anonymous
  • anonymous
Im having trouble with factoring..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
im suppose to factor x^2+4 is this prime?
anonymous
  • anonymous
or (x+2)(x+2)
hartnn
  • hartnn
it will have complex factors. else its prime.

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More answers

hartnn
  • hartnn
its NOT (x+2)(x+2)
hartnn
  • hartnn
\[a^2+b^2 = (a+ib)(a-ib)\]
anonymous
  • anonymous
ok wait a second
anonymous
  • anonymous
(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
  • hartnn
no, it cannot be any of these. its prime, if we are considering only real factors.
anonymous
  • anonymous
ok. thankyou. i thought so, just was not sure. :) i appreicaite that.
whpalmer4
  • whpalmer4
\[(x+2)(x-2) = x^2 +2x - 2x -4 = x^2 -4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]
whpalmer4
  • whpalmer4
That i in there in the (a+ib)(a-ib) does the trick of reversing the sign of the final term: \[(a+ib)(a-ib) = a^2 + aib - aib -i^2b^2 = a^2 - i^2b^2\]but \[ i^2 = -1\] so \[a^2-i^2b^2=a^2-(-b^2)=a^2+b^2\]
hartnn
  • hartnn
welcome ^_^

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