Question 745033
DISCLAIMER
There may be a procedure that you were supposedly taught, but I would not know about it.
 
If {{{16m^2+15n-40m-6mn}}} is to be the product of two factors,
both factor must include a term on {{{m}}} (so we could get terms with {{{m}}} and a term with {{{m^2}}} too),
and just one factor must include a term in {{{n}}} (so we could get twems in {{{n}}}, but no term in {{{n^2}}}).
Let's try {{{(Am+B)(Cm+Dn)=16m^2+15n-40m-6mn}}}.
{{{(Am+B)(Cm+Dn)=16m^2+15n-40m-6mn}}} --> {{{ACm^2+ADmn+BCm+BDn=16m^2+15n-40m-6mn}}}
For the expressions on the left and the right to be equivalent, all four coefficients must match:
{{{AC=16}}}
{{{AD=-6}}}
{{{BC=-40}}} and
{{{BD=15}}}
The way I would look for answers would be setting up a tic-tac-toe grid:
{{{drawing(200,200,0,9,-1,8,
line(0,5,9,5),line(0,2,9,2),
line(3,8,3,-1),line(6,8,6,-1),
locate(0.2,4,"C=?"),locate(3.2,4,AC=16),locate(6.2,4,BC=-40),
locate(0.2,1,"D=?"),locate(3.2,1,AD=-6),locate(6.2,1,BD=15),
locate(4,7,"A=?"),locate(7,7,"B=?")
)}}} My first guesses as common factors for each column are {{{A=2}}} and {{{B=5}}}.
{{{system(A=2,AC=16,AD=-6)}}} --> {{{system(C=8,D=-3)}}} but
{{{system(B=5,BC=-40,BD=15)}}} --> {{{system(C=-8,D=3)}}} does not agree
However, If I change my guess to {{{B=-5}}}, I get {{{C}}} and {{{D}}} solutions that agree with the ones from {{{A=2}}}:
{{{system(B=-5,BC=-40,BD=15)}}} --> {{{system(C=8,D=-3)}}}
So the solution is {{{system(A=2,B=-5,C=-8,D=3)}}} or
{{{16m^2+15n-40m-6mn=highlight((2m-5)(8m-3n))}}}