Question

Write the following fraction into partial fraction

$\sf \dfrac{2x + 3}{(x + 1)(x - 3)}$
​

Answer 1

Answer:

x^2 -3x + x - 3

Step-by-step explanation:

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Answer 2

$\underline{ \underline{ \sf \red{Question}}} \red{: }$

$\sf \dfrac{2x + 3}{(x + 1)(x - 3)}$

$\underline{ \underline{ \sf \red{Answer}}} \red{: }$

$\rm \red{Let}$

$\boxed{ \boxed{\rm \dfrac{2x + 3}{(x + 1)(x - 3)} = \dfrac{A }{(x + 1)} + \dfrac{B}{(x - 3)} }} - - - (1)$

$\rm \implies \dfrac{2x + 3}{ \cancel{(x + 1)(x - 3)}} = \dfrac{A (x - 3) + B(x + 1)}{ \cancel{(x + 1)(x - 3)} }$

$\rm{ \implies \: 2x + 3 =A x -3A + Bx + B }$

$\implies \rm{2x + 3 = x(A + B) + ( - 3A + B)}$

$\rm \red{Equating \: the \: corresponding \: coefficients : }$

$\rm \red{case \: i : }$

$\rm{A + B = 2}$

$\boxed{ \boxed{B = 2 - A }} - - - (2)$

$\rm \red{case \: i i: }$

$\rm{ - 3A + B = 3}$

$\rm{substituting \: \: B = 2-A }$

$\rm - 3A + 2 - A = 3$

$\rm - 4A = 3 - 2$

$\boxed{ \boxed{\rm A = \dfrac{ - 1}{4} }} - - - (3)$

$\rm \red{ From \: Equation \: (2) : }$

$\rm{substituting \: \: A \: = \dfrac{ - 1}{4} }$

$\rm B = 2 - \bigg( - \dfrac{1}{4} \bigg)$

$\rm B = 2 + \dfrac{1}{4}$

$\boxed{ \boxed{ \rm B = \dfrac{9}{4} }} - - - (4)$

$\rm \red{ From \: Equation \: (1) : }$

$\rm \dfrac{2x + 3}{(x + 1)(x - 3)} = \dfrac{ - \frac{ 1}{4} }{(x + 1)} + \dfrac{ \frac{9}{4} }{(x - 3)}$

$\boxed{ \boxed{ \rm \dfrac{2x + 3}{(x + 1)(x - 3)} = \dfrac{ - 1}{4(x + 1)} + \dfrac{ 9}{4(x - 3)} }}$