ax²+ax²y+2bx³-2bx²y
=(ax²+ax²y)+(2bx³-2bx²y)
=ax²(1+y)+2bx²(x-y)
做不下去了,可能是题目有错,检查一下,再向我追问吧.
应该还有一个,是第二项的符号为负吧,(否则就是第四个的符号为正)ax³-ax²y+2bx³-2bx²y=(ax³-ax²y)+(2bx³-2bx²y)=ax²(x-y)+2bx²(x-y)=x²(x-y)(a+2b)或者是:ax³+ax²y+2bx³+2bx²y=(ax³+ax²y)+(2bx³+2bx²y)=ax²(x+y)+2bx²(x+y)=x²(x+y)(a+2b)
ax³+ax²y-2bx³-2bx²y=(ax³+ax²y)-(2bx³+2bx²y)=ax²(x+y)-2bx²(x+y)=x²(x+y)(a-2b)