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May 10th, 2017, 06:17 PM  #1 
Member Joined: Oct 2014 From: Colorado Posts: 40 Thanks: 21  Why is $2x^2 + 4$ reducible over Z but not Q?
We can rewrite $2x^2 +4$ as $2(x^2 + 2)$ so wouldnt this polynomial be reducible over both $\mathbb{Z}$ and $\mathbb{Q}$?

May 11th, 2017, 06:22 PM  #2  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198  Quote:
It is factorable over the complex numbers $2x^2 + 4 = 2(x  i \sqrt{2})( x + i \sqrt{2}) $  
May 12th, 2017, 01:40 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 147 Thanks: 20 
Equation $\displaystyle \; \; x^2 + y^2 =0 \; \; \: \; (xiy)(x+iy)=0$ $\displaystyle \begin{cases} xiy=0 \\ x+iy=0 \end{cases}$ $\displaystyle x=y=0$ 

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$2x2, reducible 
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