Question 107879
Take the first coefficient (6) and the last coefficient (2) and multiply them to get 12. Now what two numbers multiply to 12 but add to the middle coefficient of 7?


It turns out that 4 and 3 multiply to 12 and add to 7.


{{{6m^6n+4m^5n^2+3m^5n^2+2m^4n^3}}} Now replace the middle term {{{7m^5n^2}}} with the sum {{{4m^5n^2+3m^5n^2}}} (these two are equivalent)


{{{(6m^6n+4m^5n^2)+(3m^5n^2+2m^4n^3)}}} Group like terms



{{{2m^5n(3m+2n)+(3m^5n^2+2m^4n^3)}}} Factor out the GCF {{{2m^5n}}} out of the first group



{{{2m^5n(3m+2n)+m^4n^2(3m+2n)}}} Factor out the GCF {{{m^4n^2}}} out of the second group



Notice we have a common term of 3m+2n, this means we can combine like terms



{{{(2m^5n+m^4n^2)(3m+2n)}}} Combine like terms



Notice how {{{(2m^5n+m^4n^2)(3m+2n)}}}  foils to {{{6m^6n+4m^5n^2+3m^5n^2+2m^4n^3}}}, so this verifies our answer.