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TuringTestBest ResponseYou've already chosen the best response.5
a lot, there's a fair amount to algebra, but what's the next logical step?
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Teach about polynomial.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
I have already to an extent, but where to go next...
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
That's a good point FFM so lets do simplification that you know, but a little more intense: I think I need to make sure you can simplify bigger things:\[\frac{x^2y^5z}{xy^9z^3}\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
awww, you're making me think of my cat in the hospital :(
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
How did it go with her?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
broken leg... waiting for an operation
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
You beat you cat Turing?!!!
 2 years ago

saifoo.khanBest ResponseYou've already chosen the best response.0
turing sat on her leg. :(
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
she fell off the roof. how pessimistic
 2 years ago

saifoo.khanBest ResponseYou've already chosen the best response.0
awww, sorry i was jking..
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Due to Turing's intense physics stress his cat decided to commit suicide :P
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(x^2)*(y^5)*(z)  = (x)*(y^5/9)*(z^1/3)? (x)*(y^9)*(z^3)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
why \( y^{5/9} \) ?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{ab}\]always...
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(x^2)*(y^5)*(z)  = (x)*(y^59)*(z^13)? (x)*(y^9)*(z^3)
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(x)*(y^59)*(z^13)=(x)*(y^4)*(z^2)?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
yes, do you know another way to write\[x^{a}\]???
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
\[x^{a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
Oh right, all negative exponents make the "total number" divided by one.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Btw who can explain why x^0 = 1?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
Oh dear... I have my answers about x^0=1 (not true for x=0), but I'm sure FFM would not approve of them :/ @inopeki, yes now rewrite it as 1 fraction...
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Turing that's a very important yet fundamental question.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
another one is why \( (a^b)^c = a^{bc} \) ?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
well I have a very simple proof of it and I can show why it is not true for x=0 what more do I need in your opinion?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
the last rule is easier to explain, perhaps I should...
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(x)*(1/y^4)*(1/z^2) (x)*(1/y^4*z^2)?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
yes, now write it as a fraction what goes on the bottom? what goes on the top?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
@FFMactually that's not so easy to explain now that I think about it.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Can you explain the second rule intuitively?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
no, I was thinking more of how easy it is to show x^ax^b=x^(a+b) intuitively, the other is tricky to me.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
@ inopeki you can put the x on top, it means the same thing and looks nicer
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
@Inopeki yes, no need to write the 1 though @FFM, I'd have to think about that :/
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
unspecified functions he's asking how far the rule about exponents can be extended..
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
x  Oh right y^4*z^2
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Actually we can't but I believe that is true for exponential functions though.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
it must be for simple ones\[2^x2^y=2^{x+y}\]of course
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Tha's whats Inopeki is using right ?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
basically Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
Yeah, but does that mean that it becomes x  ? yz^6
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
no because y and z are different bases notice the rule above had both base x.
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
YEah but then i dont see the connection..
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
just that dividing and multiplying are inverse operations x^a=x*x*x*...*x (a times) x^b=x*x*x*...*x (b times) so if we have a=3 b=2 we get x^a  x^b x*x*x == x x*x which is x^(ab) if we have (x^a)(x^b) we get (x*x*x)(x*x)=x*x*x*x*x=x^(a+b) so the rules for exponents here come directly from their definitions. You can count the x and see that this relationship holds.
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
I know about that, like x^2*x^8=x^10
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
right, I want you to see how that and the rule for division are inverses of each other for a reason...
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Turing you have a heck of patience :D
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
eh it's easy FFM, doesn't require too much concentration. good practice too I learn by teaching ;) ok factor \[3x^3y^3+6xy+9x^2y\]
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
The GCF of 3,6,9 is 3 The GCF of x,x^2,x^3 is x The GCF of y,y^2,y^3 is y 3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Middle step, right?
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
Yes I've stressed the importance of studying on your own as well and yes Inopeki really, FFM would have noticed a mistake I'm sure ok quick, foil (ab)(a+b)
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
When hes not here i try to go on purplemath and khan :)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Books books books!! online learning has it's own limitation :)
 2 years ago

Akshay_BudhkarBest ResponseYou've already chosen the best response.0
it doesnt @ffm i disagree
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
Foolformath, im having trouble finding books here in sweden
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
and what is the name of that form? remember?
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
The fundamental theorem of algebra? Or want that the one with the multiplex?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
Multiplex? no I think you're thinking of 'multiplicity' but this is the 'difference of squares' \[a^2b^2=(ab)(a+b)\]
 2 years ago

Akshay_BudhkarBest ResponseYou've already chosen the best response.0
with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
Thanks but you get what you put in, I think is true with all this stuff.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
The fundamental theorem of algebra: "the sum of the multiplicity of the roots of a polynomial is equal to its order" difference of squares\[a^2b^2=(ab)(a+b)\]both important, but very different
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
so remember that we can run this backwards, and I can say 'factor'\[x^24\]using difference of squares wanna try?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
run the FOIL backwards I meant*
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
no look at the form\[a^2b^2=(ab)(a+b)\]so what is a and b in\[x^24\]???
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
how about\[x^4y^4\](this is a favorite question on OS)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
good, now is that all we can do with it though? (I've seen a lot of tutors stop here too ;)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
hint:look at the first set of parentheses
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
what is the first sett of parentheses?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
yes, and can you factor that?
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
GCF of x^2 is x GCF of y^2 is y xy(x^2/xy)xy(y^2/xy)?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
no you're overthinking x^2y^2 is difference of squares again!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
\[a^2b^2=(ab)(a+b)\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
yes! so now factor\[p^4q^4\]completely!!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
you will apply difference of squares twice
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(p^2q^2)(p^2+q^2) (pq)(p+q)(p+q)+(p+q)?
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
between the parenthesis
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
(p^2q^2)(p^2+q^2) stop here we cannot factor the second set of parentheses, it is not a /difference/ of squares, it is a /sum/ of squares, and we have no formula for that. check that (a+b)(a+b) is not a^2+b^2 so only factor the one that is a /difference/ of squares
 2 years ago

InopekiBest ResponseYou've already chosen the best response.0
(pq)(p+q)(p^2+q^2) You said i had to do difference in squares 2 times? Btw, lets move to a new thread, this one is starting to lag a little bit
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.5
That is above correct, you did difference of squares twice: p^4q^4 (p^2q^2)(p^2+q^2) <once (pq)(p+q)(p^2+q^2) <twice ok new thread...
 2 years ago
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