Question

# Resolve the given expression $\dfrac{{{x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}{\text{ }}$ into partial fractions.

Hint- Here, we will be comparing the degrees of the numerator and denominator to get the given function in desired form of partial fractions.
Let the given function of $x$ be ${\text{F}} = \dfrac{{{x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}$
For converting the given function F into partial fractions, we can write
${\text{F}} = \dfrac{{{x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{\left[ {\left( {x - 1} \right) - \left( {x - 2} \right)} \right]{x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{\left( {x - 1} \right){x^4} - \left( {x - 2} \right){x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{\left( {x - 1} \right){x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} - \dfrac{{\left( {x - 2} \right){x^4}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \\ \Rightarrow {\text{F}} = \dfrac{{{x^4}}}{{\left( {x - 2} \right)}} - \dfrac{{{x^4}}}{{\left( {x - 1} \right)}} \\$
The above expression of the function F is not in a form of partial fractions as it can be further reduced.
Therefore, the function can be written as
${\text{ F}} = \dfrac{{x \times {x^3}}}{{\left( {x - 2} \right)}} - \dfrac{{x \times {x^3}}}{{\left( {x - 1} \right)}} = \dfrac{{\left( {x - 2 + 2} \right) \times {x^3}}}{{\left( {x - 2} \right)}} - \dfrac{{\left( {x - 1 + 1} \right) \times {x^3}}}{{\left( {x - 1} \right)}} \\ {\text{F}} = \dfrac{{\left( {x - 2} \right){x^3} + 2{x^3}}}{{\left( {x - 2} \right)}} - \dfrac{{\left( {x - 1} \right){x^3} + {x^3}}}{{\left( {x - 1} \right)}} = \dfrac{{\left( {x - 2} \right){x^3}}}{{\left( {x - 2} \right)}} + \dfrac{{2{x^3}}}{{\left( {x - 2} \right)}} - \dfrac{{\left( {x - 1} \right){x^3}}}{{\left( {x - 1} \right)}} - \dfrac{{{x^3}}}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = {x^3} + \dfrac{{2{x^3}}}{{\left( {x - 2} \right)}} - {x^3} - \dfrac{{{x^3}}}{{\left( {x - 1} \right)}} = \dfrac{{2{x^3}}}{{\left( {x - 2} \right)}} - \dfrac{{{x^3}}}{{\left( {x - 1} \right)}} \\$
The above expression of the function F is not in a form of partial fractions as it can be further reduced.
Therefore, the function can be written as
${\text{F}} = \dfrac{{2{x^2} \times x}}{{\left( {x - 2} \right)}} - \dfrac{{x \times {x^2}}}{{\left( {x - 1} \right)}} = \dfrac{{2{x^2}\left( {x - 2 + 2} \right)}}{{\left( {x - 2} \right)}} - \dfrac{{\left( {x - 1 + 1} \right){x^2}}}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = \dfrac{{2{x^2}\left( {x - 2} \right) + 4{x^2}}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{\left( {x - 1} \right){x^2} + {x^2}}}{{\left( {x - 1} \right)}}} \right] = \dfrac{{2{x^2}\left( {x - 2} \right)}}{{\left( {x - 2} \right)}} + \dfrac{{4{x^2}}}{{\left( {x - 2} \right)}} - \dfrac{{\left( {x - 1} \right){x^2}}}{{\left( {x - 1} \right)}} - \dfrac{{{x^2}}}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = 2{x^2} + \dfrac{{4{x^2}}}{{\left( {x - 2} \right)}} - {x^2} - \dfrac{{{x^2}}}{{\left( {x - 1} \right)}} = {x^2} + \dfrac{{4{x^2}}}{{\left( {x - 2} \right)}} - \dfrac{{{x^2}}}{{\left( {x - 1} \right)}} \\$
The above expression of the function F is not in a form of partial fractions as it can be further reduced.
Therefore, the function can be written as
${\text{F}} = {x^2} + \dfrac{{4x \times x}}{{\left( {x - 2} \right)}} - \dfrac{{x \times x}}{{\left( {x - 1} \right)}} = {x^2} + \dfrac{{4x\left( {x - 2 + 2} \right)}}{{\left( {x - 2} \right)}} - \dfrac{{x\left( {x - 1 + 1} \right)}}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = {x^2} + \dfrac{{4x\left( {x - 2} \right) + 8x}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{x\left( {x - 1} \right) + x}}{{\left( {x - 1} \right)}}} \right] = {x^2} + \dfrac{{4x\left( {x - 2} \right)}}{{\left( {x - 2} \right)}} + \dfrac{{8x}}{{\left( {x - 2} \right)}} - \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)}} - \dfrac{x}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = {x^2} + 4x + \dfrac{{8x}}{{\left( {x - 2} \right)}} - x - \dfrac{x}{{\left( {x - 1} \right)}} = {x^2} + 3x + \dfrac{{8x}}{{\left( {x - 2} \right)}} - \dfrac{x}{{\left( {x - 1} \right)}} \\$
The above expression of the function F is not in a form of partial fractions as it can be further reduced.
Therefore, the function can be written as
${\text{ F}} = {x^2} + 3x + \dfrac{{8\left( {x - 2 + 2} \right)}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{x - 1 + 1}}{{\left( {x - 1} \right)}}} \right] \\ \Rightarrow {\text{F}} = {x^2} + 3x + \dfrac{{8\left( {x - 2} \right) + 16}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{x - 1 + 1}}{{\left( {x - 1} \right)}}} \right] = {x^2} + 3x + \dfrac{{8\left( {x - 2} \right)}}{{\left( {x - 2} \right)}} + \dfrac{{16}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{x - 1}}{{\left( {x - 1} \right)}} + \dfrac{1}{{\left( {x - 1} \right)}}} \right] \\ \Rightarrow {\text{F}} = {x^2} + 3x + \dfrac{{8\left( {x - 2} \right)}}{{\left( {x - 2} \right)}} + \dfrac{{16}}{{\left( {x - 2} \right)}} - \left[ {\dfrac{{x - 1}}{{\left( {x - 1} \right)}} + \dfrac{1}{{\left( {x - 1} \right)}}} \right] = {x^2} + 3x + 8 + \dfrac{{16}}{{\left( {x - 2} \right)}} - 1 - \dfrac{1}{{\left( {x - 1} \right)}} \\ \Rightarrow {\text{F}} = {x^2} + 3x + 7 + \dfrac{{16}}{{\left( {x - 2} \right)}} - \dfrac{1}{{\left( {x - 1} \right)}} \\$
In the above expression, the given function F is represented in the form of partial fractions.

Note- In these types of problems, we have to ensure that the final representation of the function is in terms of partial fractions by simply verifying that the degree of the numerator is greater than that of the denominator in the fractional terms involved in the final representation.