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

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

Explanation

Solution

To write the above fraction 3x(x+2)(x−1)\dfrac{3x}{\left( x+2 \right)\left( x-1 \right)} in terms of partial fractions we are going to equate this expression to A(x+2)+B(x−1)\dfrac{A}{\left( x+2 \right)}+\dfrac{B}{\left( x-1 \right)} and then will take the L.C.M of the denominator of this expression in A and B. After equating the two fractions, we will get the values of A and B. And hence will find the partial fraction form of the given fraction.

Complete step by step solution:
The expression given in the above problem is as follows:
3x(x+2)(x−1)\dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}………….(1)
Now, we are asked to convert the above fraction in terms of partial fractions for that first we should know the partial form of the above expression which we have shown below:
⇒A(x+2)+B(x−1)\Rightarrow \dfrac{A}{\left( x+2 \right)}+\dfrac{B}{\left( x-1 \right)}
Taking L.C.M of the terms written in the denominator and we get,
⇒A(x−1)+B(x+2)(x+2)(x−1) =x(A+B)+2B−A(x+2)(x−1)............(2) \begin{aligned} & \Rightarrow \dfrac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)} \\\ & =\dfrac{x\left( A+B \right)+2B-A}{\left( x+2 \right)\left( x-1 \right)}............(2) \\\ \end{aligned}
Equating (1) and (2) we get,
⇒3x(x+2)(x−1)=x(A+B)+2B−A(x+2)(x−1)\Rightarrow \dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}=\dfrac{x\left( A+B \right)+2B-A}{\left( x+2 \right)\left( x-1 \right)}
Now, denominator is same on both the sides so denominator will be cancelled out and we are left with:
⇒3x=x(A+B)+2B−A\Rightarrow 3x=x\left( A+B \right)+2B-A
Equating the coefficient of x on both the sides and we get,
⇒3=A+B\Rightarrow 3=A+B ……. (3)
And also equating the constant terms on both the sides and we get,
0=2B−A ⇒2B=A ........(4) \begin{aligned} & 0=2B-A \\\ & \Rightarrow 2B=A\text{ }........(4) \\\ \end{aligned}
Substituting the above value of “A” in eq. (3) we get,
⇒3=2B+B ⇒3=3B \begin{aligned} & \Rightarrow 3=2B+B \\\ & \Rightarrow 3=3B \\\ \end{aligned}
In the above equation, 3 will be cancelled on both the sides and we get,
1=B1=B
Substituting the above value of B in eq. (4) we get,
⇒A=2(1) ⇒A=2 \begin{aligned} & \Rightarrow A=2\left( 1 \right) \\\ & \Rightarrow A=2 \\\ \end{aligned}
Hence, we got the values of “A and B” as “2 and 1” respectively so substituting these values in the partial fraction form and we get,
⇒A(x+2)+B(x−1) =2(x+2)+1(x−1) \begin{aligned} & \Rightarrow \dfrac{A}{\left( x+2 \right)}+\dfrac{B}{\left( x-1 \right)} \\\ & =\dfrac{2}{\left( x+2 \right)}+\dfrac{1}{\left( x-1 \right)} \\\ \end{aligned}
Hence, the partial fraction form of the above expression is as follows:
2(x+2)+1(x−1)\dfrac{2}{\left( x+2 \right)}+\dfrac{1}{\left( x-1 \right)}

Note: You can check the partial fraction form which you are getting is correct or not by taking the L.C.M of the denominator of the final partial form which looks as follows:
⇒2(x+2)+1(x−1) ⇒2(x−1)+x+2(x+2)(x−1) ⇒2x−2+x+2(x+2)(x−1) ⇒3x(x+2)(x−1) \begin{aligned} & \Rightarrow \dfrac{2}{\left( x+2 \right)}+\dfrac{1}{\left( x-1 \right)} \\\ & \Rightarrow \dfrac{2\left( x-1 \right)+x+2}{\left( x+2 \right)\left( x-1 \right)} \\\ & \Rightarrow \dfrac{2x-2+x+2}{\left( x+2 \right)\left( x-1 \right)} \\\ & \Rightarrow \dfrac{3x}{\left( x+2 \right)\left( x-1 \right)} \\\ \end{aligned}
As you can see that we are getting the same fraction which is given in the above problem so the partial fraction form which we have obtained is correct.