Question

已知x^2-xy-2y^2+mx+7y-3能够分解成两个整系数的一次因式的乘积,求m的值

Answer 1

x^2-xy-2y^2+mx+7y-3

=(x-2y)(x+y)+mx+7y-3

=(x-2y+a)(x+y+b)

=(x-2y)(x+y)+(a+b)x+(a-2b)y+ab

所以a+b=m (1)

a-2b=7 (2)

ab=-3 (3)

有(1),(2)

b=(m-7)/3,a=7+2b=(2m+7)/3

代入(3)

(m-7)(2m+7)/9=-3

2m²-7m-22=0

(2m-11)(m+2)=0

m=11/2,m=-2