Partial Fractions

In the presentation on algebraic fractions we saw how to add 2 algebraic fractions. To find partial fractions for an expression, we need to reverse the process of adding fractions. We will also develop a method for reducing a fraction to 3 partial fractions.

We’ll start by adding 2 fractions. e.g. The partial fractions for are

To find the partial fractions, we start with The expressions are equal for all values of x so we have an identity. The identity will be important for finding the values of A and B.

To find the partial fractions, we start with Multiply by the denominator of the l.h.s. So, If we understand the cancelling, we can in future go straight to this line from the 1st line.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. Why should I choose x = 2 ? ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. What value would you substitute next ? ANS: Any value would do but x = - 1 is good.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

e.g. 2 Express the following as 2 partial fractions. Solution: Let Multiply by : It’s very important to write this each time

So, We never leave fractions piled up like this, so The “halves” are written in the denominators ( as 2s ) and the minus sign is moved to the front of the 2nd fraction. Finally, we need to check the answer. A thorough check would be to reverse the process and put the fractions together over a common denominator.

Another check is to use the “cover-up” method: To check A, find the value of x that makes the factor under A equal to zero ( x = 3 ) Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only We get To check B, substitute x = 1 in the l.h.s. but cover-up

The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors,

The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, ( we may have to factorise to find them )

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree than the denominator The degree of a polynomial is given by the highest power of x.

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree than the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree then the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree 2

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree then the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree 2

SUMMARY To find partial fractions for expressions like Let Multiply by the denominator of the l.h.s. Substitute a value of x that makes the coefficient of B equal to zero and solve for A. Substitute a value of x that makes the coefficient of A equal to zero and solve for B. Check the result by reversing the method or using the “cover-up” method.

Exercises Express each of the following in partial fractions. 1. 2. 3. 4.

Solutions: 1. Multiply by : So, Check:

Solutions: 1. Multiply by : So, Check:

Solutions: 1. Multiply by : So, Check: ( you don’t need to write out the check in full )

Solutions: 2. Multiply by So, ( I won’t write out any more checks but it is important to do them. )

Solutions: 3. Multiply by So,

Solutions: 4. Multiply by So,

If the denominator has 3 factors, we just extend the method. e.g. Solution: Multiply by So,

The next type of fraction we will consider has a repeated linear factor in the denominator. e.g. 1 We would expect the partial fractions to be either This is wrong because the first 2 fractions just give , which is the same as having only one constant. or We will try this to see why it is also wrong.

We need 3 constants if the degree of the denominator is 3. Suppose Multiply by : However, Substituting B = 3 gives A = 1, an inconsistent result We need 3 constants if the degree of the denominator is 3.

So, for we need It would also be correct to write but the fractions are not then reduced to the simplest form

Using Multiply by :

Using Multiply by :

Using Multiply by : There is no other obvious value of x to use so we can choose any value. e.g. Subst. for A and C:

There is however, a neater way of finding B. We had Since this is an identity, the terms on each side must be the same. For example, we have on the l.h.s. so there must be on the r.h.s.

There is however, a neater way of finding B. We had Since this is an identity, the terms on each side must be the same. For example, we have on the l.h.s. so there must be on the r.h.s. So, equating the coefficients of :

There is however, a neater way of finding B. We had Since this is an identity, the terms on each side must be the same. For example, we have on the l.h.s. so there must be on the r.h.s. So, equating the coefficients of : Since We could also equate the coefficients of x ( but these are harder to pick out ) or the constant terms ( equivalent to putting x = 0 ).

So,

So, The “cover-up” method can only be used to check A and C so for a proper check we need to put the r.h.s. back over a common denominator. We get So the numerator gives:

SUMMARY To find partial fractions for expressions with repeated factors, e.g. Let Work in the same way as for type 1 fractions, using the two obvious values of x and either any other value or the coefficients of . Check the answer by using a common denominator for the right-hand side. N.B. B can sometimes be zero.

Exercises Express each of the following in partial fractions. 1. 2.

Solutions: 1. Let Multiply by : Coefficient of : So,

Solutions: 2. Let Multiply by : Coefficient of : So,

You may meet a question that combines algebraic division and partial fractions. e.g. Find partial fractions for The degree of the denominator is equal to the degree of the numerator. Both are degree 2. This is called an improper fraction. If the degree of the numerator is higher than the denominator the fraction is also improper. In an exam you are likely to be given the form of the partial fractions.

e.g. 1 Find the values of A, B and C such that Solution: We don’t need to change our method Multiply by : So,

e.g. 2 Find the values of A, B and C such that Solution: Multiply by : If you notice at the start, by looking at the terms on the l.h.s., that A = 2, the solution will be shorter as you can start with x = 0 and find C, then B. So,

If you aren’t given the form of the partial fractions, you just need to watch out for an improper fraction. You will need to divide out but you will probably only need one stage of division so it will be easy.

e.g. 2

e.g. 2

e.g. 2 So, We can now find partial fractions for We get So,

Exercise 1. Express the following in partial fractions: Solution: Dividing out:

Partial Fractions: Let Multiply by : So,

The 3rd type of partial fractions has a quadratic factor in the denominator that will not factorise. e.g. The partial fractions are of the form The method is no different but the easiest way to find A, B and C is to use the obvious value of x but then equate coefficients of the term and equate constants.

Exercise 1. Express the following in partial fractions: Solution: Multiply by : Coefficient of Constants:

So,