# Thread: Need some help with some algebra please.

1. ## 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

2. 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.

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?

3. 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

4. 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?

5. 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!

6. You're welcome. Have a good one!