Answer:
The solutions are:
and
and
![x = -6](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vucym2o7axskeoi7c52nbsu7s8ie4ekdwm.png)
Explanation:
1) Make the function equal to zero
![f(x)=x^3+4x^2-12x = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/m359j0wuxrs3q1vsiyryigphvh3lhkzo55.png)
2) Take x as a common factor
![x(x^2+4x-12) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2nfpbe0pwco7twyvavqi8bo9m499rl1c05.png)
3) Factor the expression
![x^2+4x-12](https://img.qammunity.org/2020/formulas/mathematics/high-school/ulu0hnl7fi2m7r6d6qt9kk97iglqmya7u7.png)
The sought-after factors are such numbers that when multiplying them obtain as result -12 and when adding both numbers obtain as result 4.
The numbers that meet this condition are
6 and -2
Because
![6*(-2) = -12\\\\6 -2 = 4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/du4ah2sj3rj9q0o0heub3lcwc66a6femav.png)
Then the factors are
![x^2+4x-12=(x-2)(x+6)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xfb1bjzes5qvzk13wth683qeemmyxzktqe.png)
4) Solve the equation for x
![x(x-2)(x+6) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lwfel8blykyuk1qbm4v3yuybyvofpl6aqb.png)
The solutions are:
and
and
![x = -6](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vucym2o7axskeoi7c52nbsu7s8ie4ekdwm.png)