Question

Resolve $\frac{2x^3-3x^2-8x-26}{2x^2-5x-12}$ into partial fractions.

Answer 1

$\frac{2x^3-3x^2-8x-26}{2x^2-5x-12}$

$=\frac{-2x^3+3x^2+8x+26}{-2x^2+5x+12}$

$=x+1+\frac{-9x+14}{2x^2-5x-12}$

$=x+1+\frac{9x-14}{-2x^2+5x+12}$

$=x+1+\frac{9x-14}{(x-4)(2x+3)}$....A

$\frac{9x-14}{(x-4)(2x+3)}=\frac{A}{x-4}+\frac{B}{2x+3}$

$\frac{9x-14}{(x-4)(2x+3)}=\frac{A(2x+3)+B(x-4)}{(x-4)(2x+3)}$

$9x-14=A(2x+3)+B(x-4)$

Solving the above equation, we get,

$A=2,B=5$

$\frac{9x-14}{(x-4)(2x+3)}=\frac{2}{x-4}+\frac{5}{2x+3}$

From ...A

$\frac{2x^3-3x^2-8x-26}{2x^2-5x-12}=x+1+\frac{2}{x-4}+\frac{5}{2x+3}$