Question

use the binomial theorem to expand the binomial. (3v+s)^5

Answer 1

s^5 + 4^5s^4v + 270s^3v^2 + 810s^2v^3 + 1215sv^4 + 729v^5
s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5
s^5 + 15s^4v + 90s^3 + 270s^2 + 405s + 243
s^5 – 5s^4v + 10s^3v^2 – 10s^2v^3 + 5sv^4 – v^5

Answer 2

@campbell_st please help me with this one?!

Answer 3

Do you know the Binomial Theorem?

Answer 4

kind of...

Answer 5

So you also know about binomial coefficients? Like \[\left(\begin{matrix}5 \\ 3\end{matrix}\right)=\frac{ 5! }{ 3!2! }=\frac{ 5 \cdot 4 }{ 2 }=10\]

Answer 6

yes but i dont understand them at all. how do i figure out binomial theorem to expand the binomial. (3v+s)^5

Answer 7

?

Answer 8

OK, The BT looks more difficult than it is.
You should consider it a neat way of remembering how to get all the terms of (a+b)^n without having to do the actual multiplying.

Suppose you have (a+b)^4.
You know it is (a+b)(a+b)(a+b)(a+b).
You also know you have to multiply each of the a's and b's with all the other ones.
You don't want to do that, because it is terrible to do.
What you already know is this:
The terms you get will be:
aaaa = a^4
aaab = a³b
aabb = a²b²
abbb = ab³
bbbb = b^4

So there is always a product of 4 numbers, with every combination of a and b possible.
That's the easy part.
The hard part is to know *how many* of each of these terms there are.
Of the first and last term, there is only one, because there is only one way to write four a's: aaaa.

Of the second and fourth there are ... 4, because with 3 a's and one b, the b can have four positions

The middle term, the one with 2 a's and 2 b's can be shuffled around in 6 ways, although it is already a bit of work to find them all.

So we would get:
\[a^4+4a^3b+6a^2b^2+4ab^3+b^4\]

If you can use binomial coefficients, your life is easier: instead of getting:

1 4 6 4 1 times each term, these numbers are:

\[\left(\begin{matrix}4 \\ 0\end{matrix}\right),\left(\begin{matrix}4 \\ 1\end{matrix}\right),\left(\begin{matrix}4 \\ 2\end{matrix}\right),\left(\begin{matrix}4 \\ 3\end{matrix}\right),\left(\begin{matrix}4 \\ 4\end{matrix}\right)\]

So with (a+b)^5 (almost your problem!) it will be:\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)a^5+\left(\begin{matrix}5 \\ 1\end{matrix}\right)a^4b+\left(\begin{matrix}5 \\ 2\end{matrix}\right)a^3b^2+\left(\begin{matrix}5 \\ 3\end{matrix}\right)a^2b^3+\left(\begin{matrix}5 \\ 4\end{matrix}\right)ab^4+\left(\begin{matrix}5 \\ 5\end{matrix}\right)b^5\]If you calculate the coefficients (or look them up in Pascal's Triangle (maybe you heard of it?) you will get:\[a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\]Now you only have to do one thing: substitute a=3v and b=s in this last formula, then simplify as much as possible.

Answer 9

so is the answer b?

Answer 10

Yup!