C2 BINOMIAL EXPANSION HELP!! pleaseeeee

Announcements Posted on
How helpful is our apprenticeship zone? Have your say with our short survey 02-12-2016
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can somebody show me how to solve such equations? i just cant get my head around this :-
    1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.

    2. given that
    (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.

    Thank you
    Offline

    2
    ReputationRep:
    (Original post by sharon800)
    Can somebody show me how to solve such equations? i just cant get my head around this :-
    1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.

    2. given that
    (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.

    Thank you
    I use

    Uploaded with ImageShack.us
    Offline

    0
    ReputationRep:
    For the first one you:

    1. Expand the brackets until you get ......x^3
    2. Make this coefficient equal to -720
    3. Simple algebra from here


    For the second one you:

    1. expand both to give the answer
    2. Total up the coeffients
    3. From here, work out what A, B, C are
    Offline

    3
    ReputationRep:
    Here is an example of the binomial

    (2+ax)^4 = 2^4 (1+\frac{ax}{2})^4 = 2^4[1 + 4\frac{ax}{2} + \frac{4*3}{2!}(\frac{ax}{2})^2 + \frac{4*3*2}{3!}(\frac{ax}{2})^3 + (\frac{ax}{2})^4]

    In my example the coefficient of x^2 is

    (2^4)(\frac{12a^2}{4})=48a^2
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Frodo Baggins)
    For the first one you:

    1. Expand the brackets until you get ......x^3
    2. Make this coefficient equal to -720
    3. Simple algebra from here


    For the second one you:

    1. expand both to give the answer
    2. Total up the coeffients
    3. From here, work out what A, B, C are
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong
    Offline

    3
    ReputationRep:
    (Original post by sharon800)
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2
    This is x^2

    and even if it were the correct term you have not squared your b
    Offline

    0
    ReputationRep:
    (Original post by sharon800)
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong
    I got b as -2

    Basically, to get x^3

    its 5c3 * (3)^2 * b^3x^3

    so 90b^3=-720

    so b^3=-8

    so b=-2

    i am unclear as to why i have got a neg. Is it cos im clever?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Frodo Baggins)
    I got b as -2

    Basically, to get x^3

    its 5c3 * (3)^2 * b^3x^3

    so 90b^3=-720

    so b^3=-8

    so b=-2

    i am unclear as to why i have got a neg. Is it cos im clever?
    Thank you! i did this last nigh, and got it
    how do you do the second one?
    Offline

    1
    ReputationRep:
    (Original post by sharon800)
    Thank you! i did this last nigh, and got it
    how do you do the second one?
     \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

    Expand both  \displaystyle (2+x)^5 and  \displaystyle (2-x)^5 by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by raheem94)
     \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

    Expand both  \displaystyle (2+x)^5 and  \displaystyle (2-x)^5 by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
    i got A= 0? DOESNT sound right:/
    and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer?
    Offline

    1
    ReputationRep:
    (Original post by sharon800)
    i got A= 0? DOESNT sound right:/
    and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer?
    I get A as 1024. Try again.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by raheem94)
    I get A as 1024. Try again.
    lol i got 3125...:/
    how do you do it?
    i did 5C0*(5)^5*(-X)^0 and got -3125
    Offline

    1
    ReputationRep:
    (Original post by sharon800)
    lol i got 3125...:/
    how do you do it?
    i did 5C0*(5)^5*(-X)^0 and got -3125
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by raheem94)
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
    thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
    i get it! thanks again!
    Offline

    1
    ReputationRep:
    (Original post by sharon800)
    thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
    i get it! thanks again!
    No problem, you are welcome.
    Offline

    0
    ReputationRep:
    (Original post by raheem94)
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
    You told me once never to post full solutions. :confused:
    Offline

    1
    ReputationRep:
    (Original post by Frodo Baggins)
    You told me once never to post full solutions. :confused:
    Yes, it is not allowed to post full solutions.

    Full solutions are considered a last resort.

    Actually i didn't post the full solution, i only calculated one value, of A, to show the OP how to calculate them. He also had to calculate B and C, so it isn't a full solution, just a part of the solution to let the OP get started.

    Why do you think it to be a full solution?
    Have you read the OP's question?
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: April 6, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
Would you rather have...?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.