CardiBSuperfan3
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Someone please help, i've never seen a question like this before...
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RDKGames
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(Original post by MutsaMun)
Someone please help, i've never seen a question like this before...
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CardiBSuperfan3
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just managed to figure out how to attach it
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CardiBSuperfan3
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(Original post by RDKGames)
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RDKGames
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(Original post by CardiBSuperfan3)
just managed to figure out how to attach it
Expanding \dfrac{2x^2 + 5x -10}{(x-1)(x+2)} in ascending powers of x is the same as expanding 2-\dfrac{1}{x-1}+ \dfrac{4}{x+2} in ascending powers of x which is the same as expanding each term individually in ascending powers of x and adding the common terms together.

So, since we are interested in powers up to and including x^2, you should expand \dfrac{1}{x-1} and \dfrac{4}{x+2} individually first.
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sabiha200119
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(Original post by CardiBSuperfan3)
Someone please help, i've never seen a question like this before...
use binomial expansion. i dont know what examboard you do but the formula is in the edexcel formula booklet. the 1/(x-1) would become (x-1)^-1 coz its a reciprocal. than you just use the formula and write down all the values up to x^2.
hope that helped if you cant figure out still let me know ill give it a try myself
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CardiBSuperfan3
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hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there
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RDKGames
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(Original post by CardiBSuperfan3)
hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there
Are you familiar with the binomial expansion formula

(1+x)^n = 1+nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots

??
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sabiha200119
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(Original post by CardiBSuperfan3)
hi yeah im doing edexcel
i can get up to doing the reciprocals but i dont know where to go from there

ive got
2 - (x-1)^-1 4(x 2)^-1
ill give it a try and let you know in a bit...xxx
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CardiBSuperfan3
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(Original post by RDKGames)
Are you familiar with the binomial expansion formula

(1+x)^n = 1+nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots

??
yes
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RDKGames
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(Original post by CardiBSuperfan3)
yes
So use that.

Notice that the middle term can be rewritten;
-\dfrac{1}{x-1} = \dfrac{1}{1-x} = (1-x)^{-1}
where the difference between this and (1+x)^n is that you firstly replace x in the formula I gave you by -x and set n=-1. Get the terms up to x^2. What do you get?

Repeat for 4(2+x)^{-1} which requires a slight modification before the formula above can be applied in the same way.
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CardiBSuperfan3
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(Original post by RDKGames)
So use that.

Notice that the middle term can be rewritten;
-\dfrac{1}{x-1} = \dfrac{1}{1-x} = (1-x)^{-1}
where the difference between this and (1+x)^n is that you firstly replace x in the formula I gave you by -x and set n=-1. Get the terms up to x^2. What do you get?

Repeat for 4(2+x)^{-1} which requires a slight modification before the formula above can be applied in the same way.
okay for the first bit i got 1 + x + x^2
and then i got 8 - 4x - x^2
which I'm pretty sure is completely wrong because it said i would somehow have fractions
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RDKGames
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(Original post by CardiBSuperfan3)
okay for the first bit i got 1 + x + x^2
and then i got 8 - 4x - x^2
which I'm pretty sure is completely wrong because it said i would somehow have fractions
Yep it's wrong.

The first bit looks good though, it's (1-x)^{-1} = 1+x+x^2 + O(x^3).

Then for the second bit, you need to realise that to expand (2+x)^{-1} using the binomial expansion I posted above, you first need to make sure that rather than being 2+... it's actually 1+... inside the bracket.
How do we do that? Simply factor out a 2 out of the bracket. Then it becomes 4(2+x)^{-1} = 4[2(1+\frac{x}{2})]^{-1} = 4\cdot 2^{-1}(1+\frac{x}{2})^{-1}.
So we reduce the problem to 4(2+x)^{-1} = 2(1+\frac{x}{2})^{-1}

Now you need to expand just (1+\frac{x}{2})^{-1} using the binomial formula above. Simply replace x by \frac{x}{2} and set n=-1. Expand up to x^2 and then multiply all coefficients by 2 since we seek 2(1+\frac{x}{2})^{-1}.
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CardiBSuperfan3
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(Original post by RDKGames)
Yep it's wrong.

The first bit looks good though, it's (1-x)^{-1} = 1+x+x^2 + O(x^3).

Then for the second bit, you need to realise that to expand (2+x)^{-1} using the binomial expansion I posted above, you first need to make sure that rather than being 2+... it's actually 1+... inside the bracket.
How do we do that? Simply factor out a 2 out of the bracket. Then it becomes 4(2+x)^{-1} = 4[2(1+\frac{x}{2})]^{-1} = 4\cdot 2^{-1}(1+\frac{x}{2})^{-1}.
So we reduce the problem to 4(2+x)^{-1} = 2(1+\frac{x}{2})^{-1}

Now you need to expand just (1+\frac{x}{2})^{-1} using the binomial formula above. Simply replace x by \frac{x}{2} and set n=-1. Expand up to x^2 and then multiply all coefficients by 2 since we seek 2(1+\frac{x}{2})^{-1}.
yeah that's what i did but do i not need to then multiply by 4?
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RDKGames
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(Original post by CardiBSuperfan3)
yeah that's what i did but do i not need to then multiply by 4?
I've shown you that 4(2+x)^{-1} is the same as 2(1+\frac{x}{2})^{-1}.

So when you expand (1+\frac{x}{2})^{-1} you need to multiply by 2 since that's what's in front of it, not 4.
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