bionomial expansion

[steady girder]

idk how to do it

my working is

p!/2!(p-2)! = 40

[barren furnace]

quadratic

[steady girder]

the working is wrong tho

[barren furnace]

its cuz its -2x

normally you do

nCr times (x term)^r

times other thing

[shut egret]

Damn that looks confusing
Do you have to do all of that for marks in an actual test cause I just kinda solved it simply in my head

[steady girder]

nah i just dk how to do these ones

[barren furnace]

na so like

how would you find the x^2 term

like if p was say 5

acc 5 is the right answer so not 5 but

say like 10

[steady girder]

using the formula

[barren furnace]

yeah so

in this case

[steady girder]

10cr2

[barren furnace]

times what

[steady girder]

x^2?

[barren furnace]

not quite cuz

ok so

for (1+x)^n

[steady girder]

that formula finds the coefficient of x^2 right?

[barren furnace]

it would but you havent done it quite right

[barren furnace]

so for this

the x^k term is

nCk * (x)^k * (1)^(n-k)

have you seen that before

[steady girder]

not in that form

only the basic numbers

[barren furnace]

ok so

its that basically

you have the nCr thing

and then the whole of one term gets the power of r

and the whole of the other gets the n-r power

[steady girder]

yh

[barren furnace]

ok so in your case

you have (1 - 2x)^p

so when you do this formula

the x^2 bit has

pC2 as you said

but also (-2x)^2

[steady girder]

yh

[barren furnace]

times a 1^?

so expanding that (-2x)^2 is just 4x^2

so can you see that

you have this rouge multiple of 4

that youve not included

[steady girder]

oh

[barren furnace]

does that make sense?

[steady girder]

yhh ty

il ltry to do it again

[barren furnace]

nw!

good luck

[steady girder]

hmm i figured out a way to do it

but its rly not that good

[barren furnace]

lets hear it

im interested

[steady girder]

1 sec drawing it out

[barren furnace]

i dont get

how does 10 get us to 5

[steady girder]

the questions tells you the coefficient of x^2 is 40

because of pascals triangle

the third term has to be 10

and 1 5 10 gives us the third term being 10

[barren furnace]

i see

[steady girder]

what was your way?

[barren furnace]

well so

[steady girder]

i dont mind a better method lol

[barren furnace]

really youre pretty close to my method

but so your 'y'

is actually

just pC2

[steady girder]

ah ye

[barren furnace]

that is what y represents

so really you have

pC2 = 10

then similar to waht you had before

p(p-1)/2 = 10

p^2 - p - 20 = 0

boom quadratic

solving that does give you p = 5

like really they have the same ideas

but your involves spotting the right row whereas this is

[steady girder]

how does it go to p(p-1)/2 = 10

[barren furnace]

longer but more concrete

pC2 is p(p-1)/2

[barren furnace]

which you found from before here

[steady girder]

o

i c

thx anyway ill ask my teacher to explain this ques anyway

[barren furnace]

calm nw

[steady girder]

she didnt teach us how to use the formula

so got students who do further maths to explain it

which is lame

[barren furnace]

further maths neeks 🙄

[steady girder]

they r just so arrogant even tho they are my friends xd

@barren furnace alr i see last question how do you get p(p-1)/2 = 10

because your way i tried redoing and i cant figure out how you got 10

[barren furnace]

its what y is