Express as partial fractions

(1)    \( \frac{3x+10}{(2x+4)(x-2)} \)

\begin{align} \frac{3x+10}{(2x+4)(x-2)} & = \frac{A}{2x+4}+ \frac{B}{x-2} \\ \\ \frac{3x+10}{(2x+4)(x-2)} & = \frac{A(x-2)}{(2x+4)(x-2)}+ \frac{B(2x+4)}{(x-2)(2x+4)} \\ \\ 3x+10 & = A(x-2) + B(2x+4) \\ \\ let\ x = 2\,, \quad then \quad 3(2)+10 & = A(2-2)+B(2(2)+4) \\ 16 & = 8B \\ \implies B & =2 \\ \\ let\ x =-2\,, \quad then \quad 3(-2)+10 & =A(-2-2)+B(2(-2)+4) \\ 4 & = -4A \\ \implies A & = -1 \\ \\ \therefore \quad \frac{3x+10}{(2x+4)(x-2)} & = \frac{-1}{2x+4}+ \frac{2}{x-2} \end{align}

(2)      \(\frac{9x^2+12x-6}{(x-1)(x+2)(2x+3)} \)

\begin{align} \frac{9x^2+12x-6}{(x-1)(x+2)(2x+3)} & = \frac{A}{x-1}+ \frac{B}{x+2}+\frac{C}{2x+3} \\ \\ \frac{9x^2+12x-6}{(x-1)(x+2)(2x+3)} & = \frac{A(x+2)(2x+3)}{(x-1)(x+2)(2x+3)}+ \frac{B(x-1)(2x+3)}{(x+2)(x-1)(2x+3)}+\frac{C(x-1)(x+2)}{(2x+3) (x-1)(x+2)} \\ \\ 9x^2+12x-6 & = A(x+2)(2x+3) +B(x-1)(2x+3)+ C(x-1)(x+2) \\ \\ let\ x = 1\,, \quad then \quad 9(1^2)+12(1)-6 & = A(1+2)(2(1)+3)+B(1-1)(2(1)+3)+C(1-1)(1+2) \\ 15 & = 15A \\ \implies\ A & = 1 \\ \\ let\ x =-2\,, \quad then \quad 9((-2)^2)+12(-2)-6 & =A(-2+2)(2(-2)+3)+B(-2-1)(2(-2)+3)+C(-2-1)(-2+2) \\ 6 & = 3B \\ \implies B & = 2 \\ \\ Equating\ constants\,, \quad -6 & = 6A+3B-2C \\ -6 & = 6(1)+3(2)-2C \\ -6& = -2C \\ \implies\ C= 3\\ \\ \therefore \quad \frac{9x^2+12x-6}{(x-1)(x+2)(2x+3)} & = \frac{1}{x-1}+ \frac{2}{x+2}+\frac{3}{2x+3} \\ \\ \\ \end{align}

(3)      \(\frac{5x^2+20x+4}{(x+3)(2x+3)(x-2)} \)

\begin{align} \frac{5x^2+20x+4}{(x+3)(2x+3)(x-2)} & = \frac{A}{x+3}+ \frac{B}{2x+3}+\frac{C}{x-2} \\ \\ \frac{5x^2+20x+4}{(x+3)(2x+3)(x-2)} & = \frac{A(2x+3)(x-2)}{(x+3)(2x+3)(x-2)}+ \frac{B(x+3)(x-2)}{(2x+3)(x+3)(x-2)}+\frac{C(2x+3)(x+3)}{(x-2) (2x+3)(x+3)} \\ \\ 5x^2+20x+4 & = A(2x+3)(x-2) +B(x+3)(x-2)+ C(2x+3)(x+3) \\ \\ let\ x = 2\,, \quad then \quad 5(2^2)+20(2)+4 & = A(2(2)+3)(2-2)+B(2+3)(2-2)+C(2(2)+3)(2+3) \\ 44 & = 35C \\ \implies\ C & = \frac{44}{35} \\ \\ let\ x =-3\,, \quad then \quad 5((-3)^2)+20(-3)+4 & =A(2(-3)+3)(-3-2)+B(-3+3)(-3-2)+C(2(-3)+3)(-3+3) \\ -11 & = 15A \\ \implies A & = -\frac{11}{15} \\ \\ Equating\ constants\,, \quad 4 & = -\frac{11}{15}(3)(-2)+ B(3)(-2)+ \frac{44}{35}(3)(3) \\ 4 & = -\frac{66}{15}-6B+ \frac{396}{35} \\ 4& = \frac{242}{35}-6B \\ -\frac{102}{35}= -6B \\ \implies\ B=\frac{17}{35}\\ \\ \therefore \quad \frac{5x^2+20x+4}{(x+3)(2x+3)(x-2)} & = \frac{-\frac{11}{15}}{x+3}+ \frac{\frac{17}{35}}{2x+3}+\frac{\frac{44}{35}}{x-2} \\ \\ \\ \end{align}