anonymous
  • anonymous
Factor the polynomial completely and find al zeros P(x)=x^6-729 Please help!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[P(x)= x^{6}-729\]
welshfella
  • welshfella
x^6 - 729 = (x^3)^2 - 27^2 = (x^3 - 27)(x^3 + 27)
welshfella
  • welshfella
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)

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welshfella
  • welshfella
putting a^3 = x^3 and b^3 = 27 where a= x and b = 3
welshfella
  • welshfella
to find the zeros let (x^3 - 27)( x^3 + 27) = 0 and solve for x
anonymous
  • anonymous
\[(x+3)\times (x^{2}-3x+9)\times (x-3)\times (x ^{2}+3x+9)\]
anonymous
  • anonymous
is that right?.. for the zeros.. should i just use the the quadratic equations to get the remained zeros?
welshfella
  • welshfella
yes that's it
anonymous
  • anonymous
AAAHH thank you!
welshfella
  • welshfella
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
anonymous
  • anonymous
could i solve them with the quadratic equation?
welshfella
  • welshfella
they will all be complex roots
welshfella
  • welshfella
quadratic formula yes
anonymous
  • anonymous
cool! thank you.. you were really helpful!
welshfella
  • welshfella
yw

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