# Find the coefficient of the x^2 term in the expansion of (3x - (\frac{2}{x^2})^{(8)}

## Question:

Find the coefficient of the {eq}x^2 {/eq} term in the expansion of {eq}(3x - (\frac{2}{x^2})^{(8)} {/eq}

## Coefficient of Suggested Term:

The given algebraic expression is similar to {eq}(a+b)^n {/eq} so we'll substitute the values of {eq}a {/eq} and {eq}b {/eq} according to the given expression in the formula of binomial expansion and simplify the obtained expansion to get the coefficient of suggested term.

• {eq}\displaystyle (a+b)^n=a^n+na^{n-1}b+^{n}\textrm{C}_{2}a^{n-2}b^2+^{n}\textrm{C}_{3}a^{n-3}b^3+\dots {/eq}

The given expression is:

{eq}(3x - (\frac{2}{x^2})^{(8)} {/eq}

Rewriting the above expression as {eq}(a+b)^n {/eq}, we get:

{eq}\displaystyle (3x - (\frac{2}{x^2}))^{(8)}=(3x + (-\frac{2}{x^2}))^{8} {/eq}

Here, we have:

{eq}a=3x\\ b=-\frac{2}{x^2} n=8 {/eq}

Substitute the above values in the formula of binomial expansion and simplify it.

{eq}\begin{align*} \displaystyle(3x+(-\frac{2}{x^2}))^8&=(3x)^8+8(3x)^{8-1}(-\frac{2}{x^2})+^{8}\textrm{C}_{2}(3x)^{8-2}(-\frac{2}{x^2})^2+^{8}\textrm{C}_{3}(3x)^{8-3}(-\frac{2}{x^2})^3+\dots\\ &=3^8x^8+8(3^7)x^7(-\frac{2}{x^2})+^{8}\textrm{C}_{2}3^6x^{6}(\frac{4}{x^4})+^{8}\textrm{C}_{3}3^5x^{5}(-\frac{8}{x^6})+\dots&\because (x^m)^n=x^{mn}\\ &=3^8x^8+8(3^7)x^{7-2}(-2)+^{8}\textrm{C}_{2}(4)3^6x^{6-4}+^{8}\textrm{C}_{3}3^5x^{5-6}(-8)+\dots&\because \frac{x^m}{x^n}=x^{m-n}\\ &=3^8x^8+8(3^7)x^{5}(-2)+^{8}\textrm{C}_{2}(4)3^6x^{2}+^{8}\textrm{C}_{3}3^5x^{-1}(-8)+\dots \end{align*} {/eq}

Simplifying the coefficient of {eq}x^2 {/eq} of the above expansion, we get:

{eq}\begin{align*} \displaystyle ^{8}\textrm{C}_{2}(4)3^6&=\frac{8!}{2!(8-2)!}4(729)&\because ^{n}\textrm{C}_{r}=\frac{n!}{(n-r)!r!}\\ &=\frac{8\times 7\times 6!}{2!6!}(2916)\\ &=\frac{8\times 7}{2}(2916)\\ &=(4\times 7)(2916)\\ &=81648 \end{align*} {/eq}