方程x3-6x2-15x-10=0的实根个数是（ ）
A．3
B．2
C．1
D．0
设函数f（x）=x3-6x2-15x-10，
则f'（x）=3x2-12x-15=3（x2-4x-5）=3（x+1）（x-5），
由f'（x）＞0得，x＞5或x＜-1，此时函数单调递增，
由f'（x）＜0得，-1＜x＜5，此时函数单调递减，
故当x=-1时，函数取得极大值f（-1）=-2＜0，
当x=5时，函数取得极小值f（5）=-110＜0，
∴方程x3-6x2-15x-10=0的实根个数为1个．
故选：C．
| 根据对话内容及方框中所给的句子补全对话.其中有一项是多余的。 | |
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