Question

x(x+1)(x+2)/(x+3)(x+4)(x+5) partial fractions
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Answer 1

Step-by-step explanation:

$\green{ \rightarrow\large{ \sf \: GIVEN} \leftarrow}$

$\: \sf \: \frac{x(x + 1)(x + 2)}{( x + 3)(x + 4)(x + 5)}$

$\green{ \rightarrow\large{ \sf \: TO \: \: FIND } \leftarrow}$

$\sf \: Partial \: \: fraction \: \: of \: this \: term$

$\green{ \rightarrow\large{ \sf \: SOLUTION} \leftarrow}$

$\: \sf \: let \: the \: partial \: fraction \: of \: \\ \sf\frac{x(x + 1)(x + 2)}{( x + 3)(x + 4)(x + 5)} = \frac{A}{(x + 3)} + \frac{B}{(x + 4)} + \frac{C}{(x + 5)} \\ \\ \sf \: \: \: first \: for \: finding \: the \: value \: of \: A \: put \: x = - 3 \: on \: left \: side \: \\ \sf \: but \: not \: in \: thplace \: of \: (x + 3) \\ \\ \implies\sf \: A = \frac{( - 3)( - 3 + 1)( - 3 + 2)}{( - 3 + 4)( - 3 + 5)} \\ \\ \implies \sf \: A = \frac{( - 3)( - 2)( - 1)}{(1)(2)} \\ \\ \implies \boxed{ \sf \: A = - 3}$

$\sf \: \large \: similarly \:$

$\sf \: for \: finding \: the \: value \: of \:B \: \: put \: x = - 4 \: \: on \: lest \: side \\ \sf \: but \: not \: put \: \: in \: (x + 4) \\ \\ \implies \sf \: B \: = \frac{( - 4)( - 4 + 1)( - 4 + 2)}{( - 4 + 3)( - 4 + 5)} \\ \\\implies \sf \: B = \frac{( - 4)( - 3)( - 2)}{( - 1)(1)} \\ \\ \implies \sf \: \boxed{ \sf \: B \: = 24}$

$\sf \: for \: finding \: the \: value \: of \: C \:put \: x \: = - 5 \: \\ \sf\: on \: left \: side \: \: but \: not \: in \: (x + 5) \\ \\\implies \sf \: C = \frac{( - 5)( - 5 + 1)( - 5 + 2)}{( - 5 + 3)( - 5 + 4)} \\ \\ \implies \sf C= \frac{( - 5)( - 4) (- 3)}{( - 2)( - 1)} \\ \\\implies \boxed{\sf C = - 30}$

$\sf \: Replace \: \: the \: \: value \: \: of \: \: A , B , C$

$\sf \: hence\: the \: partial \: fraction \: \: \\ \sf\frac{x(x + 1)(x + 2)}{( x + 3)(x + 4)(x + 5)} = - \frac{3}{(x + 3)} + \frac{24}{(x + 4)} - \frac{ 30}{(x + 5)}$