Question

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

Answer 1

Hint: In this question, we have to express the given algebraic term in terms of the partial fractions. The partial fraction is the method where we convert the denominator into the product of the two polynomials and then split it in terms of the addition. So, we will apply the partial fractions method to solve the problem. We will first make the check whether the given fraction is improper or proper. Since, it is improper, so we will change it into a proper fraction using the long-division method. After that, we will change the denominator into the product of two polynomials and then we will split the numerator into the factors of constant say A and B. After that, we will multiply both sides of the equation by the product of two polynomials. Then, we will calculate the value of constants A and B, which gives the required solution to the problem.

Complete step by step answer:
According to the problem, we have to express an algebraic term into partial fractions.
So, we will simply use the partial fraction method to solve the problem
The algebraic term given to us is $\dfrac{{{x}^{2}}-3x}{\left( x-1 \right)\left( x+2 \right)}$ --------- (1)
Now, we will first open the brackets in the denominator using the distributive property in equation (1), we get
$\Rightarrow \dfrac{{{x}^{2}}-3x}{x\left( x+2 \right)-1\left( x+2 \right)}$
Now, we will again apply the distributive property in the above equation, we get
$\Rightarrow \dfrac{{{x}^{2}}-3x}{{{x}^{2}}+2x-x-2}$
On further solving, we get
 $\Rightarrow \dfrac{{{x}^{2}}-3x}{{{x}^{2}}+x-2}$ -------- (2)
Since, equation (2) is an improper fraction because the power in the denominator is equal to the power in the denominator, thus we will change it into proper fraction using the long-division method, we get quotient is equal to 1, and the remainder is equal to $-4x+2$ , therefore, we get
$\Rightarrow 1+\dfrac{-4x+2}{\left( x-1 \right)\left( x+2 \right)}$ ------ (3)
Now, we will solve first $\dfrac{-4x+2}{\left( x-1 \right)\left( x+2 \right)}$ ------- (4)
Now, let the numerator of equation (4) equals the sum of two factors, say A and B that is
$-4x+2=A+B$ --------- (5)
Since the factors are linear then the numerators of the partial fractions will be constants say A and B as seen in equation (4), therefore we get

\[\Rightarrow \dfrac{-4x+2}{\left( x-1 \right)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\] --------- (6)
Now, we will take the LCM on the right-hand side of the above equation, we get
\[\Rightarrow \dfrac{-4x+2}{\left( x-1 \right)(x+2)}=\dfrac{A(x+2)+B\left( x-1 \right)}{\left( x-1 \right)(x+2)}\]
On further simplification, we get
\[\Rightarrow -4x+2=A(x+2)+B\left( x-1 \right)\]
Now, we apply the distributive property $a(b-c)=ab-ac$ in the above equation, we get
\[\Rightarrow -4x+2=Ax+2A+Bx-B\]
Therefore, we get
\[\Rightarrow -4x+2=Ax+Bx+2A-B\]
Now, we will take common x from the above equation, we get
\[\Rightarrow -4x+2=x\left( A+B \right)+2A-B\]
On comparing the left-hand side and the right-hand side of the above equation, we get
$\Rightarrow -4=A+B$ --------- (7) and
$\Rightarrow 2=2A-B$ ---------- (8)
Now, we first solve the equation (7), by subtracting A on both sides in the equation (7), we get
$\Rightarrow -4-A=A+B-A$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow -4-A=B$ ------- (9)
Now, we will substitute the value of (9) in equation (8), we get
$\Rightarrow 2=2A-\left( -4-A \right)$
On further solving, we get
$\Rightarrow 2=2A+4+A$
Therefore, we get
$\Rightarrow 2=3A+4$
Now, we will subtract 4 on both sides in the above equation, we get
$\Rightarrow 2-4=3A+4-4$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow -2=3A$
Now, we will divide 3 on both sides in the above equation, we get
$\Rightarrow \dfrac{-2}{3}=\dfrac{3}{3}A$
Therefore, we get
$\Rightarrow \dfrac{-2}{3}=A$ ------- (10)
Now, we will substitute the value of equation (10) in equation (9), we get
$\Rightarrow -4-\left( \dfrac{-2}{3} \right)=B$
On further simplification, we get
$\Rightarrow -4+\dfrac{2}{3}=B$
Now, taking the LCM in the denominator in the above equation, we get
$\Rightarrow \dfrac{-12+2}{3}=B$
On further solving, we get
$\Rightarrow \dfrac{-12+2}{3}=B$
Therefore, we get
$\Rightarrow \dfrac{-10}{3}=B$ -------- (11)
Thus, substituting the value of equation (10) and (11) in equation (6), we get
\[\Rightarrow \dfrac{-4x+2}{\left( x-1 \right)(x+2)}=\dfrac{\dfrac{-2}{3}}{x-1}+\dfrac{\dfrac{-10}{3}}{x+2}\]
On further simplification, we get
\[\Rightarrow \dfrac{-4x+2}{\left( x-1 \right)(x+2)}=\dfrac{-2}{3\left( x-1 \right)}-\dfrac{10}{3\left( x+2 \right)}\]
Now, we will substitute the above equation in equation (4), we get
$\Rightarrow 1-\dfrac{2}{3\left( x-1 \right)}-\dfrac{10}{3\left( x+2 \right)}$

Therefore, for the equation $\dfrac{{{x}^{2}}-3x}{\left( x-1 \right)\left( x+2 \right)}$ , its partial fraction is $1-\dfrac{2}{3\left( x-1 \right)}-\dfrac{10}{3\left( x+2 \right)}$

Note: While solving this problem, mention all the formulas and the method you are using to avoid mathematical error. Always first check whether the fraction is proper or not, then only proceed further in finding the partial fractions.