How many members do you have on ignore?

[the GOAT]

None

 

[the GOAT]

And to those that have me on ignore

 

[TJ the Chainsaw Mechanic]

none. lol

 

[P.M.P.]

Thinking about putting you and your other WANKERS on mine

 

[TJ the Chainsaw Mechanic]

Thinking about putting you and your other WANKERS on mine

That made me lol.

 

[Guido Salvage]

Probably not enough, took me a long time to realize you can just make the idiots go away.

 

[junkman]

And to those that have me on ignore

Ignored

 

[the GOAT]

Ignored

I wish I could see your poasting but, alas, I have you on ignore.

 

[junkman]

I wish I could see your poasting but, alas, I have you on ignore.

Ignored to the 3rd power

 

[exSW]

None. Who am I to turn a deaf ear to the pitiful cries for help from the irrelevant.

 

[the GOAT]

Ignored to the 3rd power

Ignored to the 3rd power and back again.


POW

 

[junkman]

Damm ,I'm out ,Hard to top that .

 

[junkman]

x³ + x² - x - 1

Factor the first two terms x³ + x² by taking out the
greatest common factor, x². getting x²(x+1)

Factor the last two terms -x-1 by taking out -1,
getting -1(x+1)

So we have

x²(x+1)-1(x+1)

Notice that there is a common factor, (x+1)

x²(x+1)-1(x+1)

which we can factor out leaving the x² and the -1 to put
in parentheses:

(x+1)(x²-1)

Now we notice that the (x²-1) can be factored as the difference
of two squares as (x-1)(x+1), so we now have:

(x+1)(x-1)(x+1)

Since two of the factors are the same (x+1), we write the final answer
with that factor squared:

(x+1)²(x-1)

----------------------

2x to the third power - 3x to the second power - 2x + 3.

2x³ - 3x² - 2x + 3

Factor the first two terms 2x³ - 3x² by taking out the
greatest common factor, x². getting x²(2x-3)

Factor the last two terms -2x+3 by taking out -1,
getting -1(2x-3)

So we have

x²(2x-3)-1(2x-3)

Notice that there is a common factor, (2x-3)

x²(2x-3)-1(2x-3)

which we can factor out leaving the x² and the -1 to put
in parentheses:

(2x-3)(x²-1)

Now we notice that the (x²-1) can be factored as the difference
of two squares as (x-1)(x+1), so we end up with:

(2x-3)(x-1)(x+1)



The back again wins every time

 

[merc_man]

None

 

[rburg]

One.

 

[DSS]

Three. So far.

 

[dieselfitter]

3 on ignore. Completely different site now.

 

[stihl sawing]

None, dano and slowp isn't here.

 

[cantgitright]

not one!

 

[paragonbuilder]

None


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