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endiia

factor completely: x^7-14x^6-51x^5

  • one year ago
  • one year ago

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  1. endiia
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    |dw:1336325178637:dw|

    • one year ago
  2. ParthKohli
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    Find the common factor and put that outside the bracket.

    • one year ago
  3. pokemon23
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    If I'm correct

    • one year ago
  4. ParthKohli
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    You are correct. @pokemon23

    • one year ago
  5. endiia
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    x^5(x^2-14x-51x) ?

    • one year ago
  6. optiquest
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    \[x^5(x^2-14x-51)\]

    • one year ago
  7. pokemon23
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    thanks parth nice to meet you I'ma long time user but I departed OS for three months and I'm read to rock and roll!

    • one year ago
  8. ParthKohli
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    Oops, you got the 51x incorrect @pokemon23

    • one year ago
  9. ParthKohli
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    \(\Large \color{purple}{\rightarrow x^5(51x) = 51x^6 }\)

    • one year ago
  10. amir.sat
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    \[x ^{5}\times(x-3)\times(x+17)\]

    • one year ago
  11. endiia
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    so what's the full equation?

    • one year ago
  12. endiia
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    an amir. sat has something totally different

    • one year ago
  13. endiia
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    can someone1 clarify for me plz

    • one year ago
  14. ParthKohli
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    @endiia Amir is correct, he has actually grouped the quadratic expression.

    • one year ago
  15. amir.sat
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    x^2−14x−51=(x-3)*(x+17)

    • one year ago
  16. endiia
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    so when u factor it the answer is (x-3)(x+17)

    • one year ago
  17. endiia
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    so x is 3 & 17

    • one year ago
  18. ParthKohli
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    Yes, a better and cleared answer is: \(\LARGE \color{purple}{\rightarrow x^5[(x - 3)(x + 17)] }\)

    • one year ago
  19. amir.sat
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    sorry it's x^2−14x−51=(x+3)*(x-17) so answer is x^5×(x+3)×(x-17)

    • one year ago
  20. amir.sat
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    and x is 0 & -3 & 17

    • one year ago
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