PRELIMINARY ALGEBRA
We illustrate this procedure with the following worked example.
Find the partial fraction decomposition of the function
f(x)=
x^3 +3x^2 +2x+1
x^2 −x− 6
.
Since the degree of the numerator is 3 and that of the denominator is 2, a preliminary
long division is necessary. The polynomials(x) resulting from the division will have degree
3 −2 = 1 and the remainderr(x) will be of degree 2−1 = 1 (or less). Thus we write
x^3 +3x^2 +2x+1=(s 1 x+s 0 )(x^2 −x−6) + (r 1 x+r 0 ).
From equating the coefficients of the various powers ofxon the two sides of the equation,
starting with the highest, we now obtain the simultaneous equations
1=s 1 ,
3=s 0 −s 1 ,
2=−s 0 − 6 s 1 +r 1 ,
1=− 6 s 0 +r 0.
These are readily solved, in the given order, to yields 1 =1,s 0 =4,r 1 =12andr 0 = 25.
Thusf(x) can be written as
f(x)=x+4+
12 x+25
x^2 −x− 6
.
The last term can now be decomposed into partial fractions as previously. The zeros of
the denominator are atx=3andx=−2 and the application of any method from the
previous subsection yields the respective constants asA 1 =12^15 andA 2 =−^15. Thus the
final partial fraction decomposition off(x)is
x+4+
61
5(x−3)
−
1
5(x+2)
.
Factors of the forma^2 +x^2 in the denominator
We have so far assumed that the roots ofh(x) = 0, needed for the factorisation of
the denominator off(x), can always be found. In principle they always can but
in some cases they are not real. Consider, for example, attempting to express in
partial fractions a polynomial ratio whose denominator ish(x)=x^3 −x^2 +2x−2.
Clearlyx= 1 is a zero ofh(x), and so a first factorisation is (x−1)(x^2 +2).
However we cannot make any further progress because the factorx^2 + 2 cannot
be expressed as (x−α)(x−β)foranyrealαandβ.
Complex numbers are introduced later in this book (chapter 3) and, when the
reader has studied them, he or she may wish to justify the procedure set out
below. It can be shown to be equivalent to that already given, but the zeros of
h(x) are now allowed to be complex and terms that are complex conjugates of
each other are combined to leave only real terms.
Since quadratic factors of the forma^2 +x^2 that appear inh(x) cannot be reduced
to the product of two linear factors, partial fraction expansions including them
need to have numerators in the corresponding terms that are not simply constants