PARTIAL FRACTION WITH NON REPEATED LINEAR FACTORS

Being the diligent mathematician that you are, your mind would have already been ingrained with the knowledge of adding algebraic fractions

For example, we can add this:
$\frac{1}{x+3}+\frac{3}{x+2}$
$\frac{1(x+2)+3(x+3)}{(x+3)(x+2)}$
$\frac{x+2+3x+9}{(x+3)(x+2)}$
$\frac{4x+11}{(x+3)(x+2)}$

Thus $\frac{1}{x+3}+\frac{3}{x+2}=\frac{4x+11}{(x+3)(x+2)}$

But, can we reverse this process? going from $\frac{4x+11}{(x+3)(x+2)}$ to $\frac{1}{x+3}+\frac{3}{x+2}$.

We can do this using partial fractions decomposition.

Partial fraction decomposition is the return from the single expression to the original expression.

To partial fractionize a single expression, you should observe these two things
1. The denominator must be factorizable.
2. The degree of the numerator should be lesser than the denominator

As an example, consider the algebraic fraction
$\frac{18-20x}{8x^2-18x+9}$

The above fraction meets the first and second conditions. First, its denominator is factorizable because $8x^2-18x+9$ can be factorized into $(4x-3)(2x-3)$. 

Secondly, the degree of the numerator is lesser than the denominator because the highest power of $x$ in the numerator is $1$, which is lesser than the highest power of the denominator.

Partial fraction decomposition can be solved using the substitution method(Heaviside method) and/or equating the coefficients method.

As I have observed, the first method is usually ignored by many textbooks. As you will see subsequently,  We would be solving all four examples using these two methods.

Accordingly, they are three forms of partial fraction:
1. Partial fraction with non repeated linear factors
2. Partial fraction with repeated linear factors
3. Partial fraction with irreducible quadratic expression

We would be concentrating on the first one today. We will be discussing the other ones In subsequent posts.

As the name suggests, a partial fraction with non-repeated linear factors does not have repeated factors.

Example 1
Write $\frac{6y}{(y-1)(y+2)}$ in partial fractions.

Solution:
We will separate the denominator factor and gives each symbolic character like A, B.

$\frac{6y}{(y-1)(y+2)}=\frac{A}{y-1}+\frac{B}{y+2}$

As is with additions of algebraic fractions, we take the L.C.M of the left-hand side.
$\frac{6y}{(y-1)(y+2)}=\frac{A(y+2)+B(y-1)}{(y-1)(y+2)}$.

The denominators are the same. So we equate the numerators.

$6y=A(y+2)+B(y-1)$

Always take note of the equation derived after eliminating the fraction. This is where we would choose which method to employ.

Next, we expand the fraction.
$6y=Ay+2A+By-B$

Collecting like terms
$6y=Ay+By+2A-B$
$6y=(A+B)y+2A-B$

Next, we equate those with the same coefficients
$6=A+B$
$0=2A-B$

Making B the subject of the formula in Eqn 2
$0=2A-B$
$B=2A$

Substituting the $2A$ for $B$ in eqn 1
$6=A+2A$
$6=3A$
$\frac{6}{3}=\frac{3A}{3}$
$2=A$
$A=2$

Inserting the value of $B$ in $B=2A$
$B=2(2)$
$B=4$

Therefore, the partial fraction decomposition is:
$\frac{6y}{(y-1)(y+2)}=\frac{2}{y-1}+\frac{4}{y+2}$

Alternatively, we can also use the Heaviside method. The whole idea of this method is to successfully eliminate a variable so that the other variable can be easily solved
$6y=A(y+2)+B(y-1)$

Let $y$ equals $-2$ so that $A$ is eliminated
$6(-2)=A(-2+2)+B(-2-1)$
$-12=-3B$
$\frac{-12}{-3}=\frac{-3B}{-3}$
$B=4$

Now, let $y$ be equal to $1$ so that $B$ is effectively eliminated.
$6(1)=A(1+2)+B(1-1)$
$6=3A$
$\frac{6}{3}=\frac{3A}{3}$
$A=2$

So both methods can be used to partial decompose fractions

Example 2
Express $\frac{x+4}{x^2-x-2}$ in partial fraction

Solution:
First, we would factorize the denominator
$\frac{x+4}{(x+1)(x-2)}$

As with the first example, we will use symbols to represent the numerator.

