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Bionomial Expansions C2 help

In class we moved onto binomial expansions and we learnt how to write expansions so for example: (x+y)^4. The teacher gives us homework and i'm fine up until the question asks: Find the coefficient of x^3 in the expansion of: (4+x)^4. I'm very stuck, please help!
Use pascals Triangle. the equation is ^4 so go to the 4th row down (i think) which goes 1 3 3 1

So the first number is 1x^x second 3x^x etc etc. the x^3 is 3x^3 as it goes:

1x^4 3x^3 3x^2 1x^1

Hope I helped :s have done this since last year, but im resitting C1, C2 and S1 :smile:
The 4th row is 1,4,6,4,1 :smile:
Reply 3
Original post by kandykissesxox
In class we moved onto binomial expansions and we learnt how to write expansions so for example: (x+y)^4. The teacher gives us homework and i'm fine up until the question asks: Find the coefficient of x^3 in the expansion of: (4+x)^4. I'm very stuck, please help!


There isn't a trick in this :smile:

If you know how to expand (x+y)^4 then you can just put y = 4 into this and find out what number is in front of the x^3 term.

(If you know the relevant formula you can write this down easily, but if you're just learning this topic then it's best to do it manually so you can see where the numbers come from!)
Original post by kandykissesxox
The 4th row is 1,4,6,4,1 :smile:


my bad! it will be 4x^3 then :smile:
Original post by heavyhandscott
my bad! it will be 4x^3 then :smile:


But the answer in the book is 16? how?
Original post by davros
There isn't a trick in this :smile:

If you know how to expand (x+y)^4 then you can just put y = 4 into this and find out what number is in front of the x^3 term.

(If you know the relevant formula you can write this down easily, but if you're just learning this topic then it's best to do it manually so you can see where the numbers come from!)


i don't understand
Original post by kandykissesxox
But the answer in the book is 16? how?


the 4 in (4+x^4 must times by the 4 from Pascals triangle) 4X4 = 16
Original post by heavyhandscott
the 4 in (4+x^4 must times by the 4 from Pascals triangle) 4X4 = 16


If you use that same application for the next question which is: find the coefficient of x^3 in (1-x)^5 then it would be -5 but the answer is -10
Reply 9
Original post by kandykissesxox
But the answer in the book is 16? how?


When you expanded (x+y)^4 you should have had a term that was 4x^3y. So when y = 4 this becomes 16x^3.

Original post by kandykissesxox
If you use that same application for the next question which is: find the coefficient of x^3 in (1-x)^5 then it would be -5 but the answer is -10


The relevant coefficients in this case are 1, 5, 10, 10, 5, 1 - check Pascal's triangle :smile:
Original post by davros
When you expanded (x+y)^4 you should have had a term that was 4x^3y. So when y = 4 this becomes 16x^3.



The relevant coefficients in this case are 1, 5, 10, 10, 5, 1 - check Pascal's triangle :smile:


there is no Y term when you expand it -.-
Reply 11
Original post by kandykissesxox
there is no Y term when you expand it -.-


(1-x)^5 is just the same as (x+y)^5 but with x replaced with '1' and y replaced with '-x' :smile:
Original post by kandykissesxox
there is no Y term when you expand it -.-


How could there be no y when you expand (x+y)^4

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