SOLUTION: Hello, I would really appreciate if someone could show me how to factor the following equation using the ac method. 8x^2-22x+5. I have been trying but can not figure out how to do

To factor:  8x²-22x+5

Multiply the 8 by the 5 ignoring signs.  Get 40

Write down all the ways to have two positive integers
which have product 40, starting with 40*1

40*1
20*2
10*4
 8*5

Since the last sign in 8x²-22x+5 is +, ADD them,
and place the SUM out beside that:

40*1    40+1=41
20*2    20+2=22
10*4    10+4=14
 8*5     8+5=13

Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 8x²-22x+5

So we replace the number 22 by 20+2

8x²-22x+5
8x²-(20+2)x+5

Then we distribute to remove the parentheses:

8x²-20x-2x+5

Factor the first two terms 8x²-20x by taking out the
greatest common factor, getting 4x(2x-5)

Factor the last two terms -2x+5 by taking out the
greatest common factor, -1, getting -1(2x-5)

So we have

4x(2x-5)-1(2x-5)

Notice that there is a common factor, (2x-5)

4x(2x-5)-1(2x-5)

which we can factor out leaving the 4x and the -1 to put 
in parentheses:

(2x-5)(4x-1)

To see a couple of other examples just like that using the AC
 method, go here:

http://www.algebra.com/my/change_this_name32371.lesson?content_action=show_dev

Edwin