$\frac{x+4}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$
$\frac{x+4}{(x+1)(x-2)}=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$

Equating the numerator
$x+4=A(x-2)+B(x+1)$

Expanding the brackets
$x+4=Ax-2A+Bx+B$
$x+4=Ax+Bx-2A+B$
$(1)x+4=(A+B)x-2A+B$

Equating those with the same coefficient
$1=A+B$
$4=-2A+B$

$1=A+B$
$-(4=-2A+B)$
$-3=3A$
$\frac{-3}{3}=\frac{3A}{3}$
$-1=A$
$A=-1$

Substituting the value of $A$ in Eqn 1
$1=(-1)+B$
$1=-1+B$
$B=1+1$
$B=2$

Therefore,  the partial fraction decomposition is:
$\frac{x+4}{(x+1)(x-2)}=\frac{-1}{x+1}+\frac{2}{x-2}$

OR
Using the Heaviside method
$x+4=A(x-2)+B(x+1)$

Let $x$ be $2$
$2+4=A(2-2)+B(2+1)$
$6=B(3)$
$\frac{6}{3}=\frac{3B}{3}$
$B=2$

Let $x$ be $-1$
$-1+4=A(-1-2)+B(-1+1)$
$3=A(-3)$
$\frac{3}{-3}=\frac{-3A}{-3}$
$A=-1$

Example 3
Express $\frac{5}{(2x+1)(x-2)}$ in partial fraction.

Solution:
$\frac{5}{(2x+1)(x-2)}=\frac{A}{2x+1}+\frac{B}{x-2}$
$\frac{5}{(2x+1)(x-2)}=\frac{A(x-2)+B(2x+1)}{(2x+1)(x-2)}$

Equating the numerator
$5=A(x-2)+B(2x+1)$
$5=Ax-2A+2Bx+B$
$5=Ax+2Bx-2A+B$
$5=(A+2B)x-2A+B$

Equating those with the same coefficients
$0=A+2B$
$5=-2A+B$

$-A=2B$
$\frac{-A}{-1}=\frac{2B}{-1}$
$A=-2B$

Inserting $-2B$ for $A$ in eqn 2
$5=-2(-2B)+B$
$5=4B+B$
$5B=5$
$\frac{5B}{5}=\frac{5}{5}$
$B=1$

If $B=1$
$A=-2(1)$
$A=-2$

Thus, the partial fraction decomposition is:
$\frac{5}{(2x+1)(x-2)}=\frac{-2}{2x+1}+\frac{1}{x-2}$

Using the Heaviside method.
$5=A(x-2)+B(2x+1)$

Let $x$ equal $2$
$5=A(2-2)+B(2(2)+1)$
$5=B(5)$
$\frac{5}{5}=\frac{5B}{5}$
$B=1$

Let $x$ equal $\frac{-1}{2}$
$5=A(\frac{-1}{2}-2)+B(2(\frac{-1}{2}+1)$
$5=A(\frac{-5}{2})$

Cross multiply
$5\times2=A(-5)$
$10=-5A$
$\frac{10}{-5}=\frac{-5A}{-5}$
$-2=A$
$A=-2$

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Example 4
Write $\frac{5-y}{y^2-5y+6}$ in partial fraction.

Solution
$\frac{5-y}{(y-3)(y-2)}$
$\frac{5-y}{(y-3)(y-2)}=\frac{A}{(y-3)}+\frac{B}{(y-2)}$

$\frac{5-y}{(y-3)(y-2)}=\frac{A(y-2)+B(y-3)}{(y-3)(y-2)}$
$5-y=A(y-2)+B(y-3)$
$5-y=Ay-2A+By-3B$
$5-y=Ay+By-2A-3B$
$5-(1)y=(A+B)y-2A-3B$

Equating the coefficients
$5=-2A-3B$
$-1=A+B$

Making $A$ the subject of the formula in Eqn 2
$-1=A+B$
$-1-B=A$
$A=-1-B$

Injecting $-1-B$ in eqn 1
$5=-2(-1-B)-3(B)$
$5=2+2B-3B$
$5-2=-B$
$3=-B$
$\frac{3}{-1}=\frac{-B}{-1}$
$B=-3$

$A=-1-(-3)$
$A=-1+3$
$A=2$

Thjs, the partial fraction decomposition is 
$\frac{5-y}{(y-3)(y-2)}=\frac{2}{(y-3)}+\frac{-3}{(y-2)}$

As we have always done, let's use the Heaviside method.
$5-y=A(y-2)+B(y-3)$

Let $y$ equal to $2$ so that $A$ is successfully eliminated
$5-2=A(2-2)+B(2-3)$
$3=-B$
$\frac{3}{-1}=\frac{-B}{-1}$
$B=-3$

Let $y$ be equal to $3$
$5-3=A(3-2)+B(3-3)$
$2=A$
$A=2$

As explained above, partial fraction decomposition can be solved using the substitution(Heaviside) method or equating the coefficient method.

That will be all on partial fractions with non-repeated factors. Next up is a partial fraction with repeated fraction, we will be learning just that here.

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