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Binomial Coefficients

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Question

Resolve $\frac{x^{2} + 5 x + 7}{{(x - 3)}^{3}}$ into partial fraction

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Solution

Given :

$\frac{x^{2} + 5 x + 7}{{(x - 3)}^{3}} = \frac{A}{x - 3} + \frac{B}{{(x - 3)}^{2}} + \frac{C}{{(x - 3)}^{3}}$

$x^{2} + 5 x + 7 = A {(x - 3)}^{2} + B (x - 3) + C$

$x^{2} + 5 x + 7 = A (x^{2} + 9 - 6 x) + B (x - 3) + C$

$x^{2} + 5 x + 7 = A x^{2} + (B - 6 A) x + (9 A - 3 B + C)$

Comparing the coefficient of same powers of $x$ on both sides, we have

$A = 1$

$B - 6 A = 5$

$\Rightarrow B = 6 + 5 = 11$

$9 A - 3 B + C = 7$

$\Rightarrow 9 - 33 + C = 7$

$\Rightarrow C = 31$

Substituting the value of $A, B$ and $C$ in equation $(1)$ ,
We get

$\frac{x^{2} + 5 x + 7}{{(x - 3)}^{3}} = \frac{1}{x - 3} + \frac{11}{{(x - 3)}^{2}} + \frac{31}{{(x - 3)}^{3}}$

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