[Stickyelmo]

Hi people.

I'm stuck on the first hurdle of this binomial question. Can someone just help me get past the first bit:

Find & simplify the term independent of x in the expansion of

(1/2x + x^3) ^8

I know you need to factorise it but I don't know how to get rid of the 1/2x bit.


Thanks in advance!

[TenOfThem]

A general term would have

where a+b = 8

can you find an a and b that allow the x's to cancel

[the bear]

(Original post by Stickyelmo)
Hi people.

I'm stuck on the first hurdle of this binomial question. Can someone just help me get past the first bit:

Find & simplify the term independent of x in the expansion of

(1/2x + x^3) ^8

I know you need to factorise it but I don't know how to get rid of the 1/2x bit.


Thanks in advance!

you only need to factorise out to make a 1 in the bracket if the power is not a positive integer... here it is 8 which is a positive integer.

[Stickyelmo]

what method do I use to get the x's to cancel though? trial and error?

[TenOfThem]

(Original post by Stickyelmo)
what method do I use to get the x's to cancel though? trial and error?

algebra

a+b = 8

and the total power will be ... ... ...

[Stickyelmo]

still no comprendo

[raheem94]

(Original post by Stickyelmo)
still no comprendo

Expand the expression and notice the third term of the expansion.

[Stickyelmo]

(Original post by raheem94)
Expand the expression and notice the third term of the expansion.

oki, when you did that was there some sort of technique or just doing it until you reach nCn?

[raheem94]

(Original post by Stickyelmo)
oki, when you did that was there some sort of technique or just doing it until you reach nCn?

I said you to notice the third term in the expansion, try to expand, it will only take moments, and then you will see what i mean.

[TenOfThem]

a+b = 8
-a+3b=0

4b=8
b=2
a=6





You will notice that the x's cancel because I used -a+3b=0

[TenOfThem]

(Original post by raheem94)
I said you to notice the third term in the expansion, try to expand, it will only take moments, and then you will see what i mean.

The algebraic method that I have used has more general application

[raheem94]

(Original post by TenOfThem)

[raheem94]

(Original post by TenOfThem)
The algebraic method that I have used has more general application

In this question it was simple to expand the terms. I did get the same values using an algebraic method but thought it would be easier to do this question by expanding.