2
$\begingroup$

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

$${x-(\alpha+\beta+\gamma-a-b)}+{(a-\alpha)(a-\beta)(a-\gamma)\over(x-a)(a-b)}+{(b-\alpha)(b-\beta)(b-\gamma)\over(x-b)(b-a)}$$

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

$${(a-\alpha)(a-\beta)(a-\gamma)\over(x-a)(a-b)}+{(b-\alpha)(b-\beta)(b-\gamma)\over(x-b)(b-a)}$$

But this is not the right answer.

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

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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

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.