若多项式x2-y2+3x-7y+k可以分解成两个一次因式的乘积,则k=_.

问题描述:

若多项式x2-y2+3x-7y+k可以分解成两个一次因式的乘积,则k=______.

∵x2-y2+3x-7y+k=(x+

3
2
2-(y+
7
2
2=(x+
3
2
+y+
7
2
)(x+
3
2
-y-
7
2
)=(x+y+5)(x-y-2),
又∵(x+y+5)(x-y-2)=x2-y2+3x-7y-10,
∴k=-10.
故答案为:-10.