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endiia

  • 3 years ago

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

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  1. endiia
    • 3 years ago
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    |dw:1336325178637:dw|

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

  3. pokemon23
    • 3 years ago
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    If I'm correct

  4. ParthKohli
    • 3 years ago
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    You are correct. @pokemon23

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

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

  7. pokemon23
    • 3 years ago
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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!

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

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

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

  11. endiia
    • 3 years ago
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    so what's the full equation?

  12. endiia
    • 3 years ago
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    an amir. sat has something totally different

  13. endiia
    • 3 years ago
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    can someone1 clarify for me plz

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

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

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

  17. endiia
    • 3 years ago
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    so x is 3 & 17

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

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

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

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