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itsjustme_lol Group Title

Im having trouble with factoring..

  • one year ago
  • one year ago

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  1. itsjustme_lol Group Title
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    im suppose to factor x^2+4 is this prime?

    • one year ago
  2. itsjustme_lol Group Title
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    or (x+2)(x+2)

    • one year ago
  3. hartnn Group Title
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    it will have complex factors. else its prime.

    • one year ago
  4. hartnn Group Title
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    its NOT (x+2)(x+2)

    • one year ago
  5. hartnn Group Title
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    \[a^2+b^2 = (a+ib)(a-ib)\]

    • one year ago
  6. itsjustme_lol Group Title
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    ok wait a second

    • one year ago
  7. itsjustme_lol Group Title
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    (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
  8. hartnn Group Title
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    no, it cannot be any of these. its prime, if we are considering only real factors.

    • one year ago
  9. itsjustme_lol Group Title
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    ok. thankyou. i thought so, just was not sure. :) i appreicaite that.

    • one year ago
  10. whpalmer4 Group Title
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    \[(x+2)(x-2) = x^2 +2x - 2x -4 = x^2 -4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]

    • one year ago
  11. whpalmer4 Group Title
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    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 \[a^2-i^2b^2=a^2-(-b^2)=a^2+b^2\]

    • one year ago
  12. hartnn Group Title
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    welcome ^_^

    • one year ago
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