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itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.0im suppose to factor x^2+4 is this prime?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2it will have complex factors. else its prime.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\[a^2+b^2 = (a+ib)(aib)\]

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.0ok wait a second

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.0(x + 2)(x – 2) (x + 2)(x + 2) (x + 1)(x + 4) ao it couldnt be any of these? im so confused with this one question

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2no, it cannot be any of these. its prime, if we are considering only real factors.

itsjustme_lol
 one year ago
Best ResponseYou've already chosen the best response.0ok. thankyou. i thought so, just was not sure. :) i appreicaite that.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1\[(x+2)(x2) = x^2 +2x  2x 4 = x^2 4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1That i in there in the (a+ib)(aib) does the trick of reversing the sign of the final term: \[(a+ib)(aib) = a^2 + aib  aib i^2b^2 = a^2  i^2b^2\]but \[ i^2 = 1\] so \[a^2i^2b^2=a^2(b^2)=a^2+b^2\]
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