Question 161994
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Use FOIL to multiply
{{{(5x-1)(3x-2)}}}

The letters of the word "FOIL" is an acronym for 

"F=FIRSTS", "O=OUTERS", "I=INNERS", "L=LASTS"

The FIRSTS:
The terms {{{5x}}} and {{{3x}}} are the FIRSTS, because
{{{5x}}} comes FIRST in {{{(5x-1)}}} and {{{3x}}} comes 
FIRST in {{{(3x-2)}}}

So we multiply the two FIRSTS, {{{5x*3x}}} and get
{{{15x^2}}}, so we write that down under the original
problem:

{{{(5x-1)(3x-2)}}}
{{{15x^2}}}

The OUTERS:
The terms {{{5x}}} and {{{-2}}} are the OUTERS, because
{{{5x}}} and {{{-2}}} are on the outside of {{{(5x-1)(3x-2)}}},
the ones located the farthest apart. 

So we multiply the two OUTERS, {{{5x*-2}}} and get
{{{-10x}}}, so we write that down under the original
problem next to the {{{-10x}}}. like this::

{{{(5x-1)(3x-2)}}}
{{{15x^2-10x}}}

The INNERS:
The terms {{{-1}}} and {{{3x}}} are the INNERS, because
{{{5x}}} and {{{-2}}} are on the inside of {{{(5x-1)(3x-2)}}},
the ones located in the middle. 

So we multiply the two INNERS, {{{-1*3x}}} and get
{{{-3x}}}, so we write that down under the original
problem:

{{{(5x-1)(3x-2)}}}
{{{15x^2-10x-3x}}}

The LASTS:
The terms {{{-1}}} and {{{-2}}} are the LASSTS, because
{{{-1}}} comes LAST in {{{(5x-1)}}} and {{{-2}}} comes 
FIRST in {{{(3x-2)}}}

So we multiply the two LASTS, {{{(-1)*(-2)}}} and get
{{{matrix(1,2,"+",2)}}}, so we tack that on down under 
the original problem:

{{{(5x-1)(3x-2)}}}
{{{15x^2-10x-3x+2}}}

Now that we have finished the "FOIL" process, we look
to see if there are any like terms that will combine:

We see that the middle two terms {{{-10x-3x}}} will
combine to give {{{-13x}}}, so underneath we replace
the {{{-10x-3x}}} by {{{-13x}}} and get

{{{(5x-1)(3x-2)}}}
{{{15x^2-10x-3x+2}}}
{{{15x^2-13x+2}}}

Edwin</pre>