I was going through the solution of a problem when I encountered this expression.
$${{f(x)} \over {(x - a)(x - b)(x - c)}}$$ Its given that $f(x)$ is a polynomial of degree $ < 3$.
This was simplified as follows -
By partial fractions, $${{f(x)} \over {(x - a)(x - b)(x - c)}} = {A \over {x - a}} + {B \over {x - b}} + {C \over {x - c}}$$ Then, $$A = {\left[ {{{f(x)} \over {(x - b)(x - c)}}} \right]_{x = a}} = {{f(a)} \over {(a - b)(a - c)}}$$ Similarly, $$B = {{f(b)} \over {(b - a)(b - c)}}\,\,{\rm{and}}\,\,C = {{f(c)} \over {(c - a)(c - b)}}$$
I think that the division algorithm for polynomials is involved here. But I could not understand how does the use of algorithm lead to substitution of $a,\,b\,{\rm{and}}\,c$ in place of $x$.
How does one get these expressions for $A,\,B\,{\rm{and}}\,C$?
$${{f(x)} \over {(x - a)(x - b)(x - c)}} = {A \over {x - a}} + {B \over {x - b}} + {C \over {x - c}}$$
Multiply by the denominator of lhs to get $$f(x)=A(x - b)(x - c)+B(x - a)(x - c)+C(x - a)(x - b)$$
Since this is true for all $x$, then
$$f(a)=A(a-b)(a-c)\implies A= ???$$ $$f(b)=B(b-a)(b-c)\implies B= ???$$ $$f(c)=C(c-a)(c-b)\implies C= ???$$