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Postby TSchultz » Mon Oct 29, 2007 4:26 am

If you ever want me to factor out something crazy or simplify something or integrate something or take the derivative and stuff like that, you can always send me a PM or reply in this thread though...
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Postby Ikerous » Mon Oct 29, 2007 4:33 am

JustcallmeDrago wrote:X|

This is the internet... all software is free
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:33 am

TSchultz wrote:If you ever want me to factor out something crazy or simplify something or integrate something or take the derivative and stuff like that, you can always send me a PM or reply in this thread though...


lol ok thanks.

@Ike: Piracy!! :( i have not been corrupted to the point i will rip off new software.

I'm trying to find the steps, so i worked backwards but still got stuck.

Original: x4 - a2x2 + 16a4

...

16a4 - (9a2x2 + x4)

(4a2 - 3ax + x2)(4a2 + 3ax + x2)

It appears Diff of squares was the last step, but what came before it?
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Postby Ikerous » Mon Oct 29, 2007 4:43 am

x^4 + 16a^2 - (ax)^2
x^4 + 16a^2 = (x^2 + 4a^2)^2 - 8(ax)^2
(x^2 + 4a^2) - 9(ax)^2
(x^2 + 4a^2) - (3ax)^2
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:51 am

Ikerous wrote:x^4 + 16a^2 - (ax)^2
x^4 + 16a^2 = (x^2 + 4a^2)^2 - 8(ax)^2
(x^2 + 4a^2) - 9(ax)^2
(x^2 + 4a^2) - (3ax)^2


could you elaborate on why you did (x^2 + 4a^2)^2 - 8(ax)^2 ?

I'm almost seeing it.
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Postby Ikerous » Mon Oct 29, 2007 4:56 am

I saw that both terms were squares. Kind of like a^2 + b^2 which goes to (a+b)^2 - 2ab. Which comes from (a+b)^2 = a^2 + b^2 + 2ab

Although without the answer i probably would never have done that. I really doubt you'll run across a problem that hard on an exam
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:59 am

Ikerous wrote:I saw that both terms were squares. Kind of like a^2 + b^2 which goes to (a+b)^2 - 2ab. Which comes from (a+b)^2 = a^2 + b^2 + 2ab

Although without the answer i probably would never have done that. I really doubt you'll run across a problem that hard on an exam


wow. i just looked at the paper we got with the packet and it says to skip this one! This is interesting though.

changing problems: should x3 - 1 become (x - 1)(x2 + 2x + 1)?
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Postby Ikerous » Mon Oct 29, 2007 5:01 am

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

Its a basic formula but i always forget it so I just use (x+1) or (x-1) as my first term and use long division to find the second one
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Postby JustcallmeDrago » Mon Oct 29, 2007 5:03 am

Ikerous wrote:x^3 - 1 = (x-1)(x^2 + x + 1)

Its a basic formula but i always forget it so I just use (x+1) or (x-1) as my first term and use long division to find the second one


lol whoops for got to edit the 2x out... ok, that sucks, then i have more problems that aren't totally factored :evil:
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Postby NickyNick » Mon Oct 29, 2007 10:49 am

Бляяяяя...
Пазор и пиздец. Ктулху зохавал ваши мазги!

1 minute of work:
x^4 - a^2x^2 + 16a^4 = x^4 + 8a^2x^2 + 16a^4 - 9a^2x^2 = (x^2 + 4a^2)^2 - 9a^2x^2 = (x^2 + 3ax + 4a^2)(x^2 - 3ax + 4a^2)
AR & W:A forever
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Postby JustcallmeDrago » Mon Oct 29, 2007 12:18 pm

NickyNick wrote:Бляяяяя...


Watch your language!
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Postby NickyNick » Mon Oct 29, 2007 3:34 pm

JustcallmeDrago wrote:Watch your language!

Cthulhu fhtagn! :)
AR & W:A forever
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Postby McGinge » Mon Oct 29, 2007 8:04 pm

NickyNick wrote:Бляяяяя...
Пазор и пиздец. Ктулху зохавал ваши мазги!

1 minute of work:
x^4 - a^2x^2 + 16a^4 = x^4 + 8a^2x^2 + 16a^4 - 9a^2x^2 = (x^2 + 4a^2)^2 - 9a^2x^2 = (x^2 + 3ax + 4a^2)(x^2 - 3ax + 4a^2)


Err yeah... what he said...
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Postby JustcallmeDrago » Tue Nov 06, 2007 2:05 am

x^2 + 4ax + 4a^2 - b^2

:evil:
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Postby TSchultz » Tue Nov 06, 2007 2:26 am

JustcallmeDrago wrote:x^2 + 4ax + 4a^2 - b^2

:evil:

In[1]:= Factor[x^2 + 4 ax + 4 a^2 - b^2]

Out[1]= 4 a^2 + 4 ax - b^2 + x^2

But
x^2 + 4ax + 4a^2 - b^2=
(x+2a)^2-b^2=
(x+2a+b)(x+2a-b)
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