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    Kind of stuck on how to factorise this:

    \frac{1}{s^2 + 2cs + c^2 + d^2}

    I reckon the factors are complex but I too tired to see the "obvious" factorisation. Any help please?
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    (Original post by BAD AT MATHS)
    Kind of stuck on how to factorise this:

    \frac{1}{s^2 + 2cs + c^2 + d^2}

    I reckon the factors are complex but I too tired to see the "obvious" factorisation. Any help please?
    If that is correct ( even with the change) it is not going to factorise
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    You can simplify the denominator (if you just want to stick to reals - assuming they are real to begin with) to:

    (s+c)^{2}+d^2
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    (Original post by TenOfThem)
    If that is correct ( even with the change) it is not going to factorise
    Oh right, would have been nice if you would have expanded on why instead of leaving me hanging lol.


    Doesn't matter, I worked it out anyway:

    \frac{1}{s^2 + 2cs + c^2 + d^2} = \frac{1}{(s + c - di)(s + c + di)}

    Which looks like a factorisation, unless I've got the definition of factorisation wrong. :dontknow:
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    (Original post by BAD AT MATHS)
    Oh right, would have been nice if you would have expanded on why instead of leaving me hanging lol.


    Doesn't matter, I worked it out anyway:

    \frac{1}{s^2 + 2cs + c^2 + d^2} = \frac{1}{(s + c - di)(s + c + di)}

    Which looks like a factorisation, unless I've got the definition of factorisation wrong. :dontknow:
    *hangs head*
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    (Original post by Hasufel)
    You can simplify the denominator (if you just want to stick to reals - assuming they are real to begin with) to:

    (s+c)^{2}+d^2
    Yeah it's ok, I worked it out in the end, thanks.
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    (Original post by Hasufel)
    You can simplify the denominator (if you just want to stick to reals - assuming they are real to begin with) to:

    (s+c)^{2}+d^2
    Yeah - this was where I went

    Then did not think Difference of 2 Squares - *shamed*
 
 
 
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