Resolve partial fraction

1.	Resolve partial fraction
Answer» $\large\underline{\bold{Given \:Question - }}$ $\sf \: Resolve \: in \: to \: partial \: fraction : \: \dfrac{x - 1}{ {(x - 3)}^{2} }$ $\large\underline{\bold{Solution-}}$ $\bf \: Let \: \sf \: \dfrac{x - 1}{ {(x - 3)}^{2} } = \dfrac{a}{x - 3} + \dfrac{b}{ {(x - 3)}^{2} } - - (1)$ On TAKING LCM, we get $\rm :\longmapsto\:x - 1 = a(x - 3) + b - - (2)$ On substituting 'x = 3' in equation (2), we get $\rm :\longmapsto\:3 - 1 = b$ $\bf\implies \: b \: = \: 2 \: - - (3)$ On substituting 'x = 0' in equation (2), we get $\rm :\longmapsto\:0 - 1 = a {(0 - 3)} + b$ $\rm :\longmapsto\: - 1 = - 3a + 2$ $\rm :\longmapsto\: - 3a = - 3$ $\bf\implies \:a \: = \: 1 - - (4)$ On substituting the values of a and b, in equation (1), we get $\bf \: \dfrac{x - 1}{ {(x - 3)}^{2} } = \dfrac{ 1}{(x - 3)} + \dfrac{2}{ {(x - 3)}^{2} }$ Additional Information :- Partial Fraction :- Partial fraction decomposition is the breaking down of a RATIONAL expression into simplier parts. It is the opposite of adding rational EXPRESSIONS. When adding two rational expressions, there has to be a common denominator. $\begin{gathered}\boxed{\begin{array}{c\|c} \sf factor \: in \: denominator & \sf partial \: fraction \: decomposition \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \frac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$

Answer»

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Resolve \: in \: to \: partial \: fraction : \: \dfrac{x - 1}{ {(x - 3)}^{2} }$

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

$\bf \: Let \: \sf \: \dfrac{x - 1}{ {(x - 3)}^{2} } = \dfrac{a}{x - 3} + \dfrac{b}{ {(x - 3)}^{2} } - - (1)$

On TAKING LCM, we get

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

On substituting 'x = 3' in equation (2), we get

$\rm :\longmapsto\:3 - 1 = b$

$\bf\implies \: b \: = \: 2 \: - - (3)$

On substituting 'x = 0' in equation (2), we get

$\rm :\longmapsto\:0 - 1 = a {(0 - 3)} + b$

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

$\rm :\longmapsto\: - 3a = - 3$

$\bf\implies \:a \: = \: 1 - - (4)$

On substituting the values of a and b, in equation (1), we get

$\bf \: \dfrac{x - 1}{ {(x - 3)}^{2} } = \dfrac{ 1}{(x - 3)} + \dfrac{2}{ {(x - 3)}^{2} }$

Additional Information :-

Partial Fraction :-

Partial fraction decomposition is the breaking down of a RATIONAL expression into simplier parts. It is the opposite of adding rational EXPRESSIONS. When adding two rational expressions, there has to be a common denominator.

$\begin{gathered}\boxed{\begin{array}{c|c} \sf factor \: in \: denominator & \sf partial \: fraction \: decomposition \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \frac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$