Answer:
The expression is given below as
[tex]\frac{x^2}{(x-2)^2(x+3)}[/tex]Concept:
We will create a template using the denominator
[tex]\begin{gathered} \frac{x^2}{\left(x-2\right)^2\left(x+3\right)}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x+3)} \\ \end{gathered}[/tex]By cross multiplying, we will have
[tex]\begin{gathered} \frac{x^{2}}{(x-2)^{2}(x+3)}=\frac{A}{(x-2)}+\frac{B}{(x-2)^{2}}+\frac{C}{(x+3)} \\ x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \end{gathered}[/tex]Step 1:
Put x=2
[tex]\begin{gathered} x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \\ 2^2=A(2-2)(2+3)+B(2+3)+C(2-2)^2 \\ 4=5B \\ divide\text{ both sides by 5} \\ \frac{5B}{5}=\frac{4}{5} \\ B=\frac{4}{5} \end{gathered}[/tex]Step 2:
Put x=-3
[tex]\begin{gathered} x^2=A(x-2)(x+3)+B(x+3)+C(x-2)^2 \\ (-3)^2=A(-3-2)(-3+3)+B(-3+3)+C(-3-2)^2 \\ 9=25C \\ divide\text{ both sides by 25} \\ \frac{25C}{25}=\frac{9}{25} \\ C=\frac{9}{25} \end{gathered}[/tex]Step 3:
Expand the brackets
[tex]\begin{gathered} x^{2}=A(x-2)(x+3)+B(x+3)+C(x-2)^{2} \\ x^2=A(x^2+3x-2x-6)+Bx+3B+C(x^2-4x+4) \\ x^2=Ax^2+3Ax-6A+Bx+3B+Cx^2-4Cx+4C \\ By\text{ comparing coefficient, we will have} \\ A+C=1 \\ A+\frac{9}{25}=1 \\ A=1-\frac{9}{25} \\ A=\frac{25-9}{25} \\ A=\frac{16}{25} \end{gathered}[/tex]Hence,
The partial fraction decomposition will be
[tex]\Rightarrow\frac{16}{25(x-2)}+\frac{4}{5(x-2)^2}+\frac{9}{25(x+3)}[/tex]