Question

Resolve $\frac{23x-11x^2}{(2x-1)(9-x^2)}$ into partial fractions

• A. $\frac{1}{2x-1}+\frac{4}{3+x}-\frac{1}{3-x}$
• B. $\frac{1}{2x+1}-\frac{4}{3-x}+\frac{1}{3+x}$
• C. $\frac{2}{x-1}+\frac{1}{3-x}-\frac{2}{3+x}$
• D. $\frac{5}{x+1}-\frac{4}{3-x}-\frac{1}{3+x}$

Answer 1

The correct option is A $\frac{1}{2x-1}+\frac{4}{3+x}-\frac{1}{3-x}$

$\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{x(23-11x)}{(2x-1)(9-x^2)}$
$=\frac{x\left(\right(21-7x)+(2-4x\left)\right)}{(2x-1)(9-x^2)}=\frac{x(21-7x)}{(2x-1)(9-x^2)}+\frac{x(2-4x)}{(2x-1)(9-x^2)}$
$=\frac{7x}{(2x-1)(3+x)}-\frac{2x}{9-x^2}=\frac{6x-3+x+3}{(2x-1)(x+3)}-\frac{3+x-(3-x)}{9-x^2}$
$=\frac{6x-3}{(2x-1)(x+3)}+\frac{x+3}{(2x-1)(x+3)}-\frac{3+x}{9-x^2}+\frac{(3-x)}{9-x^2}$
$=\frac{3}{x+3}+\frac{1}{2x-1}-\frac{1}{3-x}+\frac{1}{3+x}$
$=\frac{4}{x+3}+\frac{1}{2x-1}-\frac{1}{3-x}$