Question

If $\frac{x-4}{x^2-5x+6}$ can be expanded in the ascending powers of $x$, then the coefficient of $x^3$:

• A. $\frac{-73}{648}$
• B. $\frac{73}{648}$
• C. $\frac{71}{648}$
• D. $\frac{-71}{648}$

Answer 1

The correct option is A $\frac{-73}{648}$

$\frac{x-4}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{x-3}$
solving for A,B
$x-4=A(x-3)+B(x-2)$
$A=+2;B=-1$
$=\frac{2}{(x-2)}-\frac{1}{(x-3)}$
$(x+a{)}^n=a^n+n(a^{n-1})x^1+\frac{n(n-1)}{2}{x}^2{a}^{n-2}$
$+\frac{n(n-1)(n-2)}{3!}-x^3.a^{n-3}+....$
So, co-efficient $x^3$ of in $\frac{+2}{x+2}$ is and that in
$=6\left[\frac{(-1)(-2)(-3)}{3!}\right[2^{-4}\left]x^3\right]$
$\frac{-1}{x+3}=7\left[\frac{(-1)(-2)(-3)}{3!}\right[3^{-4}\left]x^3\right]$
$=-\frac{2}{2^4}+\frac{1}{3^4}=\frac{-73}{648}$