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endiia Group Title

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

  • 2 years ago
  • 2 years ago

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  1. endiia Group Title
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    |dw:1336325178637:dw|

    • 2 years ago
  2. ParthKohli Group Title
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    Find the common factor and put that outside the bracket.

    • 2 years ago
  3. pokemon23 Group Title
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    If I'm correct

    • 2 years ago
  4. ParthKohli Group Title
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    You are correct. @pokemon23

    • 2 years ago
  5. endiia Group Title
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    x^5(x^2-14x-51x) ?

    • 2 years ago
  6. optiquest Group Title
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    \[x^5(x^2-14x-51)\]

    • 2 years ago
  7. pokemon23 Group Title
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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!

    • 2 years ago
  8. ParthKohli Group Title
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    Oops, you got the 51x incorrect @pokemon23

    • 2 years ago
  9. ParthKohli Group Title
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    \(\Large \color{purple}{\rightarrow x^5(51x) = 51x^6 }\)

    • 2 years ago
  10. amir.sat Group Title
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    \[x ^{5}\times(x-3)\times(x+17)\]

    • 2 years ago
  11. endiia Group Title
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    so what's the full equation?

    • 2 years ago
  12. endiia Group Title
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    an amir. sat has something totally different

    • 2 years ago
  13. endiia Group Title
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    can someone1 clarify for me plz

    • 2 years ago
  14. ParthKohli Group Title
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    @endiia Amir is correct, he has actually grouped the quadratic expression.

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

    • 2 years ago
  16. endiia Group Title
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    so when u factor it the answer is (x-3)(x+17)

    • 2 years ago
  17. endiia Group Title
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    so x is 3 & 17

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

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

    • 2 years ago
  20. amir.sat Group Title
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    and x is 0 & -3 & 17

    • 2 years ago
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