Partial Fractions


Partial Fractions are what we use in order to seperate fraction which have 1 large denominator into multiple fractions of different denominators
We do this as it is alot easier to deal with these smaller fractions than 1 large one for instance in Method of DDifference where we want to have fractions that are taken away from another however we may not notice if there is a deduction under 1 large denominator

There are 3 types of partial fractions which we will discuss in this lesson
First is a fraction which has a linear factors:

Linear Factors

Example 1


First we need to seperate out the denominators and give them placeholder numerators as this is what we are trying to discover

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

Next we need a way in order to equate these two equations together
To do this we must multiply the entire equation by our original denominator [e.g.(x+1)(x-4)]

$$x = A(x-4)+B(x+1)$$

Now in order to find our values of A and B we must first choose an appropriate value for x such that one of the brackets equals 0 such as x=4:

$$(4) = A(4-4)+B(4+1)$$

$$4 = A(0)+B(5) = 5B$$

$$B = \frac{4}{5}$$

Now we need to find a new value for x where we are only left with A this is of course x=-1:

$$(-1) = A(-1-4)+B(-1+1)$$

$$-1 = A(-5)+B(0) = -5A$$

$$A = \frac{1}{5}$$

Now all we need to do is input our values for A and B into the equation which gives us:

$$\frac{A}{(x+1)}+\frac{B}{(x-4)} = \frac{1}{5(x+1)}+\frac{4}{5(x-4)}$$

Therefore we have now split our original equation into its smaller fractions and have hence found the solution

Example 2

Below is another example that has linear factors that looks more complicated at a glance but is really the same problem


First we need to seperate the first bracket out into its factors as we want to get the fraction into its linear factors:

$$\frac{2x+3}{(x^2-1)(x+3)} = \frac{2x+3}{(x-1)(x+1)(x+3)}$$

As you can see we are now confronted with the same problem as before but this time with 3 variables on the denomianator
As before we need to seperate these denominators out like so:

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

Now we once again multiply by the original denominator to get the equation in a form we can easily solve

$$2x+3 = A(x+1)(x+3) + B(x-1)(x+3) + C(x-1)(x+1)$$

Now we have 2 brackets after each unknown value butt we can see the brackets are repeated so with one value of x we can remove two unknown variables and be left with an equation we can easily solve for instance x=-1

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

$$-2+3 = A(0)(2) + B(-2)(2) + C(-2)(0)$$

$$1 = -4B$$

$$B = -\frac{1}{4}$$

Now we need to solve for A and C by using new values of x for instance x=-3

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

$$-6+3 = A(-2)(0) + B(-4)(0) + C(-4)(-2)$$

$$-3 = 8C$$

$$C = -\frac{3}{8}$$

Finally we need to find A which we can do with a value of x=1

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

$$2+3 = A(2)(4) + B(0)(4) + C(0)(2)$$

$$5 = 8A$$

$$A = \frac{5}{8}$$

Finally we need to input our values of A,B and C into our equation and we will have the answer:

$$\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3} = \frac{5}{8(x-1)}-\frac{1}{4(x+1)}-\frac{3}{8(x+3)}$$

Repeated Linear Factors

A repeated linear factor is where the denominator has a repeated pair for instance something is squared like below:


As you can see the denominator has a x² symbol this is the same as (x)(x) and is hance a repeated linear factor
This means we can not deal with this in the same way as usual as we woud never be able to find a value for x where either A or B is left
Instead we seperate our variables out like this:
(reducing the power on the repeated factor each time until the repeated factor has a power of 1 and adding them all)

$$\frac{x^2+3x-1}{x^2(x+3)} = \frac{A}{x^2}+\frac{B}{x}+\frac{C}{(x+3)}$$

Now when we multiply out by our original denominator they will have different ways of removing them from the equation:

$$x^2+3x-1 = A(x+3) + B(x)(x+3) + C(x^2)$$

As you can see we can eliminate B and C using x=0

$$-1 = A(3)$$

$$A = -\frac{1}{3}$$

We can now eliminate A and B using x=-3

$$(-3)^2+3(-3)-1 = A(-3+3) + B(-3)(-3+3) + C([-3]^2)$$

$$-1 = C(9)$$

$$C = -\frac{1}{9}$$

Now we are left trying to find B however there is no value for x that will remove A and C
However since we know the values for A and C this does not matter
We simply need to give x a simple value like x=1 and fill in the values for A and C to find B

$$(1)^2+3(1)-1 = A(1+3) + B(1)(1+3) + C([1]^2)$$

$$1+3-1 = A(4) + B(1)(4) + C(1)$$

$$3 = 4A + 4B + C$$

Inputting our values for A and C

$$3 = 4(-\frac{1}{3}) + 4B + (-\frac{1}{9})$$

$$\frac{27}{9}) = 4(-\frac{3}{9}) + 4B + (-\frac{1}{9})$$

$$\frac{27}{9}) = -\frac{12}{9} + 4B -\frac{1}{9}$$

$$\frac{40}{9}) = 4B$$

$$B = \frac{10}{9})$$

Inputting the values of A,B and C gives us the partial fraction:

$$\frac{A}{x^2}+\frac{B}{x}+\frac{C}{(x+3)} = -\frac{1}{3(x^2)}+\frac{10}{9x}-\frac{1}{9(x+3)}$$

Quadratic Factors that cannot be Factorised

What about if we have denominator that can not be factorised any further without using complex numbers like:


We do this because the numerator needs to be at 1 power of x less than the denominator so we must have Ax+b otherwise the partial fraction wouldnt work

Now we continue as normal where we multiply by the denominator


We now need a values for x to remove A and B which is x=-5




Now we need a way to remove either A or B
Looking at the equation we can see A is directly multiplied by x therefore setting x=0 will remove A from the equation






Now we need to find a value for x we can do this by giving x a simple value like x=1 and solving the equation




We now need to input the value of B and C into the equation






Now we just need to put these values back into our original equation: