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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    $\begingroup$ $1-ab-cd-abcd$ does not factor to $(1-cd)(1-ab)$, because $(1-cd)(1-ab)=1-ab-cd+abcd$. $\endgroup$ – Florian Mar 21 '12 at 12:32
  • $\begingroup$ @Florian: Woops, typo. Thanks. (fixed) $\endgroup$ – stariz77 Mar 21 '12 at 12:34
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    $\begingroup$ how about 1(1-ab) - cd(1-ab); then factor out a (1-ab) to get (1-ab)(1-cd)? $\endgroup$ – Ben Mar 21 '12 at 12:41
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    $\begingroup$ I guess practice more problems then you'll see $\endgroup$ – Kirthi Raman Mar 21 '12 at 12:43
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    $\begingroup$ The pattern to observe would probably be $(1-x)(1-y)=1-x-y+xy$. $\endgroup$ – Dejan Govc Mar 21 '12 at 12:54

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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