標題:

Factorization

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發問:

Factorize y^12-4096

最佳解答:

y^12 - 4096 = (y^6)^2 - 64^2 = (y^6 - 64)(y^6 + 64) = [ (y^3)^2 - 8^2 ] (y^6 + 64) = (y^3 - 8)(y^3 + 8) (y^6 + 64) = (y - 2)(y^2 + 2y + 4)(y^3 + 8) (y^6 + 64) = (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4)(y^6 + 64) = (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4) [(y^2)^3 + 4^3] = (y - 2) (y^2 + 2y + 4) (y + 2) (y^2 - 2y + 4) (y^2 + 4)(y^4 - 4y^2 + 16)

其他解答:

y^12-4096 =(y^4)^3-16^3 =(y^4-16)((y^4)^2+y^4*16+16^2) =(y^4-16)(y^8+16y^4+256). p,s, i cannot solve(y^8-16y^4+256)!!!!!|||||y^12 - 4096 =y^12 - (4*1024) =y^12 - (4*4*256) =y^12 - (4*4*4*64) =y^12 - (4*4*4*4*4*4) =y^12 - 4^6 =y^12 - 2^12 =(y^6)^2 - (2^6)^2 =(y^6+2^6) (y^6-2^6) =(y^6+64) (y^3+2^3) (y^3-2^3) =(y^6+64) (y+3) (y^2 - 2y + 4) (y -3) (y^2 + 2y + 4) =(y + 3) (y - 3) (y^2 - 2y + 4) (y^2 + 2y + 4) (y^6 + 64) 2007-12-22 13:40:52 補充: The last two steps should be corrected as=(y^6 64) (y 2) (y^2 - 2y 4) (y - 2) (y^2 2y 4) =(y 2) (y - 2) (y^2 - 2y 4) (y^2 2y 4) (y^6 64) 2007-12-22 13:43:57 補充: (y^6 64) can be further factorized into ( y^2 4)(y^4 - 4y^2 16)|||||y^12-4096 =y^12-2^12 =(y^6)^2-(2^6)^2 =(y^6+2^6)(y^6-2^6)
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