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

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

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

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

itsjustme_lolBest 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
 one year ago

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

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

whpalmer4Best ResponseYou've already chosen the best response.1
\[(x+2)(x2) = x^2 +2x  2x 4 = x^2 4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
That 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 \[a^2i^2b^2=a^2(b^2)=a^2+b^2\]
 one year ago
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