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lol I like the attitude, I better eat though, give me a bit what time is it where you are?
9:45 pm, you? Then go eat :)
alright, just so you know it's easy to send messages via fan message now, so use that instead of a question to hail me

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Other answers:

Oh, ill use that!
see you in a bit
Alright
How do you use those fan messages Turing?
Go to the profile and click "write a fan message"
factor 21x^2+14x
GCF of 21 and 14=7 GCF of x and x^2 is x 21x^2/7x+14x/7x=x(3x+2)?
almost but 1)be careful of notation 21x^2/7x+14x/7x is not equal x(3x+2) 2) remember to keep the whole GCF outside the parentheses
But i thought that if the 7 was on the LHS then it couldnt be on the RHS?
21x^2/7x+14x/7x=7x(3x+2)?
but they are not equal 21x^2/7x+14x/7x=3x+2 not 7x(3x+2) so you must write 21x^2+14x=7x(21x^2/7x+14x/7x)=?
I dont know...
7(3x^2+2)?
you almost have it but you need to put the GCF outside the parentheses what is the GCF?
7x
so that's what goes outside the parentheses
7x(3x^2+2) Cant be right.
what is 21x^2/7x=? rewrite it in a way more comfortable to you if it looks strange
21/7 * x^2/x?
yes...
Im pretty sure thats 3x man
it is, so why did you write 3x^2 above?
Cause you said the other one wasnt right :/ What is the correct way to write it?
here was your first answer x(3x+2) here was the next 21x^2/7x+14x/7x=7x(3x+2) the answer here is correct, I just pointed out that the two expressions above are not equal, I think that may have confused you. next you wrote 7x(3x^2+2) so I think we started to get off track... the answer is 21x^2+14x=7x(3x+2) I just made a point about how you show your work, and what it means.
Ohhhhh :)
write out the process as 21x^2+14x=7x(3x)+7x(2)=7x(3x+2) that is really the best way to show factoring
Oh ok :D Another one?
factor 3t^3+9t^2
sorry, so you know how to divide t^3/t^2 ???
x*x*x ----- right? It should jjust become x? x*x
t, sorry
you got it, so my question stands: factor 3t^3+9t^2
GCF of 3 and 9 is 3. GCF of t^2 and t^3 is t. 3t^3+9t^2=3t(3t^3/3t)+3t(9t^2/3t)=3t(t^3+3t^2)?
check your answer and distribute to see if you get the original: \[3t(t^3+3t^2)=3t(t^3)+3t(3t^2)=3t^4+9t^3\neq3t^3+9t^2\]you didn't divide the terms in the middle right ...but I think more importantly the GCF of t^3 and t^2 is t^2, not t that is because each term can be evenly divided by t^2.
Hm
Oh right!
what is the GCF of t^3+t^5+t^7 ??
Something to the power of something doesnt follow the rules of usual numbers, forgot that. It should be t^5
no it's t^3 and it does follow the rules of regular numbers if you think about it, otherwise it wouldn't be true! look at 16,32,8 what is their GCF ?
8
now rewrite those three numbers as powers of 2
can you do that?
256,1024,64
?
no I meant like 8=2^3 16=2^? 32=2^?
Oh! 8=2^3 16=2^4 32=2^5
...and what is their GCF? 2^3 so the GCF of a set of variables is the highest common power to each term -in this case 3 so if we have y^4+y^7+y^9 the GCF is...?
y^4?
right :)
:D
good job :D so now back to our question... factor 3t^3+9t^2 what is the GCF?
3t^2?
right now can you factor it?
3t^3+9t^2=3t^2(3t^3/3t^2)+3t^2(9t^2/3t^2)=3t^2(t+3)?
very nice!!!!!!
I got it!! :D
you totally did :D that is very cool! so now lets see what's so great about factoring say we have f(t)=3t^3+9t^2 and we want to know the 'zeros' of the function that means when the function touches the x-axis, i.e. when f(t)=0 so to answer this we must solve 0=3t^3+9t^2 how can we solve that? by factoring... 0=3t^2(t+3) now you can solve it quickly, any idea which fundamental rule of algebra tells us how?
Nope, sorry
the rule is called the 'zero factor property' it states that if\[ab=0\]then either\[a=0\]or\[b=0\]or both does this rule make logical sense to you?
