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

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Teach about polynomial.

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5That'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:\[\frac{x^2y^5z}{xy^9z^3}\]

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

Inopeki
 3 years ago
Best ResponseYou've already chosen the best response.0How did it go with her?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5broken leg... waiting for an operation

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1You beat you cat Turing?!!!

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5she fell off the roof. how pessimistic

saifoo.khan
 3 years ago
Best ResponseYou've already chosen the best response.0awww, sorry i was jking..

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

Inopeki
 3 years ago
Best 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)

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

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

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

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

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

TuringTest
 3 years ago
Best 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.

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5Oh 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...

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5well 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?

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

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

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

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

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Can you explain the second rule intuitively?

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

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

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

TuringTest
 3 years ago
Best 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 :/

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

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

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

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

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

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

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

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

Inopeki
 3 years ago
Best ResponseYou've already chosen the best response.0YEah but then i dont see the connection..

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5just 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.

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

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

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

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

Inopeki
 3 years ago
Best ResponseYou've already chosen the best response.0The 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?

Inopeki
 3 years ago
Best ResponseYou've already chosen the best response.03xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5Yes 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)

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

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

Akshay_Budhkar
 3 years ago
Best ResponseYou've already chosen the best response.0it doesnt @ffm i disagree

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

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5The 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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5run the FOIL backwards I meant*

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5what is the first sett of parentheses?

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5you will apply difference of squares twice

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

Inopeki
 3 years ago
Best ResponseYou've already chosen the best response.0between the parenthesis

TuringTest
 3 years ago
Best 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

Inopeki
 3 years ago
Best 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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.5That 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...
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