Question 149446
{{{(x-3)(x+2)}}}
Let a=(x-3).
{{{(x-3)(x+2)=a*(x+2)}}}
Using the distributive property on the right hand side, this becomes,
{{{(x-3)(x+2)=ax+2a)}}}
Now substitute for a,
{{{(x-3)(x+2)=ax+2a)}}}
{{{(x-3)(x+2)=(x-3)x+2(x-3))}}}
Use the distributive property again on the right hand side,
{{{(x-3)(x+2)=(x-3)x+2(x-3))}}}
{{{(x-3)(x+2)=(x^2-3x)+(2x-6))}}}
{{{(x-3)(x+2)=x^2-x-6)}}}
This is the FOIL Method, in disguise. 
FOIL stands for First, Outer, Inner, Last.
First {{{(highlight(x)-3)(highlight(x)+2))}}}={{{x*x=x^2}}}
Outer {{{(highlight(x)-3)(x+highlight(2)))}}}={{{x*2=2x}}}
First {{{(x+highlight(-3))(highlight(x)+2))}}}={{{-3*x=-3x}}}
Last {{{(x+highlight(-3))(x+highlight(2)))}}}={{{(-3)*2=-6}}}
{{{(x-3)(x+2)=x^2+2x-3x-6)}}}
{{{(x-3)(x+2)=x^2-x-6)}}}