# Need some help with some algebra please.

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• Jan 13th 2011, 10:37 AM
breitling
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 (Angry)
• Jan 14th 2011, 01:11 AM
Ackbeet
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

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

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

Quote:

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?
• Jan 14th 2011, 04:59 AM
breitling
Quote:

Originally Posted by Ackbeet
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

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

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
• Jan 18th 2011, 05:28 AM
Ackbeet
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:

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

Can you continue?
• Jan 25th 2011, 04:11 AM
breitling
Quote:

Originally Posted by Ackbeet
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:

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

Can you continue?

Just got around to doing it and got there, substitued back in and everything works out, so many thanks!
• Jan 25th 2011, 04:13 AM
Ackbeet
You're welcome. Have a good one!