Question

How do you write the partial fraction decomposition of the rational expression $\dfrac{{5{x^3} - 9x + 6}}{{{x^3} - {x^2}}}$

Answer 1

Hint: As we know that when any fraction is split into a sum or difference of two or more fractions then it is called partial fractions. Also any number that can be represented in the form of $\left( {\dfrac{p}{q}} \right)$, where p and q are integers and $q \ne 0$, then it is called a rational number. First we will take the numerator and denominator and then split it into factors form and then simplify.

Complete step by step solution:
According to the question, we have, $\dfrac{{5{x^3} - 9x + 6}}{{{x^3} - {x^2}}}$.
The denominator ${x^3} - {x^2}$ can be written as the form of ${x^2}\left( {x - 1} \right)$.
Similarly we can break the numerator as $5{x^3} - 9x + 6 = 5{x^3} - 5{x^2} + 5{x^2} - 9x + 6$ , Also we can write the numerator as $5\left( {{x^3} - {x^2}} \right) + 5{x^2} - 9x + 6$.
By putting these simplifications in the fraction, we get,
$ \Rightarrow \dfrac{{5\left( {{x^3} - {x^2}} \right) + 5{x^2} - 9x + 6}}{{{x^3} - {x^2}}}$
Now, on simplifying further, we get,
$ \Rightarrow 5 + \dfrac{{5{x^2} - 9x + 6}}{{{x^3} - {x^2}}}$
Now, on putting in the factored form of the denominator, we get,
$ \Rightarrow 5 + \dfrac{{5{x^2} - 9x + 6}}{{{x^2}\left( {x - 1} \right)}}$
On separating denominators, we get,
$ \Rightarrow 5 + \dfrac{{5{x^2}}}{{{x^2}\left( {x - 1} \right)}} + \dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow 5 + \dfrac{5}{{\left( {x - 1} \right)}} + \dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}}$
Now, we only need to simplify and express $\dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}}$ in partial fraction form.
Now let $\dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}} = \dfrac{A}{{{x^2}}} + \dfrac{B}{x} + \dfrac{C}{{x - 1}}$ .
Now, by taking L.C.M of the right hand side of the equation and then comparing the numerator of both sides of the equation, we can get the values of constants A,B and C.
$ \Rightarrow \dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}} = \dfrac{{A\left( {x - 1} \right) + Bx\left( {x - 1} \right) + C{x^2}}}{{{x^2}\left( {x - 1} \right)}}$
Opening brackets and simplifying, we get,
$ \Rightarrow \dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}} = \dfrac{{Ax - A + B{x^2} - Bx + C{x^2}}}{{{x^2}\left( {x - 1} \right)}}$
Grouping like terms together,
$ \Rightarrow \dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}} = \dfrac{{\left( {B + C} \right){x^2} + \left( {A - B} \right)x - A}}{{{x^2}\left( {x - 1} \right)}}$
Now, comparing both sides of the equation,
$B + C = 0$, $A - B = - 9$ and $ - A = 6$
So, we get, $A = - 6$.
Putting value of A in $A - B = - 9$, we get,
$ \Rightarrow - 6 - B = - 9$
$ \Rightarrow B = 3$
Putting value of B in $B + C = 0$, we get,
$ \Rightarrow C = - B$
$ \Rightarrow C = - 3$
So, we get,
$\dfrac{{ - 9x + 6}}{{{x^2}\left( {x - 1} \right)}} = \dfrac{{ - 6}}{{{x^2}}} + \dfrac{3}{x} + \dfrac{{ - 3}}{{x - 1}}$
So, $\dfrac{{5{x^3} - 9x + 6}}{{{x^3} - {x^2}}} = 5 + \dfrac{5}{{x - 1}} - \dfrac{6}{{{x^2}}} + \dfrac{3}{x} - \dfrac{3}{{x - 1}}$
Therefore, we get $\dfrac{{5{x^3} - 9x + 6}}{{{x^3} - {x^2}}}$ as $\left( {5 + \dfrac{2}{{x - 1}} - \dfrac{6}{{{x^2}}} + \dfrac{3}{x}} \right)$.
So, the correct answer is “$\left( {5 + \dfrac{2}{{x - 1}} - \dfrac{6}{{{x^2}}} + \dfrac{3}{x}} \right)$.”.

Note: We should note that partial fractions can be done only 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. One must be careful while doing calculations.