The question is to develop $$(x-\alpha)(x-\beta)(x-\gamma)\over(x-a)(x-b)$$ into partial fractions.

Someone challenged me to solve this question and said the answer is


I tried and what I've done is

$$(x-\alpha)(x-\beta)(x-\gamma) = A(x-b)+B(x-a)$$

If $x=a$ then $$ A= \frac{(a-\alpha)(a-\beta)(a-\gamma)}{(a-b)}$$

If $x=b$ then $$B= \frac{(b-\alpha)(b-\beta)(b-\gamma)}{(b-a)}$$

Then the answer should be


But this is not the right answer.

I dont know what's going wrong, I could use some help.

  • $\begingroup$ What would you like to do / prove with the expression you wrote in your post? It is unclear to me... $\endgroup$ – Avitus Dec 22 '14 at 11:36
  • $\begingroup$ yeah i have to prove but now i realize that the numerator > that of the denominator gonna solve it after division $\endgroup$ – Sundeep Dec 22 '14 at 11:39
  • $\begingroup$ I am sorry but, what do you want to solve? You never state any problem in your post: when you write "QUESTION" you just introduce a fraction. Do you want to find a partial fraction decomposition for your expression? $\endgroup$ – Avitus Dec 22 '14 at 11:43
  • 1
    $\begingroup$ yes i do want partial fraction decomposition for my expression $\endgroup$ – Sundeep Dec 22 '14 at 11:45
  • $\begingroup$ Thank you for clarifying! In addition to the answer below, please have a look at this page en.wikipedia.org/wiki/Partial_fraction_decomposition $\endgroup$ – Avitus Dec 22 '14 at 11:48

You have reached at $(x-\alpha)(x-\beta)(x-\gamma)$ = $A(x-b)+B(x-a)$

The left hand side is cubic whereas the right hand side is a linear polynomial.

As the degree of the numerator$(D_n)>$ that of the denominator$(D_d)$, in fact, $D_n-D_d=1$

using Partial Fraction Decomposition formula, $$\frac{(x-\alpha)(x-\beta)(x-\gamma)}{(x-a)(x-b)}=x^1+A+\dfrac B{x-a}+\dfrac C{x-b}$$

where $A,B,C$ are arbitrary constants.

  • $\begingroup$ that's very strange i didn't notice it thanks buddy $\endgroup$ – Sundeep Dec 22 '14 at 11:40
  • $\begingroup$ I didn't get results in this form after division can you help me out. $\endgroup$ – Sundeep Dec 22 '14 at 11:48
  • $\begingroup$ @sanddy1911, Set $x=a,b$ to find $C,B$ respectively. Then comparing the constants of both sides of the identity, $-\alpha\beta\gamma=Aab-aC-bB$ $\endgroup$ – lab bhattacharjee Dec 22 '14 at 11:51
  • $\begingroup$ that is not my problem look what i get as remainder $x\alpha\gamma + x\alpha\beta + x\beta\gamma – ax\alpha – ax\beta – ax\gamma + a^2x + abx – bx\alpha + b^2x – \alpha\beta\gamma + ab\alpha + ab\beta + ab\gamma – a^2b – ab^2 $ do you think with this i can get right answer, One good this is the quotient is match with answer $x - (\alpha + \beta + \gamma - a -b)$ $\endgroup$ – Sundeep Dec 22 '14 at 12:01
  • $\begingroup$ @sanddy1911, Why remainder? Do you know how to find arbitrary constants from Partial Fraction Decomposition? $\endgroup$ – lab bhattacharjee Dec 22 '14 at 12:19

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