Hi, Im really stuck with resolving this problem into partial fractions. Please help!!!!
x^2-x-13
(x^2+7)(x-2)
$\dfrac{x^2-x-13}{(x-2)(x^2+7)} = \dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+7}$
$x^2-x-13 = A(x^2+7) + (Bx+C)(x-2)$
$x^2-x-13 = Ax^2+7A + Bx^2+Cx-2Bx-2C$
$x^2-x-13 = (A+B)x^2+(C-2B)x + (7A-2C)$
match coefficients ...
$A+B=1$
$C-2B = -1$
$7A-2C = -13$
solve the system for $A$, $B$, and $C$ ...