I can see that $1 - ab - cd + abcd$ factors to $(1-cd)(1-ab)$ but only because I tried a lot of different factors in a trial and error method, so it took me a while. I was wondering what the pattern to observe is and what the general/efficient method for factoring something like this is?
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$\begingroup$The steps are usually called "factoring by grouping." As all factoring is basically educated trial-and-error, you should be encouraged that you found the correct factorization. Anyway, here's more detail about factoring by grouping:
$$ \begin{align*} 1 - ab - cd + abcd &= (1-ab) - (cd - abcd) \\ &= (1-ab) - cd(1 - ab) \\ &= (1-ab)(1-cd). \end{align*} $$ The crucial steps are: finding groups that look very similar -- with a goal of factoring a gcd from each group. If each group now contains the same expression in parentheses, then that expression can be factored out.
Hope this helps!
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