find the coeffiecient

**maptorren** · July 29th 2008, 09:41 AM

Find the coefficient of x4 in the expansion of (3x - 2)6

**Soroban** · July 29th 2008, 10:18 AM

Hello, maptorren!

I must assume you're familiar with the Binomial Expansion.

Find the coefficient of $x^4$ in the expansion of $(3x-2)^6$

The term we seek is: . ${6\choose4}(3x)^4(\text{-}2)^2 \;=\;15(81x^4)(4) \;=\;4860x^4$

The coefficient is 4860.

**masters** · July 29th 2008, 10:23 AM

Originally Posted by maptorren

Find the coefficient of x4 in the expansion of (3x - 2)6

Use the binomial theorem:

$(a-b)^4=1a^6(-b)^0+\frac{6}{1}a^5(-b)^1+\boxed{\frac{6\cdot5}{1\cdot2}a^4(-b)^2}$

Substitute 3x for a, and 2 for b

Soroban is correct.. 4860 is your coefficient.

**masters** · July 29th 2008, 10:48 AM

Using the factorial form of the binomial theorem, you want to find the 3rd term of $(3x-2)^6$ .

$(3x-2)^6=\sum_{k=0}^{6} \frac{6!}{(6-k)!k!}\ \ (3x)^{6-k}(-2)^k$

Find the 3rd term when k=2,

$\frac{6!}{(6-2)!2!}\ \ (3x)^{6-2}(-2)^2=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot 3\cdot2\cdot1\cdot2\cdot1}(3x)^4(4)=15\cdot81x^4\c dot4=4860x^4$

**maptorren** · July 29th 2008, 11:14 AM

what about when you have something like find the coefficient of x^4 in the expansion of (3x -1/x)^6.

**maptorren** · July 29th 2008, 11:58 AM

how would i solve this porblem

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