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Revision:Partial Fractions

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Partial Fractions

It is possible to split many fractions into the sum or difference of two or more fractions. This has many uses (such as in integration). At GCSE level, we saw how:

\displaystyle \frac{1}{x+1} + \frac{4}{x+6} = \frac{5(x+2)}{(x+1)(x+6)}

The method of partial fractions allows us to split the right hand side of the above equation into the left hand side.


Contents

Linear Factors in Denominator

This method is used when the factors in the denominator of the fraction are linear (in other words do not have any square or cube terms etc).


Example

Split \displaystyle \frac{5(x + 2)}{(x + 1)(x + 6)} into partial fractions.


We can write this as:

\displaystyle \frac{5(x + 2)}{(x + 1)(x + 6)} \equiv \frac{A}{x + 1} + \frac{B}{x + 6}


So now, all we have to do is find A and B:

\displaystyle \frac{5(x + 2)}{(x + 1)(x + 6)} \equiv \frac{A(x + 6) + B(x + 1)}{(x + 1)(x + 6)}

\displaystyle 5(x + 2) \equiv A(x + 6) + B(x + 1)


The above expression is an identity (hence \equiv rather than =). An identity is true for every value of x.


This means that we can substitute values of x into both sides of the expression to help us find A and B.


when x = -6,

5(-4) = B(-5)

\Rightarrow B = 4


when x = -1,

5(1) = 5A

\Rightarrow A = 1


Since \displaystyle \frac{5(x + 2)}{(x + 1)(x + 6)} \equiv \frac{A}{x + 1} + \frac{B}{x + 6}


So, the answer is:

[Unparseable or potentially dangerous latex formula. Error 2 ] or (x + 4)^3, the following method is used.


Example

Split \displaystyle \frac{x - 2}{(x + 1)(x - 1)^2} into partial fractions.


This time we write:

\displaystyle \frac{x - 2}{(x + 1)(x - 1)} \equiv \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{(x + 1)^2}


As before, all we do now is find the values of A, B and C.

\displaystyle x - 2 \equiv A(x - 1)^2 + B(x - 1)(x + 1) + C(x + 1)


So, let x = 1:

\Rightarrow -1 = 2C

\Rightarrow C = -\frac{1}{2}


Then, let x = -1

\Rightarrow -3 = 4A

\Rightarrow A = -\frac{3}{4}


Finally, let x = 0

\Rightarrow -2 = A - B + C

\Rightarrow -2 = -\frac{3}{4} - B -\frac{1}{2}

\Rightarrow B = \frac{3}{4}


Therefore the answer is:

\displaystyle \frac{-3}{4(x + 1)} + \frac{3}{4(x - 1)} - \frac{1}{2(x - 1)^2}.


Quadratic Factor in the Denominator

This method is for when there is a square term in one of the factors of the denominator.


Example

\displaystyle \frac{2x - 1}{(x + 1)(x^2 + 1)} \equiv \frac{A}{x + 1} +\frac{Bx + C}{x^2 + 1}.


Find A, B and C in the same way as above.


Note that it is Bx + C on the numerator of the fraction with the squared term in the denominator.


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