To calculate: The partial decomposition of \frac{2x^{3}-x^{2}+8x-16

Formula used:
Decomposition of ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ Partial Fractions:
Consider a rational expression ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$, where f(x) and g (x) are polynomial with real coefficients, $g(x)\ne 0$, and the degree of f (x) is less than degree of g(x).
Step1: Factor the denominator g (x) completely into linear factors of the form ${\displaystyle {(ax+b)}^m}$ and quadratic factors of the form ${\displaystyle {(a{x}^2+bx+c)}^n}$ that are not further factorable over the integers.
‘Step2: Set up the form of decomposition. That is, write the original rational expression ${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ as a sum of simpler fractions using these guidelines. Note that A1,A2,.....,Am,B1,B2,....,Bm and C1,C2,.....,Cm are constants.
Linear Factors of g(x):
For each linear factor of g(x), the partial fraction decomposition must include the sum:
${\displaystyle \frac{A1}{{(ax+b)}^1}+\frac{A2}{{(ax+b)}^2}+\dots .+\frac{Am}{{(ax+b)}^m}}$
Quadratic Factors of g (x):
For each quadratic factor of g (x), the partial fraction decomposition must include the sum:
${\displaystyle \frac{B_{1}x+C_1}{{(a{x}^2+bx+c)}^1}+\frac{B_{2}x+C_2}{{(a{x}^2+bx+c)}^2}+\dots +\frac{B_{n}x+C_n}{{(a{x}^2+bx+c)}^n}}$
Step3: With the form of the partial fraction decomposition set up, multiply both sides of the equation by the LCD to clear fractions.
‘Step4: Use the equation from step3, set up a system of linear equations by equating the constant terms and equating the coefficients of like powers of x.
Step 5: Solve the system of equations from step4 and substitute the solutions to the system into the partial fraction decomposition.
Calculation:
Consider the provided expression: 

${\displaystyle \frac{2x^3-x^2+8x-16}{x^4+5x^2+4}}$
Here, ${\displaystyle f\left(x\right)=2x^3-x^2+8x—l6\text{and}g\left(x\right)=x^4+5x^2+4}$.
Factorize g(x):
${\displaystyle g\left(x\right)=x^4+5x^2+4}$
${\displaystyle =x^4+x^2+4x^2+4}$
${\displaystyle =x^2(x^2+1)+4(x^2+1)}$
${\displaystyle =(x^2+4)(x^2+1)}$
So, the expression becomes:
${\displaystyle \frac{2x^3-x^2+8x-16}{(x^2+4)(x^2+1)}}$
Here, in g (x), there are two factors, both are quadratic. So, the expression can be decomposed as: