Question

resolve partial fractions 1/(x+1)(x+2)(x+3)

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{(x + 1)(x + 2)(x + 3)}$

Let assume that

$\rm :\longmapsto\:\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{a}{x + 1} + \dfrac{b}{x + 2} + \dfrac{c}{x + 3}$

So, On taking LCM, we get

$\rm :\longmapsto\:1 = a(x + 2)(x + 3) + b(x + 1)(x + 3) + c(x + 1)(x + 2)$

On substituting x = - 1, we get

$\rm :\longmapsto\:1 = a( - 1+ 2)( - 1 + 3) +0 + 0$

$\rm :\longmapsto\:1 = a(1)(2)$

$\bf\implies \:a = \dfrac{1}{2}$

Now, On substituting x = - 2, we get

$\rm :\longmapsto\:1 = 0 + b( - 2+ 1)( - 2 + 3) + 0$

$\rm :\longmapsto\:1 = b( -1)(1)$

$\bf\implies \:b \: = \: - \: 1$

Now, On substituting x = - 3, we get

$\rm :\longmapsto\:1 = 0 + 0 + c( - 3 + 1)( - 3 + 2)$

$\rm :\longmapsto\:1 =c( - 2)( -1)$

$\bf\implies \:c = \dfrac{1}{2}$

So, on substituting the values of a, b and c, we get

$\rm :\longmapsto\:\dfrac{1}{(x + 1)(x + 2)(x + 3)} = \dfrac{1}{2(x + 1)} - \dfrac{1}{x + 2} + \dfrac{1}{2(x + 3)}$

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LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \: Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)} & \sf \dfrac{p}{ax + b} + \dfrac{q}{cx + d} \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} } & \sf \dfrac{p}{ax + b} + \dfrac{q}{ {(ax + b)}^{2} } \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2} + d)} & \sf \dfrac{p}{ax + b} + \dfrac{qx + r}{c {x}^{2} + d} \end{array}} \\ \end{gathered}$