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Partial fractions from a single fraction? Watch

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    I've just started to study this and I'm stuck as hell on a question in my text book. I cannot for the life of me see how they get the answer they do. Can someone please explain the procedure?

    The algebraic fraction is:

    3x+9 / x2 + 8x + 12

    When I work this out I get partial fractions of:

    0.75 / x+2 + -0.25 / x+6 (the numerators are fractions themselves)

    The answer given in the book is:

    1/4 (3/x+2 + 9/x+6)


    Oh... and can someone point me towards a tutorial for writing mathematical formula, fractions etc. correctly please? I know there's one on here somewhere but I can't find it now.
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    (Original post by Darwinion)
    I've just started to study this and I'm stuck as hell on a question in my text book. I cannot for the life of me see how they get the answer they do. Can someone please explain the procedure?

    The algebraic fraction is:

    3x+9 / x2 + 8x + 12

    When I work this out I get partial fractions of:

    0.75 / x+2 + -0.25 / x+6 (the numerators are fractions themselves)

    The answer given in the book is:

    1/4 (3/x+2 + 9/x+6)


    Oh... and can someone point me towards a tutorial for writing mathematical formula, fractions etc. correctly please? I know there's one on here somewhere but I can't find it now.
    Well you've got the first fraction right anyway, so I assume you're making a mistake at some point when it comes to finding the numerator of the other fraction.

    \displaystyle \frac{3x+9}{x^2+8x+12}=\frac{3x+  9}{(x+2)(x+6)}= \frac{A}{x+2}+ \frac{B}{x+6}=\frac{Ax+6A+Bx+2B}  {(x+2)(x+6)}

    Means Ax+Bx=3x and 6A+2B=9

    And since you know that A=\frac{3}{4} then you can find B since you know that B=3-A from eq 1


    http://www.thestudentroom.co.uk/wiki/LaTex
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    Thanks for that! I dunno what I was doing for calculating B. The way you laid it out quite clearly shows that B=2\frac{1}{4}

    Yay... and I got the LaTex right That's gonna need some practice though.
 
 
 
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