Re: [代數] 多元方程式因式分解

[Desperato (Farewell)]

※ 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

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嗯嗯ow o

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推 LiamIssac   : 這種題目沒辦法 就是看自己對於這些polys的敏感度
→ LiamIssac   :  只能兜兜湊湊土法煉鋼
→ Desperato   : 了解 感謝
→ recorriendo : 所以similar method就是wolframalpha? XD
推 Ben40       : 非常感謝 我會再多練習點
→ Desperato   : 不是啦www 是跟(a)一樣 假設可以拆 然後矛盾 基本
→ Desperato   : 只有二次式可以這樣用