1. If $2 x^{2}-6 y^{2}+x y+k x+6$ can be factored into the product of two linear factors, then the value of $k$ is
$1 . \pm 7$
Analysis: Since $2 x^{2}-6 y^{2}+x y=(x-2 y)(2 x+3 y)$, we can assume without loss of generality that $2 x^{2}-6 y^{2}+x y+k x+6=(x-2 y+a)\left(2 x+3 y+\frac{6}{a}\right)$, matching the coefficients of $x, y$ we get $3 a-\frac{12}{a}=0$, $\frac{6}{a}+2 a=k$, solving these gives $a= \pm 2$, thus $k= \pm 7$.