We must factorise the denominator
Let #f(x)=2x^3-5x^2+x+2#
#f(1)=2-5+1+2=0#
Therefore, #(x-1)# is a factor of #f(x)#
To find the other factors, we have to do a long division
#color(white)(aaaa)##2x^3-5x^2+x+2##color(white)(aaaa)##∥##x-1#
#color(white)(aaaa)##2x^3-2x^2##color(white)(aaaaaaaaaaa)##∥##2x^2-3x-2#
#color(white)(aaaaaa)##0-3x^2+x#
#color(white)(aaaaaaaa)##-3x^2+3x#
#color(white)(aaaaaaaaaaa)##0-2x+2#
#color(white)(aaaaaaaaaaaaa)##-2x+2#
#color(white)(aaaaaaaaaaaaaa)##-0+0#
Therefore,
#2x^3-5x^2+x+2=(x-1)(2x^2-3x-2)#
#=(x-1)(2x+1)(x-2)#
Now, we can perform our decomposition into partial fractions
#(7x^2-12x+11)/(2x^3-5x^2+x+2)=(7x^2-12x+11)/((x-1)(2x+1)(x-2))#
#=A/(x-1)+B/(2x+1)+C/(x-2)#
#=(A(2x+1)(x-2)+B(x-1)(x-2)+C(2x+1)(x-1))/((x-1)(2x+1)(x-2))#
Therefore,
#7x^2-12x+11=A(2x+1)(x-2)+B(x-1)(x-2)+C(2x+1)(x-1)#
Let #x=1#, #=>#, #6=-3A#, #=>#, #A=-2#
Let #x=2#, #=>#, #15=5C#, #=>#, #C=3#
Coefficients of #x^2#, #=>#, #7=2A+B+2C#
#B=7-2A-2C=7+4-6=5#
So,
#(7x^2-12x+11)/(2x^3-5x^2+x+2)=-2/(x-1)+5/(2x+1)+3/(x-2)#
