Answer
\[\frac{3}{11(2x-1)}+\frac{89}{33(x+5)}+\frac{-4}{3(x+2)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{3x^2-2x+14}{(2x-1)(x +5)(x+2)}=\frac{A}{2x-1}+\frac{B}{x+5}+\frac{C}{x+2}\;\;\;...(*)\]
\[\Rightarrow 3x^2-2x+4=A(x+5)(x+2)+B(2x-1)(x+2)+C(2x-1)(x+5) \;\;\;...(1)\]
Put $x=\frac{1}{2}$ in (1)
\[\frac{3}{4}-1+4=A\left(\frac{11}{2}\right)\left(\frac{5}{2}\right)\]
\[\frac{15}{4}=\frac{55A}{4}\]
\[\Rightarrow A=\frac{3}{11}\]
Put $x=-5$ in (1)
\[3(25)+10+4=B(-11)(-3)\]
\[\Rightarrow 89=33B\]
\[\Rightarrow B=\frac{89}{33}\]
Put $x=-2$ in (1)
\[12+4+4=C(-5)(3)\]
\[20=C(-15)\]
\[C=\frac{-4}{3}\]
From (*)
\[\frac{3x^2-2x+14}{(2x-1)(x +5)(x+2)}=\frac{3}{11(2x-1)}+\frac{89}{33(x+5)}+\frac{-4}{3(x+2)}\]
Hence, \[\frac{3x^2-2x+14}{(2x-1)(x +5)(x+2)}=\frac{3}{11(2x-1)}+\frac{89}{33(x+5)}+\frac{-4}{3(x+2)}.\]
