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Proof of the Partial Fractions Theorem for Quadratic Factors

Date: 10/14/2001 at 00:54:30
From: William
Subject: Proof of the Partial Fractions Theorem for Quadratic Factors


I was reviewing a chapter in my calculus book about Integration using 
Partial Fractions.

The concept seems to be clear for linear factors and repeated linear 
factors, but why is it that when you have a non-reducible quadratic 
factor, you have to let the numerator of the partial fraction be 

Could you please show me a proof or explain me why it is Ax+B instead 
of just A?


Date: 10/14/2001 at 12:30:54
From: Doctor Fenton
Subject: Re: Proof of the Partial Fractions Theorem for Quadratic 

Hi William,

Thanks for writing to Dr. Math.  Essentially, the reason you have to
allow such terms is that the theorem isn't true unless you do. 

Partial fractions "undoes" the operation of adding rational functions 
by finding a common denominator. To decompose P(x)/Q(x), you first 
make sure the rational function is "proper": P(x)/Q(x) is proper if 
the degree of P(x) is strictly less than the degree of Q(x). If the 
degree of P(x) is greater than or equal to the degree of Q(x), you 
carry out long division

          Q(x) | P(x)

getting a quotient of A(x) (a polynomial) and a remainder R(x), whose
degree is (strictly) less than the degree of Q(x), so that

   P(x)          R(x)
   ---- = A(x) + ----
   Q(x)          Q(x) ,

and you apply partial fractions decomposition to R(x)/Q(x).

When decomposing a proper fraction into partial fractions, you must
allow enough proper fractions to occur in the decomposition. For 


is a proper fraction, and you must allow for its occurrence. There is 
no way to write 

       x           A
    --------- = -------
     x^2 + 1    x^2 + 1  .

If this equation were true for all x, you could multiply by x^2+1 
(which is never 0) and get x = A, which means that x can have only one 
value, contradicting the claim that the equation is true for ALL x.

So, if you don't allow for enough types of proper fractions in the
decomposition into partial fractions, you won't be able to decompose
the original rational function. However, you can prove that if you
do allow partial fraction terms of the form

      Ax+B              Ex+F
    --------   and   ----------  (for repeated irreducible factors)
    x^2+Cx+D         (x^2+Cx+D)^k

then the decomposition is always possible.

Does this answer your question adequately?  If you still have 
questions, please write back and I will ry to explain further. 

- Doctor Fenton, The Math Forum   

Date: 10/14/2001 at 20:02:07
From: (anonymous)
Subject: Re: Proof of the Partial Fractions Theorem for Quadratic 


Thanks for your explanation.

I asked my Calculus teacher the same question and he told me that in 
order to understand a formal Proof of the 'Partial Fractions Theorem 
for Irreducible Quadratic Factors' I must have knowledge of Abstract 
Algebra. He told me this proof went beyond the scope of our Calculus 

I would like to see that theorem. I have searched the Internet and I 
haven't been able to find it. Do you think you could show it to me?


Date: 10/15/2001 at 21:01:46
From: Doctor Fenton
Subject: Re: Proof of the Partial Fractions Theorem for Quadratic 

Hi again William,

I looked up the proof in B. L. van der Waerden's _Modern Algebra_, and
I also found one in Birkhoff and MacLane's _A Survey of Modern 
Algebra_. I don't think I had ever actually read a proof of the 
result. Every calculus book I have is content to show the method and 
state that there is a proof. Actually, it's not too hard to follow, 
but it depends upon some division properties of polynomials that may 
not be familiar.  

The first is the Division Algorithm, which says that given two 
polynomials f(x) and g(x), you can divide f(x) by g(x) to get a 
quotient q(x) and a remainder r(x) whose degree is strictly less than 
the degree of the divisor g(x): that is,

    f(x) = g(x)q(x) + r(x), and 0 <= deg(r(x)) < deg(g(x) .

