Question 959657
(FIXED MISTAKE)


{{{27p^2-48}}}


{{{cross(3*3*p*p-12*4)}}}
should be {{{3*3*3*p^2-12*4}}}


{{{3*3*3*p*p-2*2*3*2*2}}}


{{{highlight((3)(3^2p^2-2^4))}}}
which allows factoring difference of squares:
{{{highlight(3(3p-4)(3p+4))}}}


The idea is to break every term into its prime factorizations.  This allows you to see all possible distributed factors.