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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

Category:Advanced Math

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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