Next, by repeated applications of this algorithm, you obtain the 
Euclidean Algorithm for finding the greatest common divisor d(x) of 
two polynomials f(x) and g(x). The key idea we need is that d(x) can
be written

   d(x) = f(x)r(x) + g(x)s(x) .

If you're not familiar with these ideas, you'll need to read up on 
them. See

   Euclidean and Division Algorithms   

You might also want to review these ideas for integers first: see

   The Official Euclidean Algorithm   


   Explaining the Euclidean Algorithm   

Now, suppose we have a rational function

   ---- ,  where the degree of f is less than the degree of a(x),

where the denominator a(x) factors into a(x) = g(x)h(x), where g(x) 
and h(x) have no common factors. If they have no common factors, then 
their greatest common divisor must be [the constant polynomial] 1, and 
by the Euclidean Algorithm, we can find polynomials c(x) and d(x) such 

         1 = c(x)g(x) + d(x)h(x) .

Multiplying by f(x) gives

       f(x) = [f(x)c(x)]g(x) + [f(x)d(x)]h(x) .

Let G =degree of g(x), H =degree(h), so degree(f) < G + H.  If the 
degree of f(x)c(x) is H or larger, then use the Division Algorithm
to write

         f(x)c(x) = q(x)h(x) + r(x)   where degree(r(x)) < H .

      f(x) = r(x)g(x) + [q(x)g(x) + f(x)d(x)]h(x) .

The degree of f(x) < H + G, and the degree of r(x)g(x) < H + G, so 
the remaining term [q(x)g(x) + f(x)d(x)]h(x) must also have degree 
< H + G . Since h(x) has degree H, [q(x)g(x) + f(x)d(x)] must have 
degree less than G.

All this shows us that we can write

     f(x) = r(x)g(x) + s(x)h(x) , 

with degree(r(x)) < H and degree(s(x)) < G .


    f(x)   r(x)g(x) + s(x)h(x) 
    ---- = -------------------
    a(x)        g(x)h(x) 

           r(x)   s(x)
         = ---- + ----
           h(x)   g(x)     .

That's the basic decomposition. If h(x) or g(x) factors into a product
of polynomials with no common factors, you can repeat this process. 
Notice that the two partial fractions are proper: the numerator has
degree less than the denominator.

You can continue the decomposition until all your fractions are of the 
       ------    ,

because as long as you have different factors in the denominator, you
can keep applying the above decomposition. For repeated factors like
this, suppose degree(p(x))=P, so since this rational function is 
proper, degree(t(x)) < P*k .

Using the Division Algorithm, divide t(x) by p(x)^(k-1) to get

        t(x) = s1(x)p(x)^(k-1) + r1(x), and  degree(r1(x)) < P*(k-1);

then divide r1(x) by p(x)^(k-2):

       r1(x) = s2(x)p(x)^(k-2) + r2(x), with degree(r2(x)) < P*(k-2);

and so on, until you reach

       r(k-2)(x) = s(k-1)(x)p(x) + r(k-1)(x) ,  degree(r(k-1)(x))< P .

Let sk(x) = r(k-1)(x) . Note that all the quotients (the s1(x), s2(x),
,..., etc.) must have degree less than P (compare the degrees on the 
two sides of each equation for the s's).


 t(x) = s1(x)*p(x)^(k-1) + s2(x)*p(x)^(k-2) + ... + s(k-1)(x)p(x)+


  t(x)    s1(x)    s2(x)    s3(x)         s(k-1)(x)   s(k)(x)
 ------ = ----- + ------ + ------ + ... + --------- + -------
 p(x)^k    p(x)   p(x)^2   p(x)^3         p(x)^(k-1)  p(x)^k  .

For real polynomials, the polynomials p(x) must be linear or 
irreducible quadratics.

So there's the proof. If you want to study these issues in more 
detail, the Birkhoff and MacLane book is very readable. If you have 
further questions, please write again.

- Doctor Fenton, The Math Forum   
Associated Topics:
College Modern Algebra

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