Question

多项式kx^2+7xy-3y^2+x+7y-2分解成两个()次因式的乘积,且其中一个因式为2x+

Answer 1

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

=kx^2+(7y+1)x+(3y-1)(-y+2)

设k=2m,根据十字相乘法

2x (3y-1)

mx (-y+2)

2x(-y+2)+mx(3y-1)=(7y+1)x

解得m=3

∴k=2m=6

∴6x^2+7xy-3y^2+x+7y-2

=6x^2+(7y+1)x+(3y-1)(-y+2)

=(2x+3y-1)(3x-y+2)