Question

How do you express $\dfrac{x+4}{{{x}^{2}}+2x+5}$ in partial fractions?

Answer 1

Hint: We are given a fraction $\dfrac{x+4}{{{x}^{2}}+2x+5}$which we have to transform into partial fractions. We will start by transforming the denominator into the simpler form. Then we can consider the term as sum of two different terms. Next, we will consider both equations and compare them to get the needed values.

Complete step-by-step answer:
According to the question, we are to express $\dfrac{x+4}{{{x}^{2}}+2x+5}$in partial fractions.
To transform it into partial fractions, we are to transform the denominator into the simplified form.
So, we have a denominator as, ${{x}^{2}}+2x+5$.
Now, the discriminant of the denominator is negative, so it only has Complex zeros:
$\Delta ={{2}^{2}}-4.1.5=4-20=-16$
The numerator is already linear, so if you want Real coefficients then your expression is already as simple as possible.
On the other hand, you can decompose into simpler fractions using Complex coefficients.
First let's factor the denominator:
 ${{x}^{2}}+2x+5$ can be simplified by the middle term factor.
Now, ${{x}^{2}}+2x+5={{x}^{2}}+2x+1+4$
Again, we know, ${{\left( x+1 \right)}^{2}}={{x}^{2}}+2x+1$ .
So, ${{x}^{2}}+2x+5$can be written as, ${{x}^{2}}+2x+5={{\left( x+1 \right)}^{2}}+4$
Again 4 can be written as, $4=-4.-1=-4.{{i}^{2}}$ where $i=\sqrt{-1}$
Thus, $4=-{{\left( 2i \right)}^{2}}$
So, ${{x}^{2}}+2x+5={{\left( x+1 \right)}^{2}}-{{\left( 2i \right)}^{2}}$
From, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ ,
We get,${{x}^{2}}+2x+5=\left( x+1-2i \right)\left( x+1+2i \right)$
So, we are now, trying to find the solution of the form,
 \[\dfrac{x+4}{{{x}^{2}}+2x+5}=\dfrac{A}{x+1-2i}+\dfrac{B}{x+1+2i}\]
So, from the right hand side,
\[\dfrac{A}{x+1-2i}+\dfrac{B}{x+1+2i}=\dfrac{A\left( x+1+2i \right)+B\left( x+1-2i \right)}{{{x}^{2}}+2x+5}\]
Simplifying the numerator,
\[\dfrac{A\left( x+1+2i \right)+B\left( x+1-2i \right)}{{{x}^{2}}+2x+5}=\dfrac{Ax+A+2Ai+Bx+B-2Bi}{{{x}^{2}}+2x+5}=\dfrac{\left( A+B \right)x+\left( A+B \right)+\left( 2A-2B \right)i}{{{x}^{2}}+2x+5}\]
Now, by comparing this with, $\dfrac{x+4}{{{x}^{2}}+2x+5}$, we get,
$A+B=1$ and $A+B+2\left( A-B \right)i=4$
Putting, $A+B=1$in the second equation,
$\Rightarrow A+B+2\left( A-B \right)i=4$
$\Rightarrow 1+2\left( A-B \right)i=4$
$\Rightarrow 2\left( A-B \right)i=3$
Dividing both sides with 2,
$\Rightarrow \left( A-B \right)i=\dfrac{3}{2}$
Thus, $A-B=\dfrac{3}{2i}=-\dfrac{3{{i}^{2}}}{2i}=-\dfrac{3}{2}i$
Now, adding $A+B=1$and $A-B=\dfrac{3}{2i}=-\dfrac{3{{i}^{2}}}{2i}=-\dfrac{3}{2}i$
We get, $2A=1-\dfrac{3}{2}i$
So, $A=\dfrac{1}{2}-\dfrac{3}{4}i$
Similarly putting the value in $A+B=1$,
We get, $B=1-\dfrac{1}{2}+\dfrac{3}{4}i=\dfrac{1}{2}+\dfrac{3}{4}i$
Then, we can write it as, \[\dfrac{x+4}{{{x}^{2}}+2x+5}=\dfrac{\dfrac{1}{2}-\dfrac{3}{4}i}{x+1-2i}+\dfrac{\dfrac{1}{2}+\dfrac{3}{4}i}{x+1+2i}\]
This is the partial form of our given equation.

Note: In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.