Re: [代數] 多元方程式因式分解

看板Math作者Desperato (Farewell)時間7年前 (2018/04/29 20:30)推噓2(2推 0噓 5→)

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※ 引述《Ben40 (來自巴哈的魚酥)》之銘言： : https://i.imgur.com/81uw4DD.jpg

: 想要請教一下這類方程式因式分解有沒有一定方式可以依循 : 另外還有怎麼確定該多項式是不可再分解的我不知道是不是有跡可循，總之就做吧ow o (a) consider a = x, b = 2y, c = 3 then by the formula a^3 + b^3 + c^3 - 3abc we have x^3 + 8y^3 - 18xy + 27 = (x + 2y + 3) (x^2 + 4y^2 + 9 - 2xy - 3x - 6y) the latter term is x^2 - (2y+3)x + (4y^2-6y+9) in R[x], where R=Z[y] if it is reducible, then it can factorize into (x - p(y))(x - q(y)) thus p(y)q(y) = 4y^2-6y+9 and p(y) + q(y) = 2y+3, which has no solution (b) x^3 + 3x^2y - 2y^3 consider t^3 + 3t^2 - 2 = (t+1)(t^2+2t-2), we have x^3 + 3x^2y - 2y^3 = (x + y)(x^2 + 2xy - 2y^2) a similar method can prove that the latter term is irreducible (c) x^2 - 3xy^2 + 2y^2 irreducible by similar method (d) x^2 - 2zx - xy - 2yz = x^2 - (2z+y)x - 2yz in T[x], where T = Z[y, z] irreducible by similar method (e) x^3 + 2(y-z)x^2 + (x-z)y^2 + (x+y)z^2 - 3xyz = x^3 + 2(y-z)x^2 + (y^2 -3yz + z^2)x - yz(y-z) in T[x] if it is reducible, then it must has a factor out of x +- y, x +- z, x +- (y-z), x +- yz, ... etc. try x = y gives nonzero polynomial try x = -y gives 0, thus x + y is a factor continue we have = (x + y)(x - z)(x + y - z) reference: wolframalpha :D -- 嗯嗯ow o -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.25.4 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1525005044.A.002.html

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LiamIssac

04/29 20:44, 7年前 , 1^F

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LiamIssac

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Desperato

04/29 22:30, 7年前 , 3^F

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