The given equation is
![4x^2-16=0](https://img.qammunity.org/2023/formulas/mathematics/college/fp6n401ahf6s8xb1mlqakc00rtx6bg23us.png)
We will factor the given equation using the difference of squares formula given below
![a^2-b^2=(a+b)\cdot(a-b)](https://img.qammunity.org/2023/formulas/mathematics/college/zd98c30bxfblgh5gzh6bxr8njx6xbsx3gj.png)
So, we have
![a=2x,b=4](https://img.qammunity.org/2023/formulas/mathematics/college/hs0kvujdz74iefh2pekct5yhf4tikh6lug.png)
Applying the above formula will result in,
![\begin{gathered} a^2-b^2=(a+b)\cdot(a-b) \\ (2x)^2-(4)^2=(2x+4)\cdot(2x-4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y52yl6oknhed0els5c3tx7rckd501ljosi.png)
Therefore, the factors of the given equation are
![(2x+4)\text{ and }\left(2x-4\right)](https://img.qammunity.org/2023/formulas/mathematics/college/dckfn74qfdordqogc44hhl2bgva01y2vkj.png)
Bonus:
You can find the possible values of x by equating the factors to zero.
![\begin{gathered} (2x+4)=0 \\ 2x=-4 \\ x=-(4)/(2) \\ x=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pzsy1f0sevuidqz3x3lac0t2myvs6r6s7t.png)
Similarly,
![\begin{gathered} (2x-4)=0 \\ 2x=4 \\ x=(4)/(2) \\ x=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ocjl89g6ycgqbloose6zxrwnyqn4n3wypb.png)
So, the possible values of x are
![x=(2,-2)](https://img.qammunity.org/2023/formulas/mathematics/college/d0ym7owj7ajcg8jfq4yddzaa58l6mouet0.png)