Question

How do I find the partial-fraction decomposition of $\dfrac{2{{x}^{3}}+7{{x}^{2}}-2x+6}{{{x}^{4}}+4}$ ?

Answer 1

Hint: We are given a fraction as $\dfrac{2{{x}^{3}}+7{{x}^{2}}-2x+6}{{{x}^{4}}+4}$ , we are asked to find the partial fraction decomposition, to do so, we will start why do we need to do such method, we learn in generally we apply it when, then we will start our solution by factoring the denominator which is ${{x}^{4}}+4$ .

Complete step by step solution:
We are given a fraction as $\dfrac{2{{x}^{3}}+7{{x}^{2}}-2x+6}{{{x}^{4}}+4}$ , we are asked to find its partial fraction decomposition.
To do so, we start by learning that partial fraction decomposition is a method to split the fraction into simpler numerator and denominator such that it will be easier to work with them during integration.
As we generally use it to decompose (split) when we have a denominator of higher power, we will split them so that we have little power terms.
Now in our fraction we have ${{x}^{4}}+4$ as denominator we can see that it cannot be split into linear factor as ${{x}^{4}}+4\ge 0$ because ${{x}^{4}}\ge 0$ so, ${{x}^{4}}+4>0$ so no linear factor.
It will split into quadratic factors.
So, let ${{x}^{4}}+4=\left( a{{x}^{2}}+bx+c \right)\left( d{{x}^{2}}+ex+f \right)$
We consider ‘a’ and‘d’ as 1 so we get –
${{x}^{4}}+4=\left( {{x}^{2}}+bx+c \right)\left( {{x}^{2}}+ex+f \right)$
Now we look for the other variable to do so we will find the product and compare coefficient.
First we look for the coefficient of ${{x}^{3}}$ .
We have $e+b$ while on ${{x}^{4}}+4$ there is no ${{x}^{3}}$ , so $e+b=0$ .
$\Rightarrow e=-b$ ……………………………. (1)
$\Rightarrow {{x}^{4}}+4=\left( {{x}^{2}}+bx+c \right)\left( {{x}^{2}}-bx+f \right)$
Now we look for the coefficient of ‘x’, we have $bf+ec=0$ and from equation (1), we get,
$bf-bc=0$
$b\left( f-c \right)=0$
So, it mean either $b=0$ or $f-c=0$ because their product gives 0 as result.
If $b=0$ then
$\Rightarrow {{x}^{4}}+4=\left( {{x}^{2}}+0x+c \right)\left( {{x}^{2}}-0x+f \right)$
$\Rightarrow {{x}^{4}}+4=\left( {{x}^{2}}+c \right)\left( {{x}^{2}}+f \right)$
$\Rightarrow {{x}^{4}}+4={{x}^{4}}+\left( c+f \right){{x}^{2}}+cf$
And further we can say $f+c=0$ and $cf=-{{c}^{2}}=4$ which has no real solution.
If $c=f$ then $cf=4\Rightarrow c=f=\pm 2$ .
Now checking coefficient of ${{x}^{2}}$ , we get –
$f+c-{{b}^{2}}=0$ so if ‘b’ is real then –
${{b}^{2}}\ge 0$ so $f+c\ge 0$
$\Rightarrow f=c=2$
Then ${{b}^{2}}=4$ so, $b=\pm 2$ .
Hence we get our factor as –
${{x}^{4}}+4=\left( {{x}^{2}}+2x+2 \right)\left( {{x}^{2}}-2x+2 \right)$ .
Now we will use it to factor out fraction.
$\dfrac{2{{x}^{3}}+7{{x}^{2}}-2x+6}{{{x}^{4}}+4}=\dfrac{Ax+B}{{{x}^{2}}+2x+2}+\dfrac{Cx+D}{{{x}^{2}}-2x+2}$ .
By multiplying, we get –
$\begin{align}
  & \Rightarrow 2{{x}^{3}}+7{{x}^{2}}-2x+6=\left( Ax+B \right)\left( {{x}^{2}}-2x+2 \right)+\left( {{x}^{2}}+2x+2 \right)\left( Cx+D \right) \\
 & \Rightarrow 2{{x}^{3}}+7{{x}^{2}}-2x+6=\left( A+C \right){{x}^{3}}+\left( B-2A+D+2C \right){{x}^{2}}+2\left( A-B+C+D \right)x+2\left( B+D \right) \\
\end{align}$
Equating coefficient, we get –
$\begin{align}
  & A+C=2 \\
 & B-2A+D+2C=7 \\
 & 2\left( A-B+C+D \right)=-2 \\
 & 2\left( B+D \right)=6 \\
\end{align}$
Solving these, we will get –
Value of $A=0,C=2,D=0\text{ and }B=3$ .
So, we get –
$A=0,B=3,C=2,D=0$ .
So, our equation $\dfrac{2{{x}^{3}}+7{{x}^{2}}-2x+6}{{{x}^{4}}+4}=\dfrac{Ax+B}{{{x}^{2}}+2x+2}+\dfrac{Cx+D}{{{x}^{2}}-2x+2}$ become,
$=\dfrac{3}{{{x}^{2}}+2x+2}+\dfrac{2x}{{{x}^{2}}-2x+2}$


Note: The main problem that occurs while solving such factorization is the calculation mistake, we need to be very gentle and calm, as one calculation will lead us to incorrect solution, we do not have to show our calculation while finding A, B, C, D so we can do calculate elsewhere these will speed up solution.