f(x)=x³+2x²-5x-6=x³+2x²+x-6x-6=x(x²+2x+1)-6(x+1)=x(x+1)²-6(x+1)=(x+1)[x(x+1)-6]=(x+1)(x²+x-6)=(x+1)(x+3)(x-2)∴零点为x=-1,x=-3,x=2