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anonymous
 3 years ago
Im having trouble with factoring..
anonymous
 3 years ago
Im having trouble with factoring..

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

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

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

anonymous
 3 years 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
 3 years 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.

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

whpalmer4
 3 years 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
 3 years 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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