help

\(\dfrac{4x}{(x-5)(x-3)^2} = \dfrac{A(x-3)^2+B(x-5)(x-3)+C(x-3)}{(x-5)(x-3)^2}\)

\(4x = A(x^2-6x+9) + B(x^2-8x+15)+C(x-3)\\ 4x = x^2(A + B) + x(-6A-8B+C)+(9A+15B-3C)\)

\(A+B=0\\ (-6A-8B+C)=4\\ (9A+15B-3C) = 0\)

\(\text{I leave you to solve for }A,B,C\)

        4x                              A                   B                   C

____________  =        _______    +     _____    +    _______       

(x - 5)(x - 3)^2               (x - 5)                (x - 3)          (x - 3)^2

 Multiply through by (x - 5) ( x -3)^2

4x   =    A(x - 3)^2   +  B(x- 5)(x - 3)  + C(x - 5)      simplify

4x  =  A(x^2 -  6x + 9)  + B(x^2 - 8x + 15)  + C(x - 5)

4x  = (A + B)x^2    + (-6A - 8B + C)x + (9A + 15B - 5C)

Equating coefficients, we have

A + B  =  0     ⇒   B = -A      (1)    

-6A - 8B + C  = 4    (2)

9A + 15B - 5C =  0     (3)

Subbing  (1)  into (2)  and (3)   we have

-6A + 8A + C   = 4   ⇒   2A + C  =  4   ⇒   6A + 3C = 12    (4)

9A - 15A - 5C  = 0  ⇒  -6A - 5C = 0        (5)     

Add (4) and (5)

-2C =  12

C = -6

And using (4) to find (4)

6A + 3(-6)  = 12

6A - 18 = 12

6A = 30

A = 5

And B = -A =    -5

So   (A,B,C)  = (5, -5, - 6)

Yep, this is known as the "Partial Fraction Decomposition." 

Find constants A, B, and C so that

\(\large{\dfrac{4x}{(x - 5)(x - 3)^2} = \dfrac{A}{x - 5} + \dfrac{B}{x - 3} + \dfrac{C}{(x - 3)^2}}\)

\(\begin{array}{|lrcll|} \hline & \dfrac{4x}{(x - 5)(x - 3)^2} &=& \dfrac{A}{x - 5} + \dfrac{B}{x - 3} + \dfrac{C}{(x - 3)^2} \quad | \quad \cdot (x - 5)(x - 3)^2 \\\\ & \mathbf{4x} &\mathbf{=}& \mathbf{A(x - 3)^2 + B(x - 5)(x - 3) + C(x - 5)} \\\\ \hline x=3: & 4\cdot 3 &=& A\cdot 0 + B\cdot 0 + C(3 - 5) \\ & 12 &=& -2C \\ & \mathbf{C} &\mathbf{=}& \mathbf{-6} \\\\ x=5: & 4\cdot 5 &=& A(5 - 3)^2 + B\cdot 0 + C\cdot 0 \\ & 20 &=& A\cdot 2^2 \\ & 20 &=& 4A \\ & \mathbf{A} &\mathbf{=}& \mathbf{5} \\\\ x=0: & 4\cdot 0 &=& 5(0 - 3)^2+B(0 - 5)(0 - 3)-6(0 - 5) \\ & 0 &=& 45 +15B+30 \\ & 15B &=& -75 \\ & \mathbf{B} &\mathbf{=}& \mathbf{-5} \\ \hline \end{array}\)