MTH5P1: Improper Partial Fractions

This unit explains improper partial fractions

Improper Partial Fractions.

The rational expression to be reduced into partial fractions must have the numerator of lower degree
than the denominator. If this is not the case, we have an improper expression corresponding to a topheavy
or improper fraction in arithmetic.
To convert an improper expression into the required form, we can either try factorising the numerator,
or using long division.

This is an improper expression because the degree of the numerator is not less than that of the
denominator.
Substituting x=1 in the numerator gives a value of 0, therefore (x – 1) is a factor of the numerator by
the factor theorem.

correct form. The full working is given here in Example (2), and the solution is


The working is also identical to that in Example (1).

Occasionally, the following cases turn up in exams, depending on the board and syllabus:
One linear factor and one quadratic factor that cannot be factorised

Two factors in the denominator, but repeated.

Three linear factors in the denominator, all repeated.
The cover-up rule should not be used in any of those cases.
One linear factor and one quadratic factor that cannot be factorised

The quadratic cannot be factorised – in fact it does not even have real solutions.

The partial fraction form of the expression is therefore

(Long working)

Substituting B = 10-A and hence –B = A – 10 in equation 2 gives A-10+C = -11 or A+C = -1.

Adding the equations above gives 4C = -20 and C = -5. From Eqn.(2), A-5 = -1, so A = 4.
Finally, from Eqn.(1), B = 6.


Two factors in the denominator, but repeated.
Here, the method of equating coefficients is the only feasible one.

Three factors in the denominator, but repeated.
Here again, the method of equating coefficients is the only feasible one.

 

ASSIGNMENT : Partial Fractions Assignment MARKS : 30  DURATION : 2 hours

 

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