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Party time! :D

lol
I dunno...

What can you teach me?

a lot, there's a fair amount to algebra, but what's the next logical step?

Teach about polynomial.

I have already to an extent, but where to go next...

Ratios?

awww, you're making me think of my cat in the hospital :(

How did it go with her?

broken leg...
waiting for an operation

You beat you cat Turing?!!!

aww :(

turing sat on her leg. :(

she fell off the roof.
how pessimistic

awww, sorry i was jking..

It's all good :P

Due to Turing's intense physics stress his cat decided to commit suicide :P

(x^2)*(y^5)*(z)
------------ = (x)*(y^5/9)*(z^1/3)?
(x)*(y^9)*(z^3)

Cats are cool.

why \( y^{5/9} \) ?

no doubt. ;)

you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...

http://i.imgur.com/48wOw.jpg

Oh right

(x^2)*(y^5)*(z)
------------ = (x)*(y^5-9)*(z^1-3)?
(x)*(y^9)*(z^3)

yes, simplify...

(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?

http://i.imgur.com/6uzXf.jpg

cute^

yes, do you know another way to write\[x^{-a}\]???

No

http://i.imgur.com/eqTIb.jpg

Oh right, all negative exponents make the "total number" divided by one.

Btw who can explain why x^0 = 1?

(x)*(1/y^4)*(1/z^2)

Turing that's a very important yet fundamental question.

another one is why \( (a^b)^c = a^{bc} \) ?

the last rule is easier to explain, perhaps I should...

(x)*(1/y^4)*(1/z^2)
(x)*(1/y^4*z^2)?

yes, now write it as a fraction
what goes on the bottom?
what goes on the top?

@FFMactually that's not so easy to explain now that I think about it.

Can you explain the second rule intuitively?

1
x* --------
y^4*z^2

@ inopeki
you can put the x on top, it means the same thing and looks nicer

x*1
--------
y^4*z^2

Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?

Whats that f?

unspecified functions
he's asking how far the rule about exponents can be extended..

x
-------- Oh right
y^4*z^2

Actually we can't but I believe that is true for exponential functions though.

Umm, ok?

it must be for simple ones\[2^x2^y=2^{x+y}\]of course

Tha's whats Inopeki is using right ?

Yeah, but does that mean that it becomes
x
-------- ?
yz^6

no because y and z are different bases
notice the rule above had both base x.

YEah but then i dont see the connection..

I know about that, like x^2*x^8=x^10

x^10/x^5=x^5

Basic

Turing you have a heck of patience :D

:-)

yes exactly

3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?

yep :)
great!

Really? :D

When hes not here i try to go on purplemath and khan :)

a^2+ab-ba-b^2

simplify

Books books books!! online learning has it's own limitation :-)

a^2-b^2?

it doesnt @ffm i disagree

Foolformath, im having trouble finding books here in sweden

and what is the name of that form? remember?

The fundamental theorem of algebra? Or want that the one with the multiplex?

Thanks but you get what you put in, I think is true with all this stuff.

Oh right

run the FOIL backwards I meant*

x(x-4)?

Oh

no look at the form\[a^2-b^2=(a-b)(a+b)\]so what is a and b in\[x^2-4\]???

(x-2)(x+2)?

there ya go :)

how about\[x^4-y^4\](this is a favorite question on OS)

(x^2-y^2)(x^2+y^2)

Why?

good, now is that all we can do with it though?
(I've seen a lot of tutors stop here too ;-)

hint:look at the first set of parentheses

Ummm

what is the first sett of parentheses?

(x^2-y^2)?

yes, and can you factor that?

GCF of x^2 is x
GCF of y^2 is y
xy(x^2/xy)-xy(y^2/xy)?

no you're over-thinking
x^2-y^2 is difference of squares again!

\[a^2-b^2=(a-b)(a+b)\]

So (x-y)(x+y)?

yes!
so now factor\[p^4-q^4\]completely!!

you will apply difference of squares twice

(p^2-q^2)(p^2+q^2)
(p-q)(p+q)(p+q)+(p+q)?

Without the +

between the parenthesis

Ohhh