# C4 Binomial Expansion

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sabahshahed294

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#1

Hello! I have a minor confusion here regarding a mathematical step in this video I posted below:-

http://www.examsolutions.net/tutoria...C34&topic=1308

Now, at around 11:36 of this video, they find the range of values of x for which x is valid and here, there are two parts in the whole expression containing different validities(one of them being -2<x<2 and the other being -1<x<1). They've made a number line and according to that, they found the validity of the

I didn't exactly get the logic behind this step and as to why we do it. If someone can help me by explaining the exactly what is going on and the reasons behind it, it will be of a big help. TIA!

http://www.examsolutions.net/tutoria...C34&topic=1308

Now, at around 11:36 of this video, they find the range of values of x for which x is valid and here, there are two parts in the whole expression containing different validities(one of them being -2<x<2 and the other being -1<x<1). They've made a number line and according to that, they found the validity of the

**whole**expression. I'm confused in the part where they find the validity of the whole expression actually.In the video, it is mentioned that the value of x that is needed must work for both expressions and so, the one with the smallest interval must be chosen and needs to be contained within the other. Hence, -1<x<1 is chosen because it is contained within the other expression .i.e. -2<x<2.I didn't exactly get the logic behind this step and as to why we do it. If someone can help me by explaining the exactly what is going on and the reasons behind it, it will be of a big help. TIA!

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ghostwalker

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#2

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#2

(Original post by

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**sabahshahed294**)...

So, you have two binomial expansions.

The first is valid for x in the range -1 to <1

The second is valid for x in the range -2 to 2

Now, the whole expression is valid as long as both parts are valid, that is x is in the range -1 to 1 and it's in the range -2 to 2.

Since the smaller range is contained in the larger range, then the expression will be valid as long as x is in the smaller range.

If x is in the smaller range, than since the smaller range is in the larger range, then x is also in the larger range, and hence in both ranges.

[If you've covered sets we can call one range A, and the other B, and x must be in their intersection, AnB]

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sabahshahed294

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#3

(Original post by

I've made this somewhat verbose, in the hope that some of the phraseology will hit mark for your particular issue.

So, you have two binomial expansions.

The first is valid for x in the range -1 to <1

The second is valid for x in the range -2 to 2

Now, the whole expression is valid as long as both parts are valid, that is x is in the range -1 to 1 and it's in the range -2 to 2.

Since the smaller range is contained in the larger range, then the expression will be valid as long as x is in the smaller range.

If x is in the smaller range, than since the smaller range is in the larger range, then x is also in the larger range, and hence in both ranges.

[If you've covered sets we can call one range A, and the other B, and x must be in their intersection, AnB]

**ghostwalker**)I've made this somewhat verbose, in the hope that some of the phraseology will hit mark for your particular issue.

So, you have two binomial expansions.

The first is valid for x in the range -1 to <1

The second is valid for x in the range -2 to 2

Now, the whole expression is valid as long as both parts are valid, that is x is in the range -1 to 1 and it's in the range -2 to 2.

Since the smaller range is contained in the larger range, then the expression will be valid as long as x is in the smaller range.

If x is in the smaller range, than since the smaller range is in the larger range, then x is also in the larger range, and hence in both ranges.

[If you've covered sets we can call one range A, and the other B, and x must be in their intersection, AnB]

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sabahshahed294

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#4

(Original post by

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**ghostwalker**)....

http://pmt.physicsandmathstutor.com/...hapter%201.pdf

Exercise E, Question 1(b), could you please explain me the steps here in this question? I tend to get stuck in such kinds of questions. TIA.

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ghostwalker

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(Original post by

Hi, I have a question which I will need some help in please. Could you guide me here?

http://pmt.physicsandmathstutor.com/...hapter%201.pdf

Exercise E, Question 1(b), could you please explain me the steps here in this question? I tend to get stuck in such kinds of questions. TIA.

**sabahshahed294**)Hi, I have a question which I will need some help in please. Could you guide me here?

http://pmt.physicsandmathstutor.com/...hapter%201.pdf

Exercise E, Question 1(b), could you please explain me the steps here in this question? I tend to get stuck in such kinds of questions. TIA.

In this case the degree of the numerator is 2, and the degree of the denominator is 1+1=2.

So, first off they're doing long/polynomial division to reduce the degree of the numerator. For that they have to multiply the brackets in the denominator together to get a quadratic in this case.

Then do the long division to get to the line starting "Therefore".

Then proceed as usual.

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#6

(Original post by

The usual method of partial fractions is used when the degree of the numerator (hightest power of x) is less than the degree of the denominator.

In this case the degree of the numerator is 2, and the degree of the denominator is 1+1=2.

So, first off they're doing long/polynomial division to reduce the degree of the numerator. For that they have to multiply the brackets in the denominator together to get a quadratic in this case.

Then do the long division to get to the line starting "Therefore".

Then proceed as usual.

**ghostwalker**)The usual method of partial fractions is used when the degree of the numerator (hightest power of x) is less than the degree of the denominator.

In this case the degree of the numerator is 2, and the degree of the denominator is 1+1=2.

So, first off they're doing long/polynomial division to reduce the degree of the numerator. For that they have to multiply the brackets in the denominator together to get a quadratic in this case.

Then do the long division to get to the line starting "Therefore".

Then proceed as usual.

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