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Math Help - Need some help with some algebra please.

  1. #1
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    Need some help with some algebra please.

    Have the following question from a past exam paper, its a partial fraction question 7, now i have always found partial fractions quite simple however im finding it hard to simplify and reduce the equations down!

    Question 7
    i) i have the denominator in the first partial fraction = to (x+5) and the numerator on the final partial fraction= to C

    From this i have used the general way to obtain the A B and C
    A(x+5)(2x^2-x+3)+B(2x^2-x+3)+(cx+d)(x+5)^2

    Equating i got the following equations:
    2A+C=0 (equating x^3)
    9A+2B+10C+D=25 (equating x^2)
    -2A-B+10D+25C=71 (equating x)
    15A+25D=-96

    From here im having trouble not sure because of not having the correct equations or just further working out.


    Also having trouble in part B havent got a clue where to start if im honest.

    Many thanks in advance.

    Might need to attach it
    Attached Files Attached Files
    Last edited by mr fantastic; January 13th 2011 at 01:33 PM. Reason: Merged posts.
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  2. #2
    A Plied Mathematician
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    It looks like you changed your guess for the final numerator from C to Cx+D. Cx+D is correct, since the discriminant of the denominator is

    b^{2}-4ac=1-4(2)(3)=1-24=-23<0.

    Thus, the denominator does not factor over the reals, and you need the Cx+D up top.

    15A+25D=-96
    I think it should be

    15A+25D+3B=-96. Other than that, I think you've set up your equations correctly.

    Part (b) is way beyond me as well. It's an entirely different problem!?! Don't know why it's a part (b) to a partial fractions problem. I think an algebraist such as NonCommAlg or tonio or Swlabr would know what that's all about. Any chance you guys could weigh in on this?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    It looks like you changed your guess for the final numerator from C to Cx+D. Cx+D is correct, since the discriminant of the denominator is

    b^{2}-4ac=1-4(2)(3)=1-24=-23<0.

    Thus, the denominator does not factor over the reals, and you need the Cx+D up top.



    I think it should be

    15A+25D+3B=-96. Other than that, I think you've set up your equations correctly.

    Part (b) is way beyond me as well. It's an entirely different problem!?! Don't know why it's a part (b) to a partial fractions problem. I think an algebraist such as NonCommAlg or tonio or Swlabr would know what that's all about. Any chance you guys could weigh in on this?
    Yeah i ment to put that down for c sorry for that.
    Where im having problems is at the elimination stage, i keep coming out with values that dont work when plugging it into the original equations.

    say call
    2A+C=0 (equating x^3) equation 1
    9A+2B+10C+D=25 (equating x^2) equation 2
    -2A-B+10D+25C=71 (equating x) equation 3
    15A+25D+3b=-96 equation 4

    Now ive tried getting rid of B by multiplying (equation 3) by 3 then adding it to (equation 4) to form equation 5:
    9A+55D+75C=117

    From this i wanted to obtain another equation with A,C,D in therefore I subtracted (equation 2) from 2x(equation 3) in order to obtain equation 6:
    5A+55D+75C=117

    From this it gets very messy trying to subtract equation 1 then solve simultaneously, due to this I dont think im on the right lines with what equations ive selected,
    any help would be most appreciated.
    An y
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  4. #4
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    Sorry, I kind of lost track of this thread. I would treat this as an augmented system of equations. Use Gaussian elimination with back substitution thus:

    \left[\begin{matrix}2 &0 &1 &0\\<br />
9 &2 &10 &1\\<br />
-2 &-1 &25 &10\\<br />
15 &3 &0 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]_{\frac{1}{2}\,R_{1}\to R_{1}}\to <br />
\left[\begin{matrix}<br />
1 &0 &1/2 &0\\<br />
9 &2 &10 &1\\<br />
-2 &-1 &25 &10\\<br />
15 &3 &0 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]_{\begin{matrix}-9R_{1}+R_{2}\to R_{2}\\ 2R_{1}+R_{3}\to R_{3}\\-15R_{1}+R_{4}\to R_{4}\end{matrix}}\to
    <br />
\left[\begin{matrix}<br />
1 &0 &1/2 &0\\<br />
0 &2 &5.5 &1\\<br />
0 &-1 &26 &10\\<br />
0 &3 &-15/2 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]\dots

    Can you continue?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Sorry, I kind of lost track of this thread. I would treat this as an augmented system of equations. Use Gaussian elimination with back substitution thus:

    \left[\begin{matrix}2 &0 &1 &0\\<br />
9 &2 &10 &1\\<br />
-2 &-1 &25 &10\\<br />
15 &3 &0 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]_{\frac{1}{2}\,R_{1}\to R_{1}}\to <br />
\left[\begin{matrix}<br />
1 &0 &1/2 &0\\<br />
9 &2 &10 &1\\<br />
-2 &-1 &25 &10\\<br />
15 &3 &0 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]_{\begin{matrix}-9R_{1}+R_{2}\to R_{2}\\ 2R_{1}+R_{3}\to R_{3}\\-15R_{1}+R_{4}\to R_{4}\end{matrix}}\to
    <br />
\left[\begin{matrix}<br />
1 &0 &1/2 &0\\<br />
0 &2 &5.5 &1\\<br />
0 &-1 &26 &10\\<br />
0 &3 &-15/2 &25<br />
\end{matrix}\right \left|\begin{matrix}0\\25\\71\\-96\end{matrix}\right]\dots

    Can you continue?
    Just got around to doing it and got there, substitued back in and everything works out, so many thanks!
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  6. #6
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    You're welcome. Have a good one!
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