Question

因式分解…

当x=－1时，x³+2x²－5x－6=0,请根据这一事实，将x³+2x²-5x-6分解因式

Answer 1

解：根据当x=－1时，x³+2x²－5x－6=0, 则有x+1，一定是x³+2x²-5x-6的一个因式。
所以x³+2x²-5x-6 =x²(x+1)+x(x+1)-6(x+1)= (x+1)(x²+x-6)=(x+1)(x+3)(x-2)

Answer 2

x³+2x²-5x-6
=(x³+2x²+x-6x-6
=x(x+1)²-6(x+1)
=(x+1)*[x(x+1)-6]
=(x+1)(x²+x-6)
=(x+1)(x+3)(x-2)

追问

不好意思，题目打错了…当x=-1时，x³+2x²-5x-6=0,请根据这一事实，将x³+2x²-5x-6分解因式…

Answer 3

x³+2x²-5x-6
=x³+x²+x²-5x-6
=x²(x+1)+(x+1)(x-6)
=(x+1)(x²+x-6)
=(x+1)(x-3)(x+2)