The form of the partial fraction decomposition of a rational function is given below. 5x^2+3x+54/(x+3)(x^2+9)=A/x+3+Bx+C/x^2+9 Find A,B,and C

Question

asked Jun 27, 2020 91.2k views

1 Answer

Abhishek Dhoundiyal · Answer 1 · 2020-07-02T18:01:18+0000

Answer:

A=5, B=0,C=3

Explanation:

The partial fraction decomposition is given as:

$(5x^2+3x+54)/((x+3)(x^2+9)) \equiv (A)/(x+3)+(Bx+C)/(x^2+9)$

We collect LCD on the RHS to obtain;

$(5x^2+3x+54)/((x+3)(x^2+9)) \equiv (A(x^2+9)+(x+3)(Bx+C))/((x+3)(x^2+9))$

We expand the parenthesis in the numerator of the fraction on the RHS.

$(5x^2+3x+54)/((x+3)(x^2+9)) \equiv (Ax^2+9A+Bx^2+(3B+C)x+3C)/((x+3)(x^2+9))$

This implies that:

$(5x^2+3x+54)/((x+3)(x^2+9)) \equiv ((A+B)x^2+(3B+C)x+3C+9A)/((x+3)(x^2+9))$

This is now an identity. Since the denominators are equal, the numerators must also be equal.

5x^2+3x+54=(A+B)x^2+(3B+C)x+3C+9A

We compare coefficients of the quadratic terms to get:

$A+B=5\implies B=5-A...(1)$

Also the coefficients of the linear terms will give us:

3B+C=3...(2)

The constant terms also gives us;

3C+9A=54...(3)

Put equation (1) in to equations (2) and (3).

$3(5-A)+C=3\implies C=3A-12...(4)$

3C+9A=54...(5)

Put equation (4) into (5).

3(3A-12)+9A=54

9A-36+9A=54

9A+9A=54+36

18A=90

A=(90)/(18) =5

Do backward substitution to get:

C=3(5)-12=3

B=5-5=0

$\therefore A=5, B=0,C=3$