3 Proper fractions with repeated linear factors

Sometimes a linear factor appears more than once. For example in

1 x 2 + 2 x + 1 = 1 ( x + 1 ) ( x + 1 )  which equals  1 ( x + 1 ) 2

 the factor ( x + 1 ) occurs twice. We call it a repeated linear factor . The repeated linear factor ( x + 1 ) 2 produces two partial fractions of the form A x + 1 + B ( x + 1 ) 2 . In general, a repeated linear factor of the form ( a x + b ) 2 generates two partial fractions of the form

A a x + b + B ( a x + b ) 2

 This is reasonable since the sum of two such fractions always gives rise to a proper fraction:

A a x + b + B ( a x + b ) 2 = A ( a x + b ) ( a x + b ) 2 + B ( a x + b ) 2 = x ( A a ) + A b + B ( a x + b ) 2

Key Point 15

A repeated linear factor ( a x + b ) 2 in the denominator produces two partial fractions:

A a x + b + B ( a x + b ) 2

Once again the unknown constants are found by either equating coefficients and/or substituting specific values for x .

Task!

Express 10 x + 18 4 x 2 + 12 x + 9 in partial fractions.

First factorise the denominator:

( 2 x + 3 ) ( 2 x + 3 ) = ( 2 x + 3 ) 2

There is a repeated linear factor ( 2 x + 3 ) which gives rise to two partial fractions of the form

10 x + 18 ( 2 x + 3 ) 2 = A 2 x + 3 + B ( 2 x + 3 ) 2

 Multiply both sides through by ( 2 x + 3 ) 2 to obtain the equation to be solved to find A and B :

10 x + 18 = A ( 2 x + 3 ) + B

Now evaluate the constants A and B by equating coefficients:

Equating the x coefficients gives 10 = 2 A so A = 5 . Equating constant terms gives 18 = 3 A + B from which B = 3 .

Finally express the answer in partial fractions:

10 x + 18 ( 2 x + 3 ) 2 = 5 2 x + 3 + 3 ( 2 x + 3 ) 2

 

Exercises

Express the following in partial fractions.

(a)  3 − x x 2 − 2 x + 1 , (b)  − 7 x − 15 ( x − 1 ) 2 (c)  3 x + 14 x 2 + 8 x + 16
(d)  5 x + 18 ( x + 4 ) 2 (e)  2 x 2 − x + 1 ( x + 1 ) ( x − 1 ) 2 (f)  5 x 2 + 23 x + 24 ( 2 x + 3 ) ( x + 2 ) 2
(g)  6 x 2 − 30 x + 25 ( 3 x − 2 ) 2 ( x + 7 ) (h)  s + 2 ( s + 1 ) 2 (i)  2 s + 3 s 2 .
(a)  − 1 x − 1 + 2 ( x − 1 ) 2 (b)  − 7 x − 1 + 8 ( x − 1 ) 2 (c)  3 x + 4 + 2 ( x + 4 ) 2
(d)  5 x + 4 − 2 ( x + 4 ) 2 (e)  1 x + 1 + 1 x − 1 + 1 ( x − 1 ) 2 (f)  3 2 x + 3 + 1 x + 2 + 2 ( x + 2 ) 2
(g)  − 1 3 x − 2 + 1 ( 3 x − 2 ) 2 + 1 x + 7 (h)  1 s + 1 + 1 ( s + 1 ) 2 (i)  2 s + 3 s 2 .