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C2 BINOMIAL EXPANSION HELP!! pleaseeeee

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
Reply 1
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

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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
Reply 3
Here is an example of the binomial

(2+ax)4=24(1+ax2)4=24[1+4ax2+432!(ax2)2+4323!(ax2)3+(ax2)4](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 x2x^2 is

(24)(12a24)=48a2(2^4)(\frac{12a^2}{4})=48a^2
Reply 4
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
Reply 5
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 x2x^2

and even if it were the correct term you have not squared your b
(edited 11 years ago)
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?
(edited 11 years ago)
Reply 7
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:smile:
how do you do the second one?
Reply 8
Original post by sharon800
Thank you! i did this last nigh, and got it:smile:
how do you do the second one?


(2+x)5(2x)5=A+Bx2+Cx4 \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

Expand both (2+x)5 \displaystyle (2+x)^5 and (2x)5 \displaystyle (2-x)^5 by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
Reply 9
Original post by raheem94
(2+x)5(2x)5=A+Bx2+Cx4 \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

Expand both (2+x)5 \displaystyle (2+x)^5 and (2x)5 \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?
Reply 10
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.
Reply 11
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
Reply 12
Original post by sharon800
lol i got 3125...:/
how do you do it?
i did 5C0*(5)^5*(-X)^0 and got -3125


(2+x)5=5C025(x)0+5C124(x)1...=25+5×24x...=32+80x+... \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


(2x)5=5C025(x)0+5C124(x)1...=25+5×24(x)...=3280x+... \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

(2+x)5(2x)5=(32+80x+...)(3280x+...)=32×32+...=1024+... \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

So A=1024 \displaystyle A=1024
Reply 13
Original post by raheem94
(2+x)5=5C025(x)0+5C124(x)1...=25+5×24x...=32+80x+... \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


(2x)5=5C025(x)0+5C124(x)1...=25+5×24(x)...=3280x+... \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

(2+x)5(2x)5=(32+80x+...)(3280x+...)=32×32+...=1024+... \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

So A=1024 \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! :smile:
Reply 14
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! :smile:


No problem, you are welcome.
Original post by raheem94
(2+x)5=5C025(x)0+5C124(x)1...=25+5×24x...=32+80x+... \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


(2x)5=5C025(x)0+5C124(x)1...=25+5×24(x)...=3280x+... \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

(2+x)5(2x)5=(32+80x+...)(3280x+...)=32×32+...=1024+... \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

So A=1024 \displaystyle A=1024


You told me once never to post full solutions. :confused:
Reply 16
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?

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