1.4 PARTIAL FRACTIONS
Aibut linear functions ofx,i.e.oftheformBix+Ci. Thus, in the expansion,
linear terms (first-degree polynomials) in the denominator have constants (zero-
degree polynomials) in their numerators, whilst quadratic terms (second-degree
polynomials) in the denominator have linear terms (first-degree polynomials) in
their numerators. As a symbolic formula, the partial fraction expansion of
g(x)
(x−α 1 )(x−α 2 )···(x−αp)(x^2 +a^21 )(x^2 +a^22 )···(x^2 +a^2 q)should take the form
A 1
x−α 1+A 2
x−α 2+···+Ap
x−αp+B 1 x+C 1
x^2 +a^21+B 2 x+C 2
x^2 +a^22+···+Bqx+Cq
x^2 +a^2 q.Of course, the degree ofg(x) must be less thanp+2q; if it is not, an initial
division must be carried out as demonstrated earlier.
Repeated factors in the denominatorConsider trying (incorrectly) to expand
f(x)=x− 4
(x+1)(x−2)^2in partial fraction form as follows:
x− 4
(x+1)(x−2)^2=A 1
x+1+A 2
(x−2)^2.Multiplying both sides of this supposed equality by (x+1)(x−2)^2 produces an
equation whose LHS is linear inx, whilst its RHS is quadratic. This is clearly
wrong and so an expansion in the above form cannot be valid. The correction we
must make is very similar to that needed in the previous subsection, namely that
since (x−2)^2 is a quadratic polynomial the numerator of the term containing it
must be a first-degree polynomial, and not simply a constant.
The correct form for the part of the expansion containing the doubly repeatedroot is therefore (Bx+C)/(x−2)^2. Using this form and either of methods (i) and
(ii) for determining the constants gives the full partial fraction expansion as
x− 4
(x+1)(x−2)^2=−5
9(x+1)+5 x− 16
9(x−2)^2,as the reader may verify.
Since any term of the form (Bx+C)/(x−α)^2 can be written asB(x−α)+C+Bα
(x−α)^2=B
x−α+C+Bα
(x−α)^2,and similarly for multiply repeated roots, an alternative form for the part of the
partial fraction expansion containing a repeated rootαis
D 1
x−α+D 2
(x−α)^2+···+Dp
(x−α)p. (1.48)