The Student Room Group
Reply 1
Why don't you post the question and/or working?
Reply 2
(x^2+ 5x+7)/(x+2)^3 becomes 1/(x+2) + 1/(x+2)^2 + 1/(x+2)^3

I'd appreciate it if someone could tell me how this is done. Thank you.
Reply 3
If you have

x2+5x+7(x+2)3\frac{x^2 + 5x + 7}{(x+2)^3}

Then the partial fractions are

x2+5x+7(x+2)3=A(x+2)+B(x+2)2+C(x+2)3\frac{x^2 + 5x + 7}{(x+2)^3} = \frac{A}{(x+2)} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3}.

Likewise, if it were

x2+5x+7(x+2)4\frac{x^2 + 5x + 7}{(x+2)^4}

You'd want

x2+5x+7(x+2)4=A(x+2)+B(x+2)2+C(x+2)3+D(x+2)4\frac{x^2 + 5x + 7}{(x+2)^4} = \frac{A}{(x+2)} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3} + \frac{D}{(x+2)^4}

It's just a standard pattern you should learn.

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