The Student Room Group

Binomial Expansion (x-y)^4

Here:



The first part is trivial. However I cannot seem to use that to write x^4-y^4 in terms of p and q. The expansion from the first part has x^4 and y^4 as terms, no sight of any -y^4.

I also considered starting from x^4 - y^4 and using difference of two squares to break it down, but I always seem to end up with an expression I cannot completely put in terms of p and q.

Am I correct in starting from the expansion, or from the expression x^4 - y^4.

Thanks guys.
Reply 1
At a first glance breaking it down using the dfference of 2 squares seems to work, how far did you get with this method?
Reply 2


I think you have to simplify it down but I cant remember how sorry, hope that helped :smile:

use this formula:
Reply 3
Original post by james22
At a first glance breaking it down using the dfference of 2 squares seems to work, how far did you get with this method?


x4y4=p(p2+2q)(x+y)x^4-y^4=p(p^2+2q)(x+y)

Just that (x+y) stopping me >.<
Reply 4
Original post by SamKeene
x4y4=p(p2+2q)(x+y)x^4-y^4=p(p^2+2q)(x+y)

Just that (x+y) stopping me >.<


What is p(x+y)?
Reply 5
Original post by james22
What is p(x+y)?


x2y2x^2 - y^2

Which was where I got p(x+y)p(x+y) in the first place.

But I cant see a way to write that in terms of p and q either.
Reply 6
Original post by SamKeene
x2y2x^2 - y^2

Which was where I got p(x+y)p(x+y) in the first place.

But I cant see a way to write that in terms of p and q either.


Are you sure you got the question exactly right? The answer is not in a nice form.
Reply 7
Original post by james22
Are you sure you got the question exactly right? The answer is not in a nice form.


The question was screen-shotted from the pdf of the paper.

I'm thinking it's a misprint..
Reply 8
Original post by SamKeene
The question was screen-shotted from the pdf of the paper.

I'm thinking it's a misprint..


It is still solveable, just more difficult.
Reply 9
Original post by james22
It is still solveable, just more difficult.


Honestly I think it was supposed to be x^4 + y^4... Allowing you to use the expansion you wrote down earlier.

The same paper already has another misprint in the final question, and the difficulty of the paper itself (and the marks of the question) lend it to being a misprint.
Original post by SamKeene
Honestly I think it was supposed to be x^4 + y^4... Allowing you to use the expansion you wrote down earlier.

The same paper already has another misprint in the final question, and the difficulty of the paper itself (and the marks of the question) lend it to being a misprint.


x^4+y^4 is no easier.
Reply 11
Original post by james22
x^4+y^4 is no easier.


It is significantly easier.

p4=x44x2q+6q24y2q+y4p^4=x^4-4x^2q+6q^2-4y^2q +y^4

p46q2+4x2q+4y2q=x4+y4p^4-6q^2+4x^2q+4y^2q=x^4+y^4

p46q2+4q(x2+y2)=x4+y4p^4-6q^2+4q(x^2+y^2)=x^4+y^4

p46q2+4q(p2+2q)=x4+y4p^4-6q^2+4q(p^2+2q)=x^4+y^4

p46q2+4q(p2+2q)=x4+y4p^4-6q^2+4q(p^2+2q)=x^4+y^4

It also fits in with using what you did earlier in the question.
Original post by SamKeene
x2y2x^2 - y^2

Which was where I got p(x+y)p(x+y) in the first place.

But I cant see a way to write that in terms of p and q either.


I agree it would make more sense for + instead of -

But you know what (x+y)^2 = in terms of p and q so you just need an awkward sqquare root in there
Edit: Nevermind; I misread the question. I'll mess around with the expression a bit more and hopefully get the right answer this time.

That didn't take as long as I thought it would.

Spoiler



I guess the main part is figuring out how to express (x+y)(x+y), which you can do using a similar approach to the one you presumably used to find x2+y2x^2 + y^2 .
(edited 9 years ago)

Quick Reply