Question 662443
<pre>
2(x+2)(x-1)(x-4)

That has four factors, 2, (x+3), (x-1), and (x-4).

Take just the the first two, 2(x+2) and write it as 2x+4 and put parentheses
around it, (2x+4) and replace the 2(x+2) in the above by (2x+4).  So now we
have:

(2x+4)(x-1)(x-4)

Now there are just 3 factors, (2x+4), (x-1), and (x-4) 

As before, we take just the first two (2x+4)(x-1), use FOIL to multiply that
out as (2x²-2x+4x-4) and combine the middle two terms and get (2x²+2x-4),
leaving the parentheses around them, and replace the first two factors
with (2x²+2x-4).

(2x²+2x-4)(x-4)

Now there are just 2 factors, (2x²+2x-4) and (x-4).

The first factor has 3 terms and the second factor has 2 terms. So when
you multiply each term of (2x²+2x-4) by each term of (x-4) you will have
2×3 or 6 terms:  Let's go through them:

1.   Multiply the 1st term on the left, 2x², by the 1st term on the right,
     which is x, and get 2x³.  Write that down:

2x³

2.   Multiply the 1st term on the left, 2x², by the 2nd term on the right,
     which is -4, and get -8x². Write that down after the 2x³:

2x³-8x²

3.   Multiply the 2nd term on the left, +2x, by the 1st term on the right,
     which is x, and get +2x². Write that down after the 2x³-8x²:

2x³-8x²+2x²

4.   Multiply the 2nd term on the left, +2x, by the 2nd term on the right,
     which is -4, and get -8x. Write that down after the 2x³-8x²+2x²:

2x³-8x²+2x²-8x

5.   Multiply the 3rd term on the left, -4, by the 1st term on the right,
     which is x, and get -4x. Write that down after the 2x³-8x²+2x²-8x:

2x³-8x²+2x²-8x-4x

6.   Multiply the 3rd term on the left, -4, by the 2nd term on the right,
     which is -4, and get +16. Write that down after the 2x³-8x²-2x²-8x-4x:

2x³-8x²+2x²-8x-4x+16

Now combine the 2nd and 3rd terms as -6x², and combine the 4th and 5th terms
as -6x and you end up with

2x³-6x²-12x+16

Edwin</pre>