Yeah
so now look at our factored equation \[0=ab=3t^2(t+3)\]that means that either\[3t^2=0\]or \[t+3=0\]can you solve each of these equations?
*either or both I should say
OH so 3t^2 represents a and t+3 represents b!
exactly that's why i said to learn the rules on this page http://www.capitan.k12.nm.us/teachers/shearerk/basic_rules_of_algebra.htm almost all of algebra is in there, though sometimes it is hidden
3t^2=0 t=0-3t/t t=0-3 t=-3?
no, it's more simple than that, remember the same rule we just used: ab=0 then either a=0 or b=0 or both 3t^2=0 let 3=a t^2=b we know that a=3 cannot be zero, because 3 is never zero, it's a constant that leaves the possibility only of t^2=0 and the number number that times itself is zero is zero so\[3t^2=0\to t=0\]
Ohhhh.. Cause 3x0x0=0?
right so t has to be zero...
Yeah
what about the other possibility t+3=0 ???
How can that be possible when t is 0?
there are two answers to every quadratic equation as you may recall me saying this is cubic so it has 3 actually, we say that zero ocurrs twice in 3t^2=0 because it leads to two 0's as you showed: 3x0x0 we call that a 'multiplicity of 2' so there will be multiple answers, the other is found by solving t+3=0 remember that either a=0 or b=0 or BOTH we don't know so we have to solve them all.
Oh right! This is a quadratic equation. t=-3 on this one so its (0,-3)?
like I said, cubic read what I wrote above please about multiplicity...
Three answers
So these are cubic coordinates?
(0,0,-3)?
not cubic coordinates (I don't know what that is exactly...), it's just that we say that 3t^3+9t^2 has zeros (0,3) - that means the graph of f(x)=3t^3+9t^2 hits zero there... where 0 here has a 'multiplicity' of 2 (that means it occurs twice) and 3 has a multiplicity of 1
So why isnt -3 involved?
sorry typo, meant (0,-3)* and -3 has multiplicity 1*
good catch
Thanks
so do you see what I mean? the zero's of\[f(t)=3t^2+9t^2\]are found by factoring and setting to zero\[0=3t^2(t+3)\]then solving each possibility\[3t^2=0\to t=0\]\[t+3=0\to t=-3\]where we say that for t=0 k=2 and for t=-3 k=1 where k is the multiplicity
Right, the k is cause t is to the power of 3 there.
for the part that had 3t^2=0 we had k=2 (because zero is the answer twice: 3x0x0) for t+3=0 we have k=1 because t is only to the first power and we only have one answer If what you mean is that you noticed that adding up all the k's gave youu 3, the order of the cubic, then you have noticed what is called the Fundamental Theorem of Algebra: "The sum of the multiplicities of the roots of a function is equal to the order of the polynomial" \[k_0+k_{−3}=2+1=3\]
- a very important theorem as the name implies...
I dont really get the theorem, please explain it.
Basically for whatever the highest power variable you have in a polynomial, that is how many answers you have. Note that they may not be all different answers, but the ones that occur more than once are counted as having a higher multiplicity, so if you add up the multiplicities (the k's) that's how many zeros the polynomial has. for example 7x^5+3x^3+2x^2+5x+3=0 must have 5 answers, because it is 5th order x^2+2x+2=0 must have 2 answers, because it is second order
So x^4+5x-4=0 must have 4 answers?
exactly, though it may not be 4 different answers for instance x^4=0 has only the answer x=0, but that zero has a multiplicity of k=4 because 0x0x0x0=0 is how it must be...
Oh, now i get it
good :) do you want to try some more factoring problems? or perhaps you should learn a little about exponents first? or perhaps you are ready top get some rest.... which is it?
I wanna learn something new :D
let's see if this is news to you: simplify\[{x^{14}\over x^{12}}\]
x^2 lol
good that saves a lot of time...
simplify\[\sqrt[3]{x^{21}}\]
x^7?
good :) more time saved...
:D actually i didnt know that
the genereal rule is\[\sqrt[b]{x^{a}}=x^{a/b}\]any radical can be wriitten as a fractional exponent, for instance\[\sqrt x=x^{1/2}\]so... let's try some of that.
