∵x2+3xy+2y2=(x+y)(x+2y).
设x2+3xy+2y2+4x+5y+3=(x+y+m)(x+2y+n).
比较系数得,m+n=4,2m+n=5.解得m=1,n=3.
∴x2+3xy+2y2+4x+5y+3=(x+y+1)(x+2y+3).
解答下面问题:在有理数范围内,分解因式2x2﹣21xy﹣11y2﹣x+34y﹣3=.