Question

这5道因式分解这么做

每有一个人回答，悬赏+5，直到出现正确答案为止。
第四道（手写的第一道）-b右上角漏了个3

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Answer 1

①y(y+1)(x^2+1)+x(2y^2+2y+1)
=y(y+1)x^2+(2y^2+2y+1)x+y(y+1)
=(yx+y+1)[(y+1)x+y]

追问

还有4题

追答

①ab(x^2-y^2)-(a^2-b^2)(xy+1)-(a^2+b^2)(x+y)
=abx^2-(a^2-b^2)xy-aby^2-(a^2+b^2)(x+y)-(a^2-b^2)
=(ax+by)(bx-ay)-(a^2+b^2)(x+y)-(a-b)(a+b)
=(ax+by+a-b)(bx-ay-a-b)
②-b^3+(a^2+3a-1)b-a^2+a+2
=(b-1)a^2+(3b+1)a-(b^3+b-2)
=(b-1)a^2+(3b+1)a-[(b^3-1)+(b-1)]
=(b-1)a^2+(3b+1)a-(b-1)(b^2+b+2)
=[(b-1)a+b^2+b+2](a-b+1)
③(1-2a-a^2)b+a(a-1)(2b^2-1)
=2a(a-1)b^2+(1-2a-a^2)b-a(a-1)
=(2ab+a-1)[(a-1)b-a]
=(2ab+a-1)(ab-a-b)
④x^3+(2a+1)x^2+(a^2+2a-1)x+(a^2-1)
x=-1代入原式=0.所以(x+1)是原式的因式
∴原式=(x^3-x)+(2ax^2+2ax)+(x^2-1)+(a^2x+a^2)
=(x+1)(x^2-x+2ax+x-1+a^2)
=(x+1)(x^2+2ax+a^2-1)
=(x-1)[(x+a)^2-1]
=(x-1)(x+a+1)(x+a-1)