Question

已知多项式x²-y²-2x-4y-3，将其因式分解后的式子是什么？

过程需要写下来

Answer 1

首先进行配方
原式=x²-2x+1-y²-4y-4
=(x-1)²-(y+2)²
平方差公式
=(x-1+y+2)(x-1-y-2)
=(x+y+1)(x-y-3)

Answer 2

原式=(x²-2x-1)-(y²+4y+4)
=(x-1)²-(y+2)²
=(x-1+y+2)(x-1-y-2)
=(x+y+1)(x-y-3)