acx^2+x(ad+bc)+bd=x^2+4\]
that is
ac = 1
ad+bc =0
bd =4
then??

Answer 8

Sec I am getting my book out

Answer 9

Back to the first option.
if x =-b/a is a root, then (-b/a)^2 +4 =0 or \(\dfrac{b^2}{a^2}+4=0\) then??

Answer 10

Ok do you know about the whole f(x) = g(x)h(x)+r(x) where r(x) has degree less than or equal to g(x)

Answer 11

yes.

Answer 12

ok just trying to find out what all we can use

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @Loser66
another way to approach is open the bracket and get
\[acx^2+x(ad+bc)+bd=x^2+4\]
that is
ac = 1
ad+bc =0
bd =4
then??
\(\color{blue}{\text{End of Quote}}\)
Well, this means that a and c have the same sign, so does b and d

Answer 14

So consider the four cases.

Answer 15

a and c are positive, b and d are positive.. then?

Answer 16

ac + bd \(\neq 0\)

Answer 17

Then you can't have 0 in this case,

Answer 18

Do you know that if it is degree 2 or 3, then it is irreducible over a field F if and only if it has a zero in F?

Answer 19

This is not hard to prove

Answer 20

then we can use the theorem of no more roots than n and we are done

Answer 21

If a and c are positive and b and d are negative?

Answer 22

@wio then ac = bd

Answer 23

Then ad is negative and bc is negative, so they can't add up to 0

Answer 24

How about saying that
\[
x^2 + 4 > 0
\]
for any real x. So it cannot have real roots. So it cannot be factored over R. Am I missing something?

Answer 25

@Loser66 Do you think you can do the rest now?

Answer 26

@zzr0ck3r If approach on that way, we have to prove why reducible polynomial has a root :)

Answer 27

@wio yes, Sir. It gives contradiction when ac =1 and bd =4 , right?

Answer 28

@eliassaab yes, it is quite clear. Can I do it??? I am doubt myself :)

Answer 29

Oh, at the end up, it has many ways to solve. Thanks you all. :)

Answer 30

Oh, his way is simplest

Answer 31

It just needs to be understood that you need real roots to be reducible

Answer 32

:)

Answer 33

@Loser66 yes, you can it take it to the bank.

Answer 34

@Loser66 C Is an extension of R, if we can show that there are two roots in C\R (very easy), then we are done since there are no roots in R, the degree 2 polynomial is irreducible.

Answer 35

@eliassaab I think this is true iff degree 2 or 3

I have this proof in my notes

\(f(x)\in F(x)\) then for \(f(x) \) degree \(2\) or \(3\) we have \(f(x)\) is irreducible if and only if \(f\) has a root in \(f\).

I am not sure this is true for the other cases? Either way as I am sure they are starting from scratch, it should prob be proved.