求函数f(x)=x^3 + 2 x^2 - 5x - 6的零点

f(x)=x^3 + 2 x^2 - 5x - 6
=X^3+X^2+X^2-5X-6
=X^2(X+1)+(X+1)(X-6)
=(X+1)(X^2+X-6)
=(X+1)(X-2)(X+3)=0
所以零点为 -1,2, -3

f(x)=x^3 + 2 x^2 - 5x - 6
=(x^3 + x^2) + (x^2- 5x - 6)
=x^2(x+1)+(x+1)(x-6)
=(x+1)(x^2+x-6)
=(x+1)(x+3)(x-2)
令f(x)=0，则x=-1或x=-3或x=2。
所以x=-1，x=-3，x=2是f（x）的零点。

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

f(x)=x^3 +x^2+x^2-5x-6
=x^2 (x+1)+(x+1) (x-6)
=(x^2+x-6)(x+1)
=(x+1)(x-2)(x+3)
零点为-1 2 -3