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Author Topic: I suppose this would fall here...I need help. o-o  (Read 2561 times)

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Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #25 on: October 11, 2010, 11:06:41 PM »
Problem:

x^2-w^2
over
x^3-w^3

times

x^2+xw+w^2
over
x^2+2xw+w^2


Soooes....

How would I even begin this? O-o Should I try to factor them first, or should I go straight to multiplication? I know x^2+2xw+w^2 factors into (x+w)(x+w), but the top portion doesn't factor at all.

x^2-w^2 is a difference of two squares, so it becomes (x-w)(x+w)

And x^3-x^3 is a difference of two cubes, so it becomes (x-w)(x^2+xw+w^2)

Doesn't it?

so then...we're essentially left with:

(x-w)(x+w)(x^2+xw+w^2)

over

(x-w)(x^2+xw+w^2)(x+w)(x+w)

And, once you eliminate, you're left with...

O_o 0?

I'm not sure my math is right there...O-o..I was thinking I would be left with (x-w) over (x+w)

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #26 on: October 11, 2010, 11:10:46 PM »
Yes, you should factor.  I don't know if you should eliminate though.  I'll be 100% honest, this is one thing I never fully understood.  I don't think it's legal to eliminate a factor that would otherwise result in a division by zero error, because then you'll get a different answer for that division-by-zero portion of the function if you try plugging in values (aka, you're fixing the function at the division by zero point).

Ask your teacher what he wants; in some situations we do it in math (such as in the calculations of limits) but, truthfully I'm not sure where it's OK and where it's not.

I'll look over the rest when I get back in about an hour and a half.  AFK.

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #27 on: October 11, 2010, 11:16:48 PM »
Well..I finish the assignment..and though I got a 73.3 on it, I feel happy...my first attempt was a 41%. lol

^-^..Everyone helped a lot.

Offline Trieste

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Re: I suppose this would fall here...I need help. o-o
« Reply #28 on: October 11, 2010, 11:19:54 PM »
I think I can help with this one.

Go ahead and factor. I get the following factors:

x2-w2 becomes (x+w)(x-w)
x2+2xw+w2 becomes (x+w)(x+w)
x3-w3 becomes (x-w)(x2+xw+w2)

So now you have
(x+w)(x-w)
(x-w)(x2+xw+w2)

times

x2+xw+w2
(x+w)(x+w)

After all of the canceling, I get 1/(x+w).

Offline Oniya

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Re: I suppose this would fall here...I need help. o-o
« Reply #29 on: October 12, 2010, 12:06:14 AM »
Yes, you should factor.  I don't know if you should eliminate though.  I'll be 100% honest, this is one thing I never fully understood.  I don't think it's legal to eliminate a factor that would otherwise result in a division by zero error, because then you'll get a different answer for that division-by-zero portion of the function if you try plugging in values (aka, you're fixing the function at the division by zero point).

Ask your teacher what he wants; in some situations we do it in math (such as in the calculations of limits) but, truthfully I'm not sure where it's OK and where it's not.

I'll look over the rest when I get back in about an hour and a half.  AFK.

Just as a note on this:  Canceling out a factor that can result in a zero is a trick that is used in 'proving' that 1=2.  If you're using the simplified equation/function/relation to determine the value at a specific point, it's best to double-check by plugging that point into the original equation.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #30 on: October 12, 2010, 12:25:18 AM »
Ah, yeah, so if you have (1-w)/(1-w) if w can be 1, then that's not actually equal to 0.  In the situation of limits we'll allow fudge math because we're only really concerned with what the value is approaching, not what it is.  So yeah, you can't factor and cancel.