How to quickly expand binomials

[HmBInTheMovement]

Thanks !!!!! It is way easier to understand now :)

[Momoko4ever1]

GENIUS! THANKS

[musiclove824]

YOU PERFECT BEING. Now I might get a passing grade on my test tomorrow... haha

[jeoantapo]

Este método es uno de los mas simples, remember, cuando se tiene (x+y)^n los términos van alternados empesando siempre por el + - + - + - y el número de términos a obtener siempre es (n+1) es decir uno mas de el de la potencia. Ej: si (x+y)^2 entonces se tienen 3 téminos, si (x+y)^3 entonces 4 términos y en general si (x+y)^n entonces (n+1) esto les servirá cuando necesiten saber un determinado término. Saludos desde Ecuador.

[VivaLaDivax]

is this like the pascal's traingle expansions?

[nez191]

you sound mad excited about binomial expansion...thanks a lot

[MetaWolfakaZERO]

Who created math i really want to kick their ass

[worseize]

ГЕНИАЛЬНО!!!

[nickster1017]

what

[psychoTR2]

@psychoTR2 Oops, I forgot that the description mentions this. Sorry.

(It's not enough to read the description BEFORE watching a video, you have to remember what it says,too. lol)

[psychoTR2]

Great VIdeo! The only problem is that using a binomial with no negative coefficient as an example doesn't help us to know which coefficients in the expanded binomial should be made positive or negative.

I have the solution: The signs alternate if the binomial subtracts. If the first term in the binomial is positive, the first term in the expansion is positive. Then each term gets the alternative sign. Think of it this way:

Term 1: -

Term 2: - +

Term 3: - + -

And so on...

[zezo7hdk]

@psychoTR2 you say it goes - + - and in the description it says + - + ? which one is correct?

[psychoTR2]

@zezo7hdk I think that if the coefficient of the binomial is positive, all of the terms in the expanded binomial are positive. If the binomial's coefficient is negative, the signs alternate by negative, positive, negative, positive, and so on. It's been a while since I've done this stuff, so I'm rusty, but if I remember right, this is how it goes.

[zezo7hdk]

@psychoTR2 i get how it goes if the binomial is positive. im asking about when its negative does it go (negative-positive-negative...­) or (positive-negative-positive...­.)? or both are correct? thanks.

[cacapupu1234]

This helped a bit

[idraw90]

I LOVE YOU!!!!!!! thank you so much. Please keep making videos like these for the poor kids (like me) that for some reason always sleep in class even though they don't want to! once again thank you ^_^

[MsGrammarnazi]

Thanks so much!

[Rokisimsful]

thank you so much that really helped me a lot .. :D

[danvega18]

I'm in a college trig class and I used your rule for a HW problem. It worked. Here's the prob I had:

((7[{x+h}^3]-3)-([7x^3]-3))/h

Solution:

21x^2 + 21xh + 7h^2

thx

[mc0carrie]

dude! i wish my precalc teacher woulda taught me like this, instead of using pascals triangle or that other nonsense.

[MEGIDIOT]

1000^100000000 clearer than my math book!

[aldrinigop]

LOL!! u used paint??? OMG. I LOLED very loud

[jedah998]

You sound so sleepy =ω=

[djuwell]

what if it asks for the coef of y?

[muzteci123]

this makes no sense what if it is ( 2x +3) to the power of 4

[AmalucaandCo]

That realy helped my fast!

[000blocks000]

Great stuff. Thank you!

[de8ball]

i love you!!! u are heaven sent!

[teraniel11]

OMG!! thank you so much! i dont need to bring the pascals trinangle anymore as a cheat!!! hahahahaha!!!(JOKE) tnx!!

[valeagrl1]

thanks for this I can still use my lesson's system now but at least now, thanks to you, I can wrap my head around it.

[xXxz0om3rxXx]

oh my gosh im actually understandin it lol but ur writin sux D: lmfao try a different method like.. writing on a board ..? lol idk .. xD

[Rapidfire220]

Pascals triangle is much easier, and I can easily remember all the coefficients up to x^5.

[quns25]

similar to Pascal's triangle.. but it is very cleaver, never seen that before

[wkrepelin]

Didn't know that trick. Are you aware of the multi dimensional generalizations of Pascal's triangle called Pascal Simplexes? My way of doing this for any number of terms is this: Come up with every possible multiple of the terms where the powers of terms add to the original power "n". The coefficient of each term will be n! divided by the products of the factorials of the individual terms powers. So, the coefficient for x^3y^4Z^1 is (3+4+1)!/(3!4!1!)=420 and would come from (x+y+z)^8.

[wkrepelin]

Mistake, 280 not 420. Whoops.

[mathacademy]

Very interesting!

I will look into it more. Thank you for sharing.