Question

How do you express $\dfrac{1}{{{x}^{3}}-6{{x}^{2}}+9x}$ in partial fraction?

Answer 1

Hint: The given expression is to be expressed in partial fraction. We first have to factor the polynomial in the given expression. Then, we have to express each of the factors as the sum of the partial fraction of all the factors using variables. We then proceed to calculate the value of the variables used. Substituting the values back in the equation, we get the expression in the partial fraction form.

Complete step-by-step solution:
According to the given question, we have a fractional polynomial expression which we have to express in partial form.
We will begin with writing the given expression, we have,
$\dfrac{1}{{{x}^{3}}-6{{x}^{2}}+9x}$----(1)
\[\Rightarrow \dfrac{1}{x({{x}^{2}}-6x+9)}\]------(2)
We will first factorize the polynomial we have, we get,
 \[{{x}^{2}}-6x+9\]
\[\Rightarrow {{x}^{2}}-(3+3)x+3\times 3\]
\[\Rightarrow {{x}^{2}}-3x-3x+9\]
\[\Rightarrow x(x-3)-3(x-3)\]
Taking common terms out, we get,
\[\Rightarrow (x-3)(x-3)\]
We can rewrite the equation (2) as,
\[\dfrac{1}{x({{x}^{2}}-6x+9)}=\dfrac{1}{x{{(x-3)}^{2}}}\]-----(3)
Let us write the equation as the sum of partial fraction, we get,
\[\dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{A}{x}+\dfrac{B}{(x-3)}+\dfrac{C}{{{(x-3)}^{2}}}\]-----(4)
We have to find the value of the variable A, B and C, so we will have to solve the RHS in the equation (4), we will begin by taking the LCM, we get,
\[\Rightarrow \dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{A}{x}\times \dfrac{{{(x-3)}^{2}}}{{{(x-3)}^{2}}}+\dfrac{B}{(x-3)}\times \dfrac{x(x-3)}{x(x-3)}+\dfrac{C}{{{(x-3)}^{2}}}\times \dfrac{x}{x}\]
\[\Rightarrow \dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{A({{x}^{2}}-6x+9)+B({{x}^{2}}-3x)+Cx}{x{{(x-3)}^{2}}}\]
Multiplying the terms, we get,
\[\Rightarrow \dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{A{{x}^{2}}-6Ax+9A+B{{x}^{2}}-3Bx+Cx}{x{{(x-3)}^{2}}}\]
\[\Rightarrow \dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{(A+B){{x}^{2}}+(-6A-3B+C)x+9A}{x{{(x-3)}^{2}}}\]
We will now compare the coefficients of \[{{x}^{2}}\], \[x\] and constant, from LHS and RHS, we get,
\[A+B=0\]----(5)
\[-6A-3B+C=0\]----(6)
\[9A=1\]---(7)
We get the value of A from equation (7), we have,
\[\Rightarrow A=\dfrac{1}{9}\]
Substituting the value of A in equation (5), we get,
\[\dfrac{1}{9}+B=0\]
\[\Rightarrow B=-\dfrac{1}{9}\]
Now, we substitute the value of A and B in equation (6), we have
\[\Rightarrow -6\left( \dfrac{1}{9} \right)-3\left( -\dfrac{1}{9} \right)+C=0\]
\[\Rightarrow \dfrac{-6}{9}+\dfrac{3}{9}+C=0\]
\[\Rightarrow \dfrac{-3}{9}+C=0\]
\[\Rightarrow C=\dfrac{1}{3}\]
Now, we have the values of A, B and C, now we will substitute the values in the equation (4), we get,
\[\dfrac{1}{x{{(x-3)}^{2}}}=\dfrac{1}{9x}+\dfrac{-1}{9(x-3)}+\dfrac{1}{3{{(x-3)}^{2}}}\]
Therefore, we get,
\[\dfrac{1}{{{x}^{3}}-6{{x}^{2}}+9x}=\dfrac{1}{9x}+\dfrac{-1}{9(x-3)}+\dfrac{1}{3{{(x-3)}^{2}}}\]

Note: While expressing the given expression in partial fraction, care should be taken while writing the terms. Before starting writing the expression, make sure that the polynomial in the denominator is factorized first and then expressed as a sum of partial fraction of each of the factors.