Question

How do you express $\dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}$ in partial fractions?

Answer 1

Hint: This problem deals with reducing the given complex fraction into partial fractions. In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction is an operation that consists of expressing the fraction as a sum of a polynomial and one or several fractions with a simpler denominator.

Complete step-by-step answer:
Given a complex fraction which is as given below:
\[ \Rightarrow \dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\]
This is a simple fraction, it can be simply split into simpler fractions.
Now consider the given fraction into parts of fractions, as given below:
\[ \Rightarrow \dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} = \dfrac{A}{{\left( {x + 2} \right)}} + \dfrac{B}{{\left( {x - 1} \right)}}\]
Now simplify the right hand side of the above equation, as shown below:
\[ \Rightarrow \dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} = \dfrac{{A\left( {x - 1} \right) + B\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\]
Now in the above equation, the denominators are equal, so equating the numerators of the above equation.
Here equating the numerators of the above equation, as shown below:
\[ \Rightarrow 3x = A\left( {x - 1} \right) + B\left( {x + 2} \right)\]
\[ \Rightarrow 3x = \left( {Ax - A} \right) + \left( {Bx + 2B} \right)\]
Now opening the brackets and grouping the like terms and constants together, as shown below:
\[ \Rightarrow 3x = Ax - A + Bx + 2B\]
\[ \Rightarrow 3x = Ax + Bx + 2B - A\]
Now taking $x$ from the first two terms, as shown below:
\[ \Rightarrow 3x = \left( {A + B} \right)x + 2B - A\]
So equating the $x$ terms and the constants as shown below:
Here $A + B = 3$ and $2B - A = 0$
From here $A = 2B$, substituting this in the equation $A + B = 3$, so $B = 1$ and $A = 2$.
Substituting the values of $A$ and $B$, in the partial fractions, as shown:
\[ \Rightarrow \dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} = \dfrac{A}{{\left( {x + 2} \right)}} + \dfrac{B}{{\left( {x - 1} \right)}}\]

\[\therefore \dfrac{{3x}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} = \dfrac{2}{{\left( {x + 2} \right)}} + \dfrac{1}{{\left( {x - 1} \right)}}\]

Note:
Please note that partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. Partial fractions are a way of breaking apart fractions with polynomials in them. The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition.