因式分解：x^2+2y^2-z^2+3xy-yz赶时间,百度搜不到呀!答出并且答对追加50财富

原式＝(ax+by+cz)(a1x+b1y+c1z)
＝aa1x^2+bb1y^2+cc1z^2 +ab1xy+ac1xz+ba1xy+bc1yz+ca1xz+cb1zy
=aa1x^2+bb1y^2+cc1z^2+xy(ab1+ba1)+yz(bc1+cb1)+xz(ac1+ca1)
所以aa1=1 bb1=2 cc1=-1
ab1+ba1=3 bc1+cb1=-1 ac1+ca1=0 (a,b,c都不等于0)
a1=1/a c1=-1/c -a/c+c*1/a=0 a/c=c/a a^2=c^2 ...(1)
b1=2/b 2a/b+b*1/a=3 -b/c+2c/b=-1
(2a^2+b^2)=3ab (-b^2+2c^2)=-bc
由(1)若a=c 则 -b^2+2a^2=-ba 2a^2=b^2-ab =3ab-b^2
2b^2=4ab b(b-2a)=0 b=2a ...(2)
若a=-c 则2a^2=ab+b^2 =3ab-b^2 2b^2=2ab b=a ...(3)
这样有c=a b=2a
a1=1/a b1=2/b =1/a c1=-1/c =-1/a 代入得
原式＝(ax+2ay+az)(1/a x+1/a y-1/a z) =a(x+2y+z)*1/a (x+y-z)=(x+2y+z)(x+y-z)
或c=-a b=a a1=1/a b1=2/b=2/a c1=-1/c=1/a
原式=(ax+ay-az)(1/a x+2/a y+1/a z) =(x+2y+z)(x+y-z)
所以得原式＝(x+2y+z)(x+y-z)

x^2+2y^2-z^2+3xy-yz
=x²+3xy+9/4y²-y²/4-z²-yz
=(x+3y/2)²-(z+y/2)²
=(x+3y/2-z-y/2)(x+3y/2+z+y/2)
=(x+y-z)(x+2y+z)

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