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anonymous
 one year ago
Factor the polynomial completely and find al zeros
P(x)=x^6729
Please help!
anonymous
 one year ago
Factor the polynomial completely and find al zeros P(x)=x^6729 Please help!

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1x^6  729 = (x^3)^2  27^2 = (x^3  27)(x^3 + 27)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1now 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)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1putting a^3 = x^3 and b^3 = 27 where a= x and b = 3

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1to find the zeros let (x^3  27)( x^3 + 27) = 0 and solve for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x+3)\times (x^{2}3x+9)\times (x3)\times (x ^{2}+3x+9)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that right?.. for the zeros.. should i just use the the quadratic equations to get the remained zeros?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1hmm 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could i solve them with the quadratic equation?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1they will all be complex roots

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1quadratic formula yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cool! thank you.. you were really helpful!
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