Answer:
[tex]A=5, B=0,C=3[/tex]
Step-by-step explanation:
The partial fraction decomposition is given as:
[tex]\frac{5x^2+3x+54}{(x+3)(x^2+9)} \equiv \frac{A}{x+3}+\frac{Bx+C}{x^2+9}[/tex]
We collect LCD on the RHS to obtain;
[tex]\frac{5x^2+3x+54}{(x+3)(x^2+9)} \equiv \frac{A(x^2+9)+(x+3)(Bx+C)}{(x+3)(x^2+9)}[/tex]
We expand the parenthesis in the numerator of the fraction on the RHS.
[tex]\frac{5x^2+3x+54}{(x+3)(x^2+9)} \equiv \frac{Ax^2+9A+Bx^2+(3B+C)x+3C}{(x+3)(x^2+9)}[/tex]
This implies that:
[tex]\frac{5x^2+3x+54}{(x+3)(x^2+9)} \equiv \frac{(A+B)x^2+(3B+C)x+3C+9A}{(x+3)(x^2+9)}[/tex]
This is now an identity. Since the denominators are equal, the numerators must also be equal.
[tex]5x^2+3x+54=(A+B)x^2+(3B+C)x+3C+9A[/tex]
We compare coefficients of the quadratic terms to get:
[tex]A+B=5\implies B=5-A...(1)[/tex]
Also the coefficients of the linear terms will give us:
[tex]3B+C=3...(2)[/tex]
The constant terms also gives us;
[tex]3C+9A=54...(3)[/tex]
Put equation (1) in to equations (2) and (3).
[tex]3(5-A)+C=3\implies C=3A-12...(4)[/tex]
[tex]3C+9A=54...(5)[/tex]
Put equation (4) into (5).
[tex]3(3A-12)+9A=54[/tex]
[tex]9A-36+9A=54[/tex]
[tex]9A+9A=54+36[/tex]
[tex]18A=90[/tex]
[tex]A=\frac{90}{18} =5[/tex]
Do backward substitution to get:
[tex]C=3(5)-12=3[/tex]
[tex]B=5-5=0[/tex]
[tex]\therefore A=5, B=0,C=3[/tex]
