Question

The coefficient of $x^4$ in the expansion of

$\frac{3x-8}{4-4x+x^2}$ is

• A. $\frac{-1}{4}$
• B. $-4$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{-1}{4}$

Given, $\frac{3x-8}{x^2-4x+4}$ $=\frac{3x-8}{(x-2{)}^2}$
$=\frac{3(x-2)-2}{(x-2)(x-2)}$

$=\frac{3}{(x-2{)}^1}-\frac{2}{(x-2{)}^2}$

$=3(x-2{)}^{-1}-2(x-2{)}^{-2}$

$=\frac{-3}{2}(1-\frac{x}{2}{)}^{-1}-\frac{1}{2}(1-\frac{x}{2}{)}^{-2}$
$=\frac{-3}{2}(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+\frac{x^3}{16}...\infty )-\frac{1}{2}(1+\frac{2x}{2}+\frac{3x^2}{4}+\frac{4x^3}{8}+\frac{5x^4}{16}...\infty )$

Hence coefficient of $x^4$ will be
$\left(\frac{-3}{2}\frac{x^4}{16}\right)-\left(\frac{1}{2}\frac{5x^4}{16}\right)$

$=(\frac{-3}{32}-\frac{-5}{32})x^4$

$=\frac{-8}{32}{x}^4$
$=\frac{-x^4}{4}$
Hence, coefficient is $\frac{-1}{4}$