How do you factor ab(a+b)+bc(b+c)+ca(c+a)+3abc?

Friday, August 2, 2013

How do you factor ab(a+b)+bc(b+c)+ca(c+a)+3abc?

In order to factor this expression, first open all parenthesis by distributing the term in front of the parenthesis. The given expression will then become

$\displaystyle{{a}}^{{2}}{b}+{a}{{b}}^{{2}}+{{b}}^{{2}}{c}+{b}{{c}}^{{2}}+{a}{{c}}^{{2}}+{{a}}^{{2}}{c}+{3}{a}{b}{c}$ .

We need to group these terms in a way so that we could find some common factors. Now there are 7 terms in the expression, which is not a very convenient number. However, the coefficient 3 in front of the abc term gives us a clue that this term should be broken up into 3: abc + abc + abc.

Now we have 9 terms, which we can group into three groups of three so that each group has a common factor:

$\displaystyle{{a}}^{{2}}{b}+{{a}}^{{2}}{c}+{a}{b}{c}={a}{\left({a}{b}+{a}{c}+{b}{c}\right)}$

$\displaystyle{a}{{b}}^{{2}}+{{b}}^{{2}}{c}+{a}{b}{c}={b}{\left({a}{b}+{b}{c}+{a}{c}\right)}$

$\displaystyle{b}{{c}}^{{2}}+{a}{{c}}^{{2}}+{a}{b}{c}={c}{\left({b}{c}+{a}{c}+{a}{b}\right)}$

Note that the expressions in parenthesis are the same, so the three groups of terms have a common trinomial factor! It can be factored out, and the entire sum of nine (originally, seven) terms becomes

$\displaystyle{\left({a}+{b}+{c}\right)}{\left({a}{b}+{a}{c}+{b}{c}\right)}$ .

The given expression can be factored as $\displaystyle{\left({a}+{b}+{c}\right)}{\left({a}{b}+{a}{c}+{b}{c}\right)}$

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Friday, August 2, 2013