Answer
\[ = \frac{A}{{x - 4}} + \frac{B}{{\,{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3x + 4}}\]
Work Step by Step
\[\begin{gathered}
\frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\
\hfill \\
{\text{factor}}\,\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
\frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} = \frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 2\,\left( 4 \right)x + {4^2}} \right)\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\
\hfill \\
= \frac{{2{x^2} + 3}}{{\,{{\left( {x - 4} \right)}^2}\,\left( {{x^2} + 3x + 4} \right)}} \hfill \\
\hfill \\
we\,can\,\,use\,partial\,fraction\,of\,the\,form \hfill \\
\hfill \\
\frac{{2{x^2} + 3}}{{\,\left( {{x^2} - 8x + 16} \right)\,\left( {{x^2} + 3x + 4} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{\,{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3x + 4}} \hfill \\
\end{gathered} \]
