You multiply it out. What do you mean "the easiest way"? I guess you can do it term-by-term using the Distributive Property:

$\begin{align*}(x-1)(x+3)(x-3)(x-7) & = x(x+3)(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = x\cdot x(x-3)(x-7) + x\cdot 3(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = \vdots\end{align*}$

For $(1-a^2)(1-b^2) - 4ab$, multiply out, then factor.

For $x^4+4$, it is irreducible over the reals. Are you allowed to factor over some larger field, like the complex numbers?