the binomial theorem (1 Viewer)

[queenie]

arrrg im havin trouble with these... i dnt know how do them..

1. Find the coefficient of:
a) x^3 in (2-5x)(x^2-3)^4
b) x^5 in (x^2-3x+11)(4+x^3)^3
c) x^0 in (3-2x)^2(x +2/x)^5
d)x^9 in (x+2)^3(x-2)^7

2. Use Pascal's triangle to expand (x+h)^3
- if f(x) = x^3, simplify f(x+h)- f(x)


yes... they seem basic enough but i dnt understand this topic at all :S

 

[Xayma]

Hmm I havent done it yet so dont know how to do 1.

2.Pascal's Triangle for the power of 3 is 1 3 3 1

So we get x³+3x²h+3xh²+h³

If f(x)=x³

f(x+h)-f(x)=(x+h)³-x³
=3x²h+3xh²+h³
=h(3x²+3xh+h²)

 

[ND]

Originally posted by queenie
arrrg im havin trouble with these... i dnt know how do them..

1. Find the coefficient of:
a) x^3 in (2-5x)(x^2-3)^4
b) x^5 in (x^2-3x+11)(4+x^3)^3
c) x^0 in (3-2x)^2(x +2/x)^5
d)x^9 in (x+2)^3(x-2)^7

2. Use Pascal's triangle to expand (x+h)^3
- if f(x) = x^3, simplify f(x+h)- f(x)


yes... they seem basic enough but i dnt understand this topic at all :S

The ones in q1 you just do by inspection. for example in the 1st one:

We know that the expansion of (x^2-3)^4 is gonna have x's to the power of 0, 2, 4, 6, etc. but we want the coefficent of x^3. We can look at the other factor and see that it has x's to teh power of 0 and 1. Now it's pretty obvisous (since there are no negative powers of x) that the only way is with the x from the 1st factor and the x^2 from the 2nd. So just sum the coefficients of those two.
For b), (4+x^3)^3 will have x's to teh powers 3, 6, 9 etc, and (x^2-3x+11) has x's to the powers 2, 1 and 0. Again, it's obvious that the only way to get x^5 is from x^2 and x^3. Just do that for all of them...

 

[Heinz]

I'll just quickly do c and d

c) x^0 in (3-2x)^2(x +2/x)^5

expanding (3-2x)^2(x +2/x)^5 we get

(9-12x+4x^2)(5C0x^5 + 5C1(x)^4(2/x) + 5C2(x)^3(2/x)^2 + 5C3(x)^2(2/x)^3 +5C4(x)(2/x)^4 + 5C5(2/x)^5)
=(9-12x+4x^2)(x^5 + 10x^3 + 40x + 80/x + 80/x^3 + 32/x^5)

Upon examination, the only product that yields the term x^0 is (-12x)(80/x). Hence the coefficient of x^0 in the (3-2x)^2(x +2/x)^5 is -600.

d) x^9 in (x+2)^3(x-2)^7

expand to get
(x^3 + 6x^2 + 12x + 8)(7C0(x)^7 + 7C1(x)^6(-2) + ...) the rest of the expansion isnt needed so dont waste your time expanding any further
= (x^3 + 6x^2 + 12x + 8)(x^7 -14x^6 + ...)

In order to get x^9 we add the products up
(x^3)(-14x^6) + (6x^2)(x)^7 = -8x^9 so the coefficient is -8

 

[kpq_sniper017]

I haven't done this topic yet, and I'm not too familiar with it.

But, for Q1(a) is the answer 540??

I got the coefficient of x² as being -108 and
-108 x -5 = 540.

Hhhmm....well, someone tell me if I'm wrong.

 

[queenie]

Originally posted by pcx_demolition017
I haven't done this topic yet, and I'm not too familiar with it.

But, for Q1(a) is the answer 540??

I got the coefficient of x² as being -108 and
-108 x -5 = 540.

Hhhmm....well, someone tell me if I'm wrong.

lol yea, the answer is 540 and its rite wat u did.. -180 x -5

 

[Heinz]

Originally posted by pcx_demolition017
I haven't done this topic yet, and I'm not too familiar with it.

But, for Q1(a) is the answer 540??

I got the coefficient of x² as being -108 and
-108 x -5 = 540.

Hhhmm....well, someone tell me if I'm wrong.

Yeah, its 540 and b is 48?

 

[kpq_sniper017]

Phew, thought I might be wrong - I;ve still got a long way to go with Binomial Theorem - haven't done it in class yet.

 

[Affinity]

Originally posted by ND
The ones in q1 you just do by inspection. for example in the We know ...
Now it's pretty obvisous ...
Again, it's obvious that ...

You missed 'trivial'

 

[ND]

I don't get it..