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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 $A,\,B\,{\rm{and}}\,C$?

DoubtExpert
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  • Related: https://math.stackexchange.com/questions/20963/integration-by-partial-fractions-how-and-why-does-it-work, https://math.stackexchange.com/questions/1313454/is-this-a-valid-partial-fraction-decomposition – Hans Lundmark Oct 08 '17 at 13:03

2 Answers2

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$${{f(x)} \over {(x - a)(x - b)(x - c)}} = {A \over {x - a}} + {B \over {x - b}} + {C \over {x - c}}$$

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

$$f(a)=A(a-b)(a-c)\implies A= ???$$ $$f(b)=B(b-a)(b-c)\implies B= ???$$ $$f(c)=C(c-a)(c-b)\implies C= ???$$

Claude Leibovici
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3

$${{f(x)} \over {(x - a)(x - b)(x - c)}} = {A \over {x - a}} + {B \over {x - b}} + {C \over {x - c}}$$

$${f(x)} = {A (x-b)(x-c)} + {B (x-a)(x-c)} + {C (x-a)(x-b)}$$

Now you can put $x=a,b,c$ to get $A,B,C$ respectively.

Jaideep Khare
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