If the factors of a polyomials are
![\begin{gathered} (x+3) \\ (x+12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y09tye1fg8bfmz6y8s6hcr2k6m7towk0nw.png)
This means that the polynomial can be written as a power of thouse factors, like:
![P(x)=(x+3)^n(x+12)^m](https://img.qammunity.org/2023/formulas/mathematics/college/3ezo15z1581djgnrfmnxcatghtlfcwxrjz.png)
As we can see, if either of the factors zeros out, the hole polynomial you be zero too, because anything will be multiplied by zero.
So, to find the values of x that will make this polynomial equal to zero, we just solve the equation of each factor equal to zero.
So, the first possible x value is:
![\begin{gathered} (x+3)=0 \\ x+3=0 \\ x=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8gj8bj99e7pwp75di8ua45hldh65futsbn.png)
And the second possible value of x is:
![\begin{gathered} (x+12)=0 \\ x+12=0 \\ x=-12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i8hyxvaokotthp7oc9ld81u9h9r7n2l21c.png)
Thus, the possible values of x that make the polynomial zero are -3 and -12, which corresponds to alternative C.