2c 2009 Partial Fractions

Partial Fractions In this chapter, you should recall an appropriate form for expressing rational functions in partial fractions, and carry out the decomposition, in cases where the denominator is no more complicated than (ax + b)(cx + d)(ex + f), (ax + b)(cx + d) 2 , (ax + b)(x 2 + c 2 ) and where the degree of the numerator does not exceed that of the denominator. 1. Distinct Linear Factors To every linear factor (ax + b) in the denominator of a proper fraction, there corresponds a partial fraction of the form

A where A is a ax + b

constant (in other words do not have any square or cube terms etc). Express

5(x + 2) as partial fractions. (x + 1)(x + 6)

We can write this as 5(x + 2) A B ≡ + . So now all we have to do is finding A and B. (x + 1)(x + 6) x + 1 x + 6

5( x + 2) A( x + 6) + B ( x + 1) ≡ . By comparison or elimination, ( x + 1)( x + 6) ( x + 1)( x + 6) A = 1, B = 4. Therefore

1.

2.

5( x + 2) 1 4 = + ( x + 1)( x + 6) x + 1 x + 6

2x + 3 ( x + 1)( x − 1)( x − 2) x 2 + 3x + 5 2

x + 5x + 6

1 6( x + 1 )

−

5 2( x − 1 )

1−

+

7 3(x − 2)

5 x+3

+

3 x+2

2. Repeated Linear Factors To every linear factor (ax + b) repeated n times in the denominator, there corresponds the sum of n partial fractions. A2 An A1 + 2 + …… + (ax + b) n (ax + b) ax + b

x 2 + 6x + 9 Resolve f(x) = into partial fractions. ( x − 3) 2 ( x + 5) A B C x 2 + 6x + 9 + + ≡ . 2 2 x+5 ( x − 3) ( x + 5) ( x − 3) ( x − 3) 15 9 1 By comparison or elimination, A = , B = , C = 16 2 16

4x − 9 ( x − 3) 2 4x − 1 2. 2 2 x ( x − 4)

4

1.

x −3 −

1 x

1

+

4x

2

+

7 16( x − 2)

+

+

3 (x − 3)

2

9 16( x + 2)

3. Distinct Quadratic Factors To

every

quadratic

factor

ax 2 + bx + c in

corresponds a partial fraction of the form

Express

the

denominator,

Ax + B

ax 2 + bx + c

2x 2 + x − 3 in partial fractions. ( x + 1)(2 x 2 + 3 x − 4)

Bx + C 2x 2 + x − 3 A + ≡ 2 2 ( x + 1)(2 x + 3 x − 4) x + 1 (2 x + 3 x − 4) By comparison or elimination, A =

1.

3x − 4 ( x − 1)( x 2 − 2 x + 2)

2.

1− x x ( x 2 + 4)

2

2

2 6 7 , B = ;C = − 5 5 5 7 10 ( x + 1)

−

1 2(x − 1)

−

1 4x

+

−

x−8 2 5 ( x − 2x + 2) 1

4x

2

+

x −1 2 4 (x + 4 )

there