PLEASEEEE can someone explain binomial expansion!!!! REP AVAILABLE PLEASE HELP ME!

I really cannot get my head around binomial expansion. And what the hell has it all got to do with factorials?

I've got a list of questions like

find the term in x³ of (4-3x)⁷

i have no idea where to begin :frown:

help appreciated, + repp

jb

Reply 1

tara592

jumblebumble

help appreciated, + repp

jb

okay im gonna try and explain this...by the way i dont know how you write the numbers as a power so il just write them after a ^ sign (i.e for 4 squared il write 4^2)
so what i first do is write out the first stage:

4^0 4^1 4^2 4^3 4^4 4^5 4^6 4^7then:
(-3)^7 (-3)^6 (-3)^5 (-3)^4 (-3)^3 (-2)^2 (-2)^1 (-2)^0

so then you can write them together, but remembering the rules of indices you can simplify some of them (i.e 4^0 = 1 and 4^1=4)

*make sure you leave loads of space between each of the numbers though else it makes it hard to read...

and then you need to use the nCR button on your calculator to type it in beforehand...does that make any sense?

Reply 2

tara592

ahh sorry i left loads of space in between them so it would look easier to read but it hasnt shown up! just as long as you lay them out directly underneath it might look a bit easier :smile:

Reply 3

jumblebumble

tara592

but then how do i solve this question? because the answer is apparently -241 920x^3....?

Reply 4

TheFriendlySocialist

jumblebumble

but then how do i solve this question? because the answer is apparently -241 920x^3....?

You're only looking for the one term.

When you do the question, you get a bunch of terms. Starting with x^7, x^6, and so on. It's the term which ends with x^3 which you want.

Reply 5

jumblebumble

TheFriendlySocialist

You're only looking for the one term.

When you do the question, you get a bunch of terms. Starting with x^7, x^6, and so on. It's the term which ends with x^3 which you want.

please can you explain
i don't know how to 'do' the question

how do i get to the answer, it is given in the book and i can't get to it

Reply 6

Nfixlol

jumblebumble

help appreciated, + repp

jb

i'm not sure how to help you on this one exactly, but maybe walking you through the steps whilst doing the question myself will help:

(4 - 3x)^7

when you expand this using binomial, you get a bunch of terms, the first time being in

x^0

and the last term being in

x^7

the coefficients of terms, in order, can be found using this pyramid of coefficients, lol

since the first term is in

x^0

, the term in

x^3

is the fourth term, do you see why? so it's number coefficent is the fourth number along the eigth row of the triangle, which is 35.

i.e., the entire term is:

35*4^4*(-3x)^3

which should give you the right answer :yep:

Spoiler

Reply 7

MathsHamster

The formula is in the formula book, otherwise have you been taught about nCr buttons and Pascal's triangle?

Basically (x + y)^7 = 1(x^7)(y^0) + 7(x^6)(y^1) + 7C2(x^5)(y^2) + 7C3(x^4)(y^3) + ........ 7C7(x^0)(y^7)

Does this notation mean anything to you? ie. 7C4 is the number of ways you can choose four things out of seven things and there is a button on your calculator for working it out.

Back to the question... stick x = -3x and y = 4 into the formula find the term (and there will be only one) which has an x^3 in it, and that is the one you want. Hope this helps, quote me if it doesn't and I'll try and explain more.

Reply 8

haribo kid

I’ll try to explain binomials how it was explained to me..I hope you get it
If you were to imagine a question which asks you how many times you could rearrange letters, then factorials should makes sense.
Example
How many times can you rearrange AA? – that’s obviously just once..if you switch it you get the same answer back
AAB: AAB, ABA, BAA – the answer is 3. The way to work that out using factorials is to treat A and B as if they were individual terms. In mathematics you could rewrite AAB as A^2B. The rule is that you divide the factorial of the total number of terms, by the factorial of each group of terms. So basically using the original example there are 3 in total, so you do 3! Divided by 2! 1! As there are 2 a’s and one b.
3! Divided by 2! = 3, which is the original answer we got.
Using another example if you had AABBBBCCC, to find out how many different ways you can rearrange it you do 9! Divided by 2!4!3! which will give you your answer.
So if you had n terms in total and there were r different groups of terms then you’d divide n! By r! (n-r)!
To explain why you use (n-r)! using the first example 3!-2! = 1! and 3!-1! = 2!. So if you know the amount of terms in one of the groups i.e. r, then you can easily work out how many r in the other group be doing that number minus the total, i.e. n-r.
So using binomials, if you had (a+b)^3= a^3 + a^2b.. Then remembering that a^3 = aaa and that a^2b = aab, Then you can work it in exactly the same way as we did earlier.
Another thing about the theory behind binomial expansion is that you need to remember when you expand the power of a is always decreasing and that of b is always increasing. Both powers have to add up to the original one on the LHS of the eqn. So a quick example is t hat:
(a+b)^4 = a^4 + a^3 b + a^2b^2+ ab^3 + b^4
Just remember that you find the coefficients using Pascal’s triangle.
Urmmm this is a ridiculously long post. I’ll explain ur specific question in a different post or it will be painful to read lol