Armadillo Run forum

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...

JustcallmeDrago wrote:X|

This is the internet... all software is free

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?

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

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.

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

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)?

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

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

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)

NickyNick wrote:Бляяяяя...

Watch your language!

JustcallmeDrago wrote:Watch your language!

Cthulhu fhtagn!

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...

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

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

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)