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?
What are the steps for factorizing $1 - ab - cd + abcd$?
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$\begingroup$
factoring
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3$1-ab-cd-abcd$ does not factor to $(1-cd)(1-ab)$, because $(1-cd)(1-ab)=1-ab-cd+abcd$. – 2012-03-21
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0@Florian: Woops, typo. Thanks. (fixed) – 2012-03-21
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2how about 1(1-ab) - cd(1-ab); then factor out a (1-ab) to get (1-ab)(1-cd)? – 2012-03-21
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1I guess practice more problems then you'll see – 2012-03-21
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3The pattern to observe would probably be $(1-x)(1-y)=1-x-y+xy$. – 2012-03-21
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0Another heuristic that works here: You can factorize any expression with 4 terms if the sum of the 2nd and the 3rd terms equals the product of the first and the 4th. – 2015-06-11
1 Answers
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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!