6 Improper fractions

When calculating the partial fractions of improper fractions an extra polynomial is added to any partial fractions that would normally arise. The added polynomial has degree n − d where n is the degree of the numerator and d is the degree of the denominator. Recall that

a polynomial of degree 0 is a constant, A say,

a polynomial of degree 1 has the form A x + B ,

a polynomial of degree 2 has the form A x 2 + B x + C ,

and so on.

If, for example, the improper fraction is such that the numerator has degree 5 and the denominator has degree 3, then n − d = 2 , and we need to add a polynomial of the form A x 2 + B x + C .

Key Point 17

If a fraction is improper an additional term is included taking the form of a polynomial of degree n − d , where n is the degree of the numerator and d is the degree of the denominator.

Example 42

Express as partial fractions

2 x 2 − x − 2 x + 1

 

Solution

The fraction is improper because n = 2 , d = 1 and so d ≤ n . Here n − d = 1 , so we need to include as an extra term a polynomial of the form B x + C , in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction A x + 1 . So altogther we have

2 x 2 − x − 2 x + 1 = A x + 1 + ( B x + C )

 Multiplying both sides by x + 1 we find

2 x 2 − x − 2 = A + ( B x + C ) ( x + 1 ) = B x 2 + ( C + B ) x + ( C + A )

 Equating coefficients of x 2 gives B = 2 .

Equating coefficients of x gives − 1 = C + B and so C = − 1 − B = − 3 .

Equating the constant terms gives − 2 = C + A and so A = − 2 − C = − 2 − ( − 3 ) = 1 .

Finally, we have

2 x 2 − x − 2 x + 1 = 1 x + 1 + 2 x − 3

 

Exercise

Express each of the following improper fractions in terms of partial fractions.

  1. x + 3 x + 2 ,
  2. 3 x − 7 x − 3 ,
  3. x 2 + 2 x + 2 x + 1 ,
  4. 2 x 2 + 7 x + 7 x + 2
  5. 3 x 5 + 4 x 4 − 21 x 3 − 40 x 2 − 24 x − 29 ( x + 2 ) 2 ( x − 3 ) ,
  6. 4 x 5 + 8 x 4 + 23 x 3 + 27 x 2 + 25 x + 9 ( x 2 + x + 1 ) ( 2 x + 1 )
  1. 1 + 1 x + 2 ’
  2. 3 + 2 x − 3 ,
  3. 1 + x + 1 x + 1
  4. 2 x + 3 + 1 x + 2 ,
  5. 1 ( x + 2 ) 2 + 1 x + 2 + 1 x − 3 + 3 x 2 + x + 2 ,
  6. 2 x 2 + x + 7 + 1 2 x + 1 + 1 x 2 + x + 1