Question

Expand the binomial (4v+s)^5

Answer 1

\[(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\]

Answer 2

@hartnn

Answer 3

Lol...use this Formula...

a = x

b = 2y^2

Answer 4

I did but I get this really long formula

Answer 5

i mean equation

Answer 6

DISCO DEEWANE...

Answer 7

It is not that long:\[(x+2y^2)^3=x^3+3x^22y^2+3x(2y^2)^2+(2y^2)^3=...\]Left to do now are the 2nd, 3rd and 4th term:
\[x^3+6x^2y^2+12xy^4+8y^6\]

Answer 8

OK thanks !

Answer 9

Can you help me with this one (4v+s)^5

Answer 10

Well...yes, but that will be a lot of work, if you do not see the overall pattern in the expansion of (a+b)^n.
Have you ever heard of Pascal's Triangle?

Answer 11

It looks like this:

Answer 12

I've neber heard of it

Answer 13

never

Answer 14

That is no problem, we come to it later!

To see what (a+b)^5 would be, write it as:
(a+b)(a+b)(a+b)(a+b)(a+b).

What you have to do to expand this, is calculate all possible products of 5 numbers, each of which can be a or b:
aaaaa+aaaab+aaaba+...+abbbb+bbbbb. That is 25 possibilities, some of which can be added up: aaaab is the same product as aaaba. In fact this one occurs 5 times (it is a combination of 4 a's and one b).
Every one of these 5 same products is \[a^4b\]so we have \[5a^4b\]
as one of the terms of the expanded (a+b)^5.

Does this make sense to you (after close inspection, of course... ;)

Answer 15

yes

Answer 16

If you think about it, this are the possible terms of (a+b)^5:\[a^5\]\[a^4b\]\[a^3b^2\]\[a^2b^3\]\[ab^4\]\[b^5\]
See the pattern? It's always 5 numbers, as it should be. You can see this in advance.
The only problem we have now, is how MANY of each of these terms?
That's where Pascal's Triangle comes in.

Answer 17

ok

Answer 18

In Pascal's Triangle, the outer numbers are always 1.
To get any other number, just add up the two numbers that are on top left and top right of it: look up 10. It is 4+6, which are in the row above, to the left and right.
This way you can always expand the triangle one row further...

Now, the 5th row (beginning to count from 0) has the numbers 1, 5, 10, 10, 5, 1.
These are just the numbers we are looking for:\[(a+b)^5=1 \cdot a^5+5 \cdot a^4b+10 \cdot a^3b^2 + 10 \cdot a^2b^3+5 \cdot ab^4+1 \cdot b^5\]\[=a^5+5a^4b+10a^2b^3+10a^2b^3+5ab^4+b^5\]

Answer 19

okay i see

Answer 20

Now you "only" have to expand (4v+s)^5, so you set a=4v and b=s.
This would become:\[(4v)^5+5(4v)^4s+10(4v)^3s^2+10(4v)^2s^3+5\cdot4vs^4+s^5\]Now there is only a little work left...

Answer 21

ok

Answer 22

The options all start with s^5? and mine began with 4v^5?

Answer 23

Maybe the order of the terms has been changed?

Answer 24

i guess so i did (s+4v)^5

Answer 25

Beware, though:\[(4v)^5=4^5v^5=1024v^5\]

Answer 26

My final answer to (4v+s)^5 would be:\[1024v^5+1280v^4s+640v^3s^2+160v^2s^3+20vs^4+s^5\]Terms my be placed in any order ;)

Answer 27

ok thanks