Yeah, i assumed that :)
good guess, try to really rationalize it if you can... since you seem to know the rules let's try a trickier one simplify \[\frac{\sqrt[3]{x^2}\sqrt[5]{x^3}}{\sqrt x}\]
x^2/3 * x^3/5 ----------- Its basically this, right? If so, i just need to make the variables "suitable" to x^1/2 be merged, right?
right, just remember that\[x^ax^b=x^{a+b}\]and that\[\frac{x^a}{x^b}=x^{a-b}\]so you're gonna have and subtract to add those fractions, so they all need a common denominator.
so you're gonna have to add and subtract those fractions*
Yeah, thats what i meant by making them suitable to merge
I figured, but that's not a known term to me, just making sure
x^2/3 * x^3/5 x^10/15 * x^9/15 x^19/15 ----------- = =-------------- x^1/2 x^1/2 Now i need to get 19/15 divisible by 2
good so far :)
19/15 ------------38/30
x^38/30 -------- = x^1/2 Actually, dont i need to get them divisible by 15 so i can get 2 as the denominator?
what can you multiply the bottom by to get 1/2 over 30 ?
Oh! x^38/30 -------- = 23/30? No, that cant be it.. x^15/30
yes it can and is but you have to leave the x of course... x^(23/30)
Oh right lol
so that was good, I won't test you on that do you know how to FOIL ?
What?
i.e. simplify (a+b)(c+d)
abcd?
FOIL= First Outer Inner Last|dw:1325979609902:dw|watch the arrows they multiply the first in the brackets, the outer terms, the inner, and the last seperateyl
it's just like distribution, but you have to do a and b seperate
I think i get it. give me another one and ill solve it fast
It's okay if you don't solve it fast actually remember that solving is not simplifying, here we have no = sign so we are simplifying: (x+3)(x+2)
x^2+2x+3x+6?
nice! you can simplify the middle terms
Oh right,x^2+5x+6
nicely done! especially for never having FOILed before. simplify (2x+5)(3x-1)
5x+2x+15x-5?
6x*
careful... what's the first? what's the outer? the others are right.
yes the first is 6x the outer though...
-1?
times what?
2x, maybe thats -2x..
there ya go ;-)
so it should be...?
6x+-2x+15x-5
scratch the extra plus sign, but yes, now simplify...
29x-5?
oh my mistake, you forgot that the first term would be squared First=2x(3x)=6x^2 now what do you get?
6x^2-2x+15x-5?
yes, and now simplify
4x^2+15x-5?
like terms, x with x.... not x with x^2...
so 6x^2+13x-5?
yes. good job. now simplify (2p-3)(5p-1)
10p^2-2p-15p-3? Simplified:10p^2-17p-3
good, but watch the last term negative times negative is...?
Oh right, positive. So 10p^2-17p+3?
2r^2+2r*3t+t*r+3t^2 Simplified: 2r^2+2r^2+3t^2+3t^2
stop before your simplification (which is wrong, sorry) look at the first step: 2r^2+2r*3t+t*r+3t^2 ^^^ should everything be positive?
Oh right.. 2r^2+2r*-3t+t*r
yes but you forgot the last term you should also put parentheses around the negative terms
2r^2+2r*(-3t)+t*(-r)?
lol it got deleted, what was the original problem please?
I cant remember :(
ok another, because we started losing terms in the last one (2y+x)(4y-3x)
8y+2y*3x+x*4y(-3x^2)
there should be four terms what is your first? write out the middle step
it's close in many ways, but there are quite a few mistakes
Ill put them in prenthesis (8y)+(2y*3x)+(x*4y)-(3x^2)
better, but the first is 2y(4y) no? and the outer is a positive times a negative as well
Oh, ill solve this one and then sleep :) (8y^2)+(2y*-3x)+(x*4y)-(3x^2)
nice, so to simplify this first multiply the coefficients...
(8y^2)+(8y^2*-3x^2)-(3x^2)
not I think you're getting tired (8y^2)+(2y*-3x)+(x*4y)-(3x^2)=8y^2-6xy+4xy-3x^2=?
no*
can you simplify the last step?
Yeah, i need to sleep.
for future reference:\[(8y^2)+(2y*-3x)+(x*4y)-(3x^2)\]\[=8y^2-6xy+4xy-3x^2=8y^2-2xy-3y^2\]goodnight!
Good night! Thanks for all youve taught me :D See you tomorrow?

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