4
$\begingroup$

I'm having some trouble doing this partial fraction decomposition: $$\frac{1}{t^3-2t+1}$$ using Ruffini rule i get: $$\frac{1}{t^3-2t+1}= \frac{1}{(t-1)(t^2+t-1)}$$ i would like to decompose the previous result into partial fraction. I did in this way: $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{B}{t^2+t-1} \leftrightarrow$$ $$\leftrightarrow t^2A+t(A+B)+(-A-B)=1$$

comparing the coefficients i get the following system of equations:\begin{cases} A=0 \\ A+B=0 \\ -A-B=1 \end{cases} that are not true.. what am i doing wrong?

$\endgroup$
7
$\begingroup$

It should be $$\frac A{t-1}+\dfrac{Bt+C}{t^2+t-1}$$

See : Partial Fraction Decomposition

$\endgroup$
  • $\begingroup$ thanx... small gaps :) $\endgroup$ – Duccio Bertieri Jul 27 '15 at 10:42
1
$\begingroup$

Notice, In general $$\frac{1}{(ax+b)(px^2+qx+r)}=\frac{A}{ax+b}+\frac{Bx+C}{px^2+qx+r}$$ Now, factorizing the expression, we have $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{Bt+C}{t^2+t-1}$$ $$\implies \frac{1}{(t-1)(t^2+t-1)}=\frac{A(t^2+t-1)+(Bt+C)(t-1)}{(t-1)(t^2+t-1)}$$ $$\implies (A+B)t^2+(A-B+C)t-(A+C)=1$$ Now, comparing the corresponding coefficients of both the sides, we get $$\begin{cases} A+B=0\\ A-B+C=0\\ A+C=-1 \end{cases}$$ On solving the above three equations, we get $A=1, B=-1$ & $C=-2$ Hence, the required partial fractions are as follows $$\frac{1}{(t-1)(t^2+t-1)}=\frac{(1)}{t-1}+\frac{(-1)t+(-2)}{t^2+t-1}$$$$=\color{blue}{\frac{1}{t-1}-\frac{t+2}{t^2+t-1}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.