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:

__ 1 __+ __ 4 __ = __ 5(x + 2) __

(x + 1) (x + 6) (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.

**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 __ 5(x + 2) __ into partial fractions.

(x + 1)(x + 6)

We can write this as:

__ 5(x + 2) __ º __ A __ + __ B __

(x + 1)(x + 6) (x + 1) (x + 6)

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

__ 5(x + 2) __ º __ A(x + 6) + B(x + 1)__

(x + 1)(x + 6) (x + 1)(x + 6)

5(x + 2) º A(x + 6) + B(x + 1)

The above expression is an **identity** (hence º 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)

B = 4

when x = -1,

5(1) = 5A

A = 1

since __ 5(x + 2) __ º __ A __ + __ B __

(x + 1)(x + 6) (x + 1) (x + 6)

the answer is __ 1 __ + __ 4 __ (as we knew)

(x + 1) (x + 6)

Algebra

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Algebra

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Algebra

Algebra