一道初中数学题

一道数学题?若多项式x^2-kx

(ax+by+c)(dx+ey+f)=adx2+bey2+(ae+bd)xy+(af+cd)x+(bf+ce)y+cf ad=1 then a=1/d (1) cf=-6 then f=(-6)/c (2) be=-2 then b=(-2)/e (3) af+cd=1 then (1/d)(-6)(1/c)+cd=1 then (-6)/dc+cd=1 then (cd)^2-cd-6=0 So We have cd=3 OR cd=-2 (4) bf+ce=7 then b(-6)/c+c(-2)/b=7 then 2c^2+7cb+6b^2=0 then (2c+3b)(...全部

  (ax+by+c)(dx+ey+f)=adx2+bey2+(ae+bd)xy+(af+cd)x+(bf+ce)y+cf ad=1 then a=1/d (1) cf=-6 then f=(-6)/c (2) be=-2 then b=(-2)/e (3) af+cd=1 then (1/d)(-6)(1/c)+cd=1 then (-6)/dc+cd=1 then (cd)^2-cd-6=0 So We have cd=3 OR cd=-2 (4) bf+ce=7 then b(-6)/c+c(-2)/b=7 then 2c^2+7cb+6b^2=0 then (2c+3b)(c+2b)=0 So We have 2c=-3b OR c=2b (5) k=-(ae+bd)=-(1/d)(-2)/b-bd=(2/bd)-bd (6) When I) cd=3 2c=-3b then bd=(3/c)[(2c)/(-3)]=-2 So k=-1+2=1 II) cd=3 c=2b then bd=(c/2)(3/c)=3/2 So k=4/3-3/2=-(1/6) III) cd=-2 2c=-3b then bd=(2c/3)(c/2)=1/3 So k=6-1/3=17/3 IV) cd=-2 c=2b then bd=-(c/2)(2/c)=-1 So k=-2+1=-1 K为多项式系数，应该为正整数，所以"k=1" “这个方法可能是最朴素的一种方法，好多年没有学习数学了，一些轻巧的解决方法一时想不起来了，呵呵。
  ” 。收起