# Binomial expansion.. please help! Deduce the coefficient of x^2 in the expansion of (1 + 3x)^2 (1 - 3x)^10Please help me solve the question above.. an explanation will be appreciated.\ Thanks!

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### 1 Answer

You need to find the coefficient of `x^2` in the following expansion, such that:

`(1 + 3x)^2(1 - 3x)^10`

You may write `(1 - 3x)^10` using the exponential properties, such that:

`(1 - 3x)^10 = (1 - 3x)^(2 + 8)`

`(1 - 3x)^10 = (1 - 3x)^2*(1 - 3x)^8`

You need to substitute `(1 - 3x)^2*(1 - 3x)^8` for `(1 - 3x)^10` such that:

`(1 + 3x)^2(1 - 3x)^2*(1 - 3x)^8`

Since the binomials `1 + 3x` and `1 - 3x` have the same exponent 2, you may write ` (1 + 3x)^2(1 - 3x)^2` such that:

`(1 + 3x)^2(1 - 3x)^2 = ((1 + 3x)(1 - 3x))^2`

You need to convert the product `(1 + 3x)(1 - 3x)` into a difference of squares, such that:

`(1 + 3x)(1 - 3x) = 1^2 - (3x)^2`

`(1 + 3x)(1 - 3x) = 1 - 9x^2`

`(1 + 3x)^2(1 - 3x)^2*(1 - 3x)^8 = (1 - 9x^2)^2*(1 - 3x)^8`

You need to expand the square `(1 - 9x^2)^2` such that:

`(1 - 9x^2)^2 = 1 - 18x^2 + 81x^4`

You need to expand the binomial `(1 - 3x)^8` using binomial theorem, such that:

`(1 - 3x)^8 = C_8^0*1^8 - C_8^1*1^7*(3x) + C_8^2*1^6*(3x)^2 + ... + C_8^8 (3x)^8`

`(1 - 18x^2 + 81x^4)(C_8^0*1^8 - C_8^1*1^7*(3x) + C_8^2*1^6*(3x)^2 + ... + C_8^8 (3x)^8) = 1*C_8^0*1^8 - 1*C_8^1*1^7*(3x) + 1*C_8^2*1^6*(3x)^2 + ... - 18x^2*C_8^0*1^8 + 18x^2 C_8^1*1^7*(3x) + ...`

You need to find the coefficient of ` x^2` such that:

`x^2(1*9*C_8^2*1^6 - 18x^2*C_8^0*1^8) = ((9*8*7)/2 - 18)x^2 = 234x^2`

**Hence, evaluating the coefficient of `x^2` in the given expansion, using binomial theorem, yields 234.**