Question

if x^4/(x -a) (x-b)(x-c) = P(x) + A/(x-a ) + B/ (x-b) + C/ (x-c) then P(x) is equal to

Answer 1

Given:

$\frac{x^4}{(x -a)(x-b)(x-c) }= P(x) + \frac{A}{(x-a )} +\frac{B}{(x-b)} + \frac{C}{(x-c)}$

To find:

P(x)=?

Solution:

$\frac{x^4}{(x -a)(x-b)(x-c) }= P(x) + \frac{A}{(x-a )} +\frac{B}{(x-b)} + \frac{C}{(x-c)}$

$\to P(x) = -(\frac{A}{(x-a )} +\frac{B}{(x-b)} + \frac{C}{(x-c)})+\frac{x^4}{(x -a)(x-b)(x-c) }$

             $= -(\frac{A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x -a)(x-b)(x-c)})+\frac{x^4}{(x -a)(x-b)(x-c) }$

             $= \frac{-A(x-b)(x-c)-B(x-a)(x-c)-C(x-a)(x-b)}{(x -a)(x-b)(x-c)})+\frac{x^4}{(x -a)(x-b)(x-c) }\\\\\\= \frac{-A(x-b)(x-c)-B(x-a)(x-c)-C(x-a)(x-b) + x^4}{(x -a)(x-b)(x-c)})\\\\\\= \frac{-A(x^2-xc-bx+bc)-B(x^2-xc-cx+ac)-C(x^2-bx-ax+ab)+x^4}{(x -a)(x-b)(x-c)})$

             $= \frac{-A(x^2-xc-bx+bc)-B(x^2-2xc+ac)-C(x^2-bx-ax+ab)+x^4}{(x -a)(x-b)(x-c)})\\\\\\= \frac{-Ax^2+Axc+Abx-Abc-Bx^2+2Bxc-Bac-Cx^2+bCx+aCx-abC+x^4}{(x -a)(x-b)(x-c)})$

The P(x) value is :

$\bold{= \frac{-Ax^2+Axc+Abx-Abc-Bx^2+2Bxc-Bac-Cx^2+bCx+aCx-abC+x^4}{(x -a)(x-b)(x-c)})\\}$