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## #1 2009-05-04 03:36:57

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

### Need help on Polynomials

Find the independent term of the expansion

.

Is it a must to expand the whole thing out and look at the independent term, or is there a shortcut??

i still can't figure out how to make the "0" smaller

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## #2 2009-05-04 04:02:26

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Need help on Polynomials

If you want a power to be more than one character, you need to put it in curly braces.
eg. x^{10} -->

There is a shortcut to this, because most of the expansion is irrelevant to you.
You're only interested in the term that doesn't contain any x. This is (x^2/2)^5 * (-1/x^2)^5 = -1/32.

The coefficient of this term is 10C5 = 252, which means the independent term of your expansion is -252/32 = -7.875.

Why did the vector cross the road?
It wanted to be normal.

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## #3 2009-05-04 06:02:04

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Need help on Polynomials

Hi 1a2b3c2212;

Good answer mathysperson,  that is the way. In case you can't follow it, I think this is his reasoning. Normally to find the independent term (constant term) you would just set x=0, but that is not possible here, because of the x in the denominator, Start with the binomial expansion of.

The general term is:

with

So:

Pull out of the brackets all the constants and get them on the left

Set the exponents of

and
equal to 0

Now just plug into (A).

Last edited by bobbym (2009-05-05 01:02:33)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #4 2009-05-04 23:21:50

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

### Re: Need help on Polynomials

mathsyperson, how did u factorise the original polynomial to (x^2/2)^5 * (-1/x^2)^5 = -1/32??

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## #5 2009-05-04 23:25:56

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Need help on Polynomials

Why did the vector cross the road?
It wanted to be normal.

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## #6 2009-05-04 23:50:05

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

### Re: Need help on Polynomials

What you meant was

It doesn't make sense to me though.

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## #7 2009-05-05 00:25:09

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Need help on Polynomials

Hi 1a2b3c2212;

That statement is not true. Mathsy did not mean that.

Last edited by bobbym (2009-05-05 00:26:21)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #8 2009-05-05 00:27:03

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

### Re: Need help on Polynomials

uh then where did he get that equation from?

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## #9 2009-05-05 00:30:00

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Need help on Polynomials

Hi 1a2b3c2212;

Look at my post, it is what he/she may have been thinking about when he/she solved the problem.

Last edited by bobbym (2009-05-05 00:40:36)

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #10 2009-05-05 00:35:21

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

oh.

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## #11 2009-05-05 00:39:59

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Need help on Polynomials

Did it make sense?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #12 2009-05-05 01:06:51

1a2b3c2212
Member
Registered: 2009-04-04
Posts: 419

### Re: Need help on Polynomials

i think i get the gist of it..'cause i haven't started on binomial theorem yet.

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