Answer:
All zeros are
x=-10 , x=-5 , x=5
Explanation:
we are given function as
![f(x)=x^3+10x^2-25x-250](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jecsuim9hrz3g04a6ro2gqulv6kyy0j7gu.png)
we are given one of zero is x=-10
we have to use Remainder theorem
we can find all possible factor of 250
![250=-5* 5* -10](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cz57ebrh6gbxh1rsffi0nw5uxfpv2mjxpn.png)
so, we will check zeros at x=-5 and x=5
At x=-5:
we can plug x=-5
![f(-5)=(-5)^3+10(-5)^2-25(-5)-250](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gc99cj9hacm8xuj4i51hnahrkhu18ox18q.png)
![f(-5)=-125+250-\left(-125\right)-250](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1lof6jueey6a04pth6s4nxng6cjchb5zjv.png)
![f(-5)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hebl8evkkotgvqeh1lcfwlet0bvwcr9akb.png)
At x=5:
we can plug x=5
![f(5)=(5)^3+10(5)^2-25(5)-250](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o9m7fqysgb37xyxs1e98e4p9l0flc2uuom.png)
![f(5)=125+250-\left(125\right)-250](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4ynhqmfxsgxucdmif152ux7ua3w2eabeog.png)
![f(5)=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/zw1kv8p9qb86mgncapysi7swp373lw7xo6.png)
So, other zeros are
x=-5 and x=5
All zeros are
x=-10 , x=-5 , x=5