Partial FractionsPartial Fractions
In the presentation on algebraic fractions wesaw how to add 2 algebraic fractions.
To find partial fractions for an expression, weneed to reverse the process of adding fractions.
We will also develop a method for reducing afraction to 3 partial fractions.
The expressions are equal for all values of x sowe have an identity.
The identity will be important for finding thevalues 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 gostraight to this line from the 1st line.
To find the partial fractions, we start with
This is where the identity is important.
The expressions are equal for all values of x, so I canchoose to let x = 2.
Why should I choose x = 2 ?
ANS: x = 2 means the coefficient of B is zero, soB disappears and we can solve for A.
This is where the identity is important.
What value would you substitute next ?
ANS: Any value would do but x = 1 is good.
The expressions are equal for all values of x, so I canchoose to let x = 2.
This is where the identity is important.
So,
The expressions are equal for all values of x, so I canchoose to let x = 2.
If we chose x = 1 instead, we get 4 = 2A – B, givingthe same result.
This is where the identity is important.
So,
The expressions are equal for all values of x, so I canchoose to let x = 2.
If we chose x = 1 instead, we get 4 = 2A – B, givingthe same result.
This is where the identity is important.
So,
The expressions are equal for all values of x, so I canchoose to let x = 2.
If we chose x = 1 instead, we get 4 = 2A – B, givingthe same result.
Solution: Let
Multiply by :
It’s very important to write this each time
e.g. 2 Express the following as 2 partial fractions.
We never leave fractions piled up like this, so
•The “halves” are written in the denominators ( as2s ) and
•the minus sign is moved to the front of the 2ndfraction.
So,
Finally, we need to check the answer.
A thorough check would be to reverse the processand put the fractions together over a commondenominator.
Another check is to use the “cover-up” method:
We get
To check B, substitute x = in the l.h.s. but cover-up
Cover-up on the l.h.s. and substitute x = 3into the l.h.s. only
To check A, find the value of x that makes thefactor under A equal to zero
( x = 3 )
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
where,
the denominator has 2 linear factors,
e.g.
where,
•the denominator has 2 linear factors,
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
e.g.
( we may have to factorise to find them )
•and the numerator is a polynomial of lowerdegree than the denominator
The degree of a polynomial is given by the highestpower of x.
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
e.g.
where,
•the denominator has 2 linear factors,
The degree of a polynomial is given by the highestpower of x.
Here the numerator is of degree
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
e.g.
1
and the denominator of degree
•and the numerator is a polynomial of lowerdegree than the denominator
where,
•the denominator has 2 linear factors,
where,
the denominator has 2 linear factors,
The degree of a polynomial is given by the highestpower of x.
and the numerator is a polynomial of lowerdegree then the denominator
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
e.g.
and the denominator of degree
2
Here the numerator is of degree
1
where,
the denominator has 2 linear factors,
The degree of a polynomial is given by the highestpower of x.
and the numerator is a polynomial of lowerdegree then the denominator
The method we’ve used finds partial fractions forexpressions I’ll call Type 1
e.g.
and the denominator of degree
2
Here the numerator is of degree
1
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 thecoefficient of B equal to zero and solve for A.
•Substitute a value of x that makes thecoefficient of A equal to zero and solve for B.
•Check the result by reversing the method orusing the “cover-up” method.
Solutions:
1.
Multiply by :
So,
Check:
( you don’t need to write out the check in full )
Solutions:
Multiply by
So,
2.
( I won’t write out any more checks but it is importantto do them. )
If the denominator has 3 factors, we just extendthe method.
e.g.
Solution: Multiply by
So,
The next type of fraction we will consider has arepeated linear factor in the denominator.
e.g. 1
We would expect the partial fractions to be
or
either
This is wrong because the first 2 fractions just give
, which is the same as having only one constant.
We will try this to see why it is also wrong.
Suppose
Multiply by :
However,
Substituting B = 3 gives A = 1, an inconsistent result
We need 3 constants if the degree of thedenominator is 3.
So, for we need
It would also be correct to write
but the fractions are not then reduced to thesimplest form
Using
Multiply by :
There is no other obvious value of x to use so wecan choose any value.
e.g.
Subst. for A and C:
There is however, a neater way of finding B.
Since this is an identity, the terms on each sidemust be the same.
For example, we have on the l.h.s. so theremust be on the r.h.s.
We had
There is however, a neater way of finding B.
Since this is an identity, the terms on each sidemust be the same.
So, equating the coefficients of :
For example, we have on the l.h.s. so theremust be on the r.h.s.
We had
There is however, a neater way of finding B.
Since this is an identity, the terms on each sidemust be the same.
So, equating the coefficients of :
For example, we have on the l.h.s. so theremust be on the r.h.s.
We had
Since
We could also equate the coefficients of x ( but theseare harder to pick out ) or the constant terms( equivalent to putting x = 0 ).
So,
We get
The “cover-up” method can only be used to check Aand C so for a proper check we need to put ther.h.s. back over a common denominator.
So the numerator gives:
SUMMARY
•Let
To find partial fractions for expressions withrepeated factors, e.g.
•Check the answer by using a common denominatorfor the right-hand side.
•Work in the same way as for type 1 fractions,using the two obvious values of x and either anyother value or the coefficients of .
N.B. B can sometimes be zero.
You may meet a question that combines algebraicdivision and partial fractions.
e.g. Find partial fractions for
The degree of the denominator is equal to thedegree of the numerator. Both are degree 2.
This is called an improper fraction.
If the degree of the numerator is higher than thedenominator the fraction is also improper.
In an exam you are likely to be given the form ofthe 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 :
So,
If you notice at the start, by looking at the termson the l.h.s., that A = 2, the solution will be shorteras you can start with x = 0 and find C, then B.
You will need to divide out but you will probably onlyneed one stage of division so it will be easy.
If you aren’t given the form of the partial fractions,you just need to watch out for an improper fraction.
The 3rd type of partial fractions has a quadraticfactor 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 tofind A, B and C is to use the obvious value of x butthen equate coefficients of the term and equateconstants.
1. Express the following in partial fractions:
Exercise
Solution:
Multiply by :
Coefficient of
Constants:
