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anonymous

  • one year ago

Factor the polynomial completely and find al zeros P(x)=x^6-729 Please help!

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  1. anonymous
    • one year ago
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    \[P(x)= x^{6}-729\]

  2. welshfella
    • one year ago
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    x^6 - 729 = (x^3)^2 - 27^2 = (x^3 - 27)(x^3 + 27)

  3. welshfella
    • one year ago
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    now you can factor each parentheses using the cube identities a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a^3 - b^3 = (a - b)(a^2 + ab + b^2)

  4. welshfella
    • one year ago
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    putting a^3 = x^3 and b^3 = 27 where a= x and b = 3

  5. welshfella
    • one year ago
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    to find the zeros let (x^3 - 27)( x^3 + 27) = 0 and solve for x

  6. anonymous
    • one year ago
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    \[(x+3)\times (x^{2}-3x+9)\times (x-3)\times (x ^{2}+3x+9)\]

  7. anonymous
    • one year ago
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    is that right?.. for the zeros.. should i just use the the quadratic equations to get the remained zeros?

  8. welshfella
    • one year ago
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    yes that's it

  9. anonymous
    • one year ago
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    AAAHH thank you!

  10. welshfella
    • one year ago
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    hmm maybe its better to find the zeroes using the last expression x = 3 and -3 are obvious roots to find the other 4 roots you need to solve x^2 - 3x + 9 = 0 and x^2 + 3x + 9 = 0

  11. anonymous
    • one year ago
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    could i solve them with the quadratic equation?

  12. welshfella
    • one year ago
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    they will all be complex roots

  13. welshfella
    • one year ago
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    quadratic formula yes

  14. anonymous
    • one year ago
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    cool! thank you.. you were really helpful!

  15. welshfella
    • one year ago
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    yw

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