Question

Solve the expression \[\dfrac{{{x^4} + 24{x^2} + 28}}{{{{\left( {{x^2} + 1} \right)}^3}}} = \]

Answer 1

Hint: Here, in the given question we are asked to simplify the problem. In order to solve this question, at first we have to factorize the numerator considering\[{x^2} + 1\]as one of its factors may be by long division method. After that, we will have two terms in the problem. Separate them and use a partial fraction method to simplify the numerator parts. Continue to solve to get the desired result.
Formula used:
Partial fraction decomposition:
\[\dfrac{{p\left( x \right) + q}}{{{{\left( {x - a} \right)}^2}}} = \dfrac{{{A_1}}}{{\left( {x - a} \right)}} + \dfrac{{{A_2}}}{{{{\left( {x - a} \right)}^2}}}\]

Complete step-by-step solution:
Given problem,
\[\dfrac{{{x^4} + 24{x^2} + 28}}{{{{\left( {{x^2} + 1} \right)}^3}}}\]
By hit and trial, if we consider \[{x^2} + 1\] as one of the factors of\[{x^4} + 24{x^2} + 28\] and then dividing it by \[{x^2} + 1\], we get numerator as \[\left( {{x^2} + 1} \right)\left( {{x^2} + 23} \right) + 5\]
Then, the given problem becomes:
\[\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 23} \right) + 5}}{{{{\left( {{x^2} + 1} \right)}^3}}}\]
Separating the two terms, we get,
\[\dfrac{{\left( {{x^2} + 23} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} + \dfrac{5}{{{{\left( {{x^2} + 1} \right)}^3}}}\]
Let\[I = \dfrac{{\left( {{x^2} + 23} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} + \dfrac{5}{{{{\left( {{x^2} + 1} \right)}^3}}}\]
Before simplifying \[I\], let’s consider: \[\dfrac{{{x^2} + 23}}{{{{\left( {{x^2} + 1} \right)}^2}}}\]
Clearly, the degree of numerator here is strictly less than the degree of denominator. So, we can use a partial fraction method to solve.
\[\dfrac{{{x^2} + 23}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \dfrac{{Ax + B}}{{{x^2} + 1}} + \dfrac{{Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}}\] \[ \ldots \left( 1 \right)\]
Using LCM method to add the terms in RHS, we get,
\[\dfrac{{{x^2} + 23}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \dfrac{{\left( {Ax + B} \right)\left( {{x^2} + 1} \right) + Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}}\]
Multiplying \[{\left( {{x^2} + 1} \right)^2}\] both sides, we get,
\[{x^2} + 23 = \left( {Ax + B} \right)\left( {{x^2} + 1} \right) + Cx + D \\
   \Rightarrow {x^2} + 23 = A{x^3} + Ax + B{x^2} + B + Cx + D \]
Rearranging them,
\[{x^2} + 23 = A{x^3} + B{x^2} + \left( {A + C} \right)x + \left( {B + D} \right)\]
On comparing the coefficients, we get,
\[ A = 0 \\
  B = 1 \\
  A + C = 0 \\
  B + D = 23 \]
Now, solving the above system, we get,
\[A = 0 \\
  B = 1 \\
  C = 0 \\
  D = 22 \]
By putting these values in equation \[\left( 1 \right)\], we get,
\[\dfrac{{{x^2} + 23}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \dfrac{1}{{{x^2} + 1}} + \dfrac{{22}}{{{{\left( {{x^2} + 1} \right)}^2}}}\]
Now, we have \[I = \dfrac{{\left( {{x^2} + 23} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} + \dfrac{5}{{{{\left( {{x^2} + 1} \right)}^3}}}\]

Note: We can split the number of fractions into the sum or difference of two or more fractions, such a fraction is known as Partial fraction. We use Partial fraction many times (such as in integration problems). But one thing is important to remember that we can use partial fraction only and only if the degree of the numerator is strictly less than the degree of the denominator.