Question

How do you express $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}$ in partial fractions?

Answer 1

Hint: We first try to describe the requirement and the process of finding the partial fractions. Then we factorise the denominator of $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}$ and form ${{x}^{2}}+x+2=f\left( x \right)g\left( x \right)$. We form the numerator of $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}$ with the factors. We simplify the equation $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}=1+\dfrac{A}{f\left( x \right)}+\dfrac{B}{g\left( x \right)}$ to find the partial fraction form.

Complete step by step answer:
We have to form the partial form of $\dfrac{5x-1}{{{x}^{2}}-x-2}$.
A brief introduction to partial fractions is required to solve the problem.
The condition for the process is to break the denominator in its factor forms and rearrange the numerator in addition or subtraction form of those exact factors in the denominator.
This process is required to break the whole expression in its simpler form.
For our given problem $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}$, we first break ${{x}^{2}}+x+2$ into factors.
We use the concept of quadratic formula solving where for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part ${{b}^{2}}-4ac$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have ${{x}^{2}}+x+2$. The values of a, b, c are $1,1,2$ respectively.
We put the values and get $x$ as \[x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 2\times 1}}{2\times 1}=\dfrac{-1\pm \sqrt{-7}}{2}=\dfrac{-1\pm i\sqrt{7}}{2}\].
Therefore, the roots are imaginary. The denominator can’t be broken.
Expressing the numerator in the form of ${{x}^{2}}=\left( {{x}^{2}}+x+2 \right)-\left( x+2 \right)$.
Therefore, $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}=\dfrac{\left( {{x}^{2}}+x+2 \right)-\left( x+2 \right)}{{{x}^{2}}+x+2}=1-\dfrac{\left( x+2 \right)}{{{x}^{2}}+x+2}$.

The partial fraction form becomes $\dfrac{{{x}^{2}}}{{{x}^{2}}+x+2}=1-\dfrac{\left( x+2 \right)}{{{x}^{2}}+x+2}$.

Note: The form gets trickier when there are the same factors within its power form. The indices of the numerator and the denominator decides how to form the partial fractions. If they are equal then we start with a constant term as the additional value.