Question

换元法分解因式！！！快！！！

（x²+2x-1）²+2（x²+2x)-10

Answer 1

（x²+2x-1）²+2（x²+2x)-10令y=x²+2x-1, 则
原式=y²+2y-8
=(y+4)(y-2)
=(x²+2x+3)(x²+2x-3)
=(x+3)(x-1)(x²+2x+3)

Answer 2

（x²+2x-1）²+2（x²+2x)-10
=（x²+2x-1）²+2（x²+2x-1)-10+2
设x²+2x-1=k，
原式=k²+2k-8
=k²+2k+1-9
=(k+1)²-9
=(k+1+3)(k+1-3)
=(k+4)(k-2)
=(x²+2x-1+4)(x²+2x-1-2)
=(x²+2x+3)(x²+2x-3)
=(x²+2x+3)(x+3)(x-1)

Answer 3

令x²+2x=y，则有
原式=(y-1)²+2y-10
=y²-9
=(y-3)(y+3)
=(x²+2x-3)(x²+2x+3)
=(x+3)(x-1)(x²+2x+3)

Answer 4

x²+2x-1=t x²+2x=t+1 （x²+2x-1）²+2（x²+2x)-10=t^2+2t+2-10=t^2+2t-8=(t+4)(t-2)=(x^2+2x-1+4)(x^2+2x-1-2)=(x^2+2x+3)(x^2+2x-3）=（x^2+2x+3)(x+3)(x-1)