Solve the following equation:

$$
(1+x)^{4}=16 x^{2} .
$$
$$
(1+x)^{4}=16 x^{2}
$$

Let's move everything to the left side of the equation, then factor the resulting expression:

$$
0=(1+x)^{4}-16 x^{2}=\left[(1+x)^{2}-4 x\right]\left[(1+x)^{2}+4 x\right]=\left(x^{2}-2 x+1\right)\left(x^{2}+6 x+1\right)
$$

A product is zero if and only if one of its factors is zero. Thus, the solutions to (1) are provided by the solutions to the equations \(x^{2}-2 x+1=0\) and \(x^{2}+6 x+1=0\). From the first equation, we get \(x_{1}=1\), and from the second equation, we get \(x_{2}=-3+\sqrt{8}\) and \(x_{3}=-3-\sqrt{8}\), which are the roots of (1).