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Hello sir,

Kindly could you assist me with the question below. I need a step-by step working shown for this please.
This is where I'm stuck:

3-x/(x^2+3)(x+3)= A/x+3 + Bx+C

3-x= A(x^2+3)+(Bx+C)(x+3)

So my question is what will x be to find the values of A, B and C respectively? Kindly can you please provide me with the working step-by-step? Your assistance is much appreciated.
Many thanks.

Questioner:Sugi
Country:Selangor, Malaysia
Private:No
Subject:Partial Fractions
Question:Hello sir,

Kindly could you assist me with the question below. I need a step-by step working shown for this please.
This is where I'm stuck:

3-x/(x^2+3)(x+3)= A/x+3 + Bx+C/(x^2 + 3)   << you meant this, I think.

3-x= A(x^2+3)+(Bx+C)(x+3)

No, you want this: (You must multiply by the LCD, (x+3)(x^2 + 3)

(3-x)(x+3)(x^2 + 3) = A(x^2+3)+(Bx+C)(x+3)

Now you must find A,B,C.  You need THREE equations to solve, so you must choose THREE different values of x.  Perhaps  x = 79,  x = 116,  x = - 277.

No, those would not be so good.  How about:

x = 0  << really makes things simple.
x = -3  << kills a couple of factors.
x = 1   << can't think of anything better.

x = 0
(3-x)(x+3)(x^2 + 3) = A(x^2+3)+(Bx+C)(x+3)
becomes
27 = 3A + 3C

x = -3  << kills a couple of factors.
(3-x)(x+3)(x^2 + 3) = A(x^2+3)+(Bx+C)(x+3)
becomes
0 = 12A

x = 1   << can't think of anything better.
(3-x)(x+3)(x^2 + 3) = A(x^2+3)+(Bx+C)(x+3)
32 = 4A + 4B + 4C

OK, you can take it from there.

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