Expanding with power 4 (how)? (1 Viewer)

[Lurch]

is there an easy way of expanding (x+y)^4 without going through the process of tediously doing (x+y)(x+y)(x+y)(x+y) OR Without going (x+y)^2 multiplied by (x+y)^2 ?

 

[~*HSC 4 life*~]

binomial theorem?

edit: crap thats EXT 1, never mind

 

[Lurch]

I suppose pascals triangle but I was wondering if there is any easy method (even though pascals triangle is easy)
Just curious..

 

[:: ck ::]

ok for 2u... mayb u can think of

just expanding it as (x+y)^2
then square it again?

or u can use binomial theorem/pascals triangle...

 

[kpq_sniper017]

if you're a 3U student, then binomial theorem is the best.

(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

if you really need to remember it, then remember:
1. the powers always add up to 4 (for ^4)
2. as powers of x increase, powers of y decrease and vice versa
3. and for ^4, the coefficients are: 1, 4, 6, 4 and 1

 

[Lurch]

yeah essentially pascals triangle....

 

[kpq_sniper017]

i don't think there is anything easier than binomial theorem.

to my knowledge.

 

[Affinity]

By inspection... in other words.. do it a few(~100) times then you should remember it

 

[~*HSC 4 life*~]

Originally posted by pcx_demolition017
i don't think there is anything easier than binomial theorem.

to my knowledge.

Yeah, binomial theorem is "easy" hahaha

 

[Calculon]

Originally posted by ~*HSC 4 life*~
Yeah, binomial theorem is "easy" hahaha

Those inverted commas are unnecessary

 

[Collin]

Is pascal's triangle 2 unit? I thought it was 3 unit.

Anyway, assuming pascals' is not 2 unit, you really have no alternative but to manually expand.

Another method, if you're quick at picking up mathematical concepts, is to simply memorise the binomial expansion rules from a 3 unit textbook. But expanding to the power of 4 doesn't take too long, even by slow manual expansion. If they ask you for (p - q)^32 or something, then I'd be slightly worried.

 

[~*HSC 4 life*~]

Originally posted by Calculon
Those inverted commas are unnecessary

pppft just because you're a maths genius

 

[Calculon]

Originally posted by ~*HSC 4 life*~
pppft just because you're a maths genius

I don't claim to be a maths genius. If you knew some of the people at my school you would see why.

 

[kpq_sniper017]

just for interest....
are you allowed to use 3U methods in 2U?
e.g. if a question said:

hence or <u>otherwise</u> expand (x+y)^4 - note the "otherwise" in a 2U test

does that mean you could use binomial theorem??

 

[CM_Tutor]

In this paricular case, how would they know? Suppose a question was:

1 (a). Expand (x + y)²

(b). Hence, or otherwise, expand (x + y)⁴

and you wrote

1(a) (x + y)² = x² + 2xy + y²

1(b) (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

how would they know? - even if the question only said 'hence', could they really penalise this answer?

Obviously, they'd know you did 'hence' if you wrote:

(x + y)⁴ = [(x + y)²]² = (x² + 2xy + y²)² = (x² + 2xy)² + 2(x² + 2xy) * y² + (y²)² = ...

Similarly, they'd know you used the binomial theorem if ⁴C_k's started to appear. In practice, 'hence or otherwise' only starts to appear once you must be showing enough working for them to be able to tell.

As for your more general question, it's better to stick to 2u methods if you can, but if you can't, a 3u answer beats no answer any day.

 

[Ragerunner]

Ok.....Is it me or is the binomial theorm freaky? Or am I looking at the wrong one?

 

[Winston]

thats the one!

 

[CM_Tutor]

Actually, that's an application of the binomial theorem - the theorem itself is just ⁿC_r = n! / r!(n - r)!

 

[JamiL]

if it is (x+y)^3 this would equal x^3+3x^2y+3xy^2+y^3...
jus cheeking im in yr 11... i do 3unit but we haven learnt that yet

 

[velox]

arent we looking to the power of 4