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1. 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
2. (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

3. 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
4. Here is an example of the binomial

In my example the coefficient of is

5. (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
6. (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

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

Expand both and by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
10. (Original post by raheem94)

Expand both and 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?
11. (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.
12. (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
13. (Original post by sharon800)
lol i got 3125...:/
how do you do it?
i did 5C0*(5)^5*(-X)^0 and got -3125

So
14. (Original post by raheem94)

So
thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
i get it! thanks again!
15. (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.
16. (Original post by raheem94)

So
You told me once never to post full solutions.
17. (Original post by Frodo Baggins)
You told me once never to post full solutions.
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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Updated: April 6, 2012
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