As function in the denominator #f)x)=x^3-3x^2-4x+12# has #2, -2# and #3# as zeros, we have #x^3-3x^2-4x+12=(x-2)(x+2)(x-3)# and we can write partial fractions as

#(x+6)/(x^3-3x^2-4x+12)hArrA/(x+2)+B/(x-2)+C/(x-3)# or

#(x+6)/(x^3-3x^2-4x+12)hArr(A(x-2)(x-3)+B(x+2)(x-3)+C(x+2)(x-3))/((x-2)(x+2)(x-3))# or

#(A(x^2-5x+6)+B(x^2-x-6)+C(x^2-4))/((x-2)(x+2)(x-3))# or

#(x^2(A+B+C)+x(-5A-B)+(6A-6B-4C))/((x-2)(x+2)(x-3))#

Comparing like terms we have three terms #A+B+C=0#, #-5A-B=1# and #6A-6B-4C=6#

From second equation we get #B=-1-5A# and putting this in third equation we get #6A-6(-1-5A)-4C=6# or #6A+6+30A-4C=6# or #4C=36A# or #C=9A#.

Now putting these in first we get #A-1-5A+9A=0# or #5A=1# or #A=1/5#. Hence #B=-2# and #C=9/5#

Hence partial fractions are #1/(5(x+2))-2/((x-2))+9/(5(x-3))#