Expert: Josh - 11/23/2009

Question
QUESTION: Hi Josh,
when equating coefficients, i'm not sure where to start with this

1)   x^3+1
    ----    = Ax +B + C +D
    x^2+1          --  ---
         x  x^2+1


2)  how do i deal with the quadratic of  5x^2 +17x +15
         -------------
         (x+2)^2 (x+1)

i get -1=B  and 3=C  but can not solve for A, as my calculation for A is saying x^2(A+C)which gives me -3




3) x^2 -3x -7          answer 2x+1         3      
  ------------          -------  -  -------
(x^2+x+2)(2x-1)          x^2 +x +2     2x-1



i know some of these answers are pretty long winded,question 2 you can shortcut and just tell me how to solve for A, if i have used the right approach by getting B and C.
The others in more detail if possible.       

Thanks for all the help you can give me,and i hope you can understand this.

ANSWER: Hi Richard,

For Q1, (x^3+1)/(x^2+1) = Ax + B + C/x + D/(x^2+1),
write LHS numerator as (x^3+x-x+1) = x(x^2+1) -x+1.

This gives
[x(x^2+1)-x+1]/(x^2+1) = x + (-x+1)/(x^2+1) = Ax + B + C/x + D/(x^2+1).

Clearly, A=1.
As "(-x+1)/(x^2+1)" on the LHS contains no x^2 term, we can set B=0.
The remaining parts of interest are (-x+1)/(x^2+1) = C/x + D/(x^2+1).

Note: I don't think the RHS is correct. No further factorization is possible over the field of real numbers. We can go further if we work with complex numbers, with "i" defined as sqrt(-1). Then, the roots of x^2+1 are +i and -i.

Consider (-x+1)/(x^2+1) = (-x+1)/[(x-i)(x+i)] = C/(x-i) + D(x+i)
To find C, multiply both sides by (x-i) and evaluate at x=i (chosen as the root of (x-i)).
Substituting x=+i in (-x+1)/(x+i) = C + D(x-i)/(x+i) gives C=(1-i)/(2i)=(-1-i)/2.
(Hint: 1/i = i/(i*i) = -i)
To find D, multiply both sides by (x+i) and evaluate at x=-i.
Substituting x=-i in (-x+1)/(x-i) = C(x+i)/(x-i) + D gives D=(1+i)/(-2i)=(-1+i)/2.

Ans: (x^3+1)/(x^2+1) = Ax + B + C/(x-i) + D/(x+i), where A=1, B=0, C=(-1-i)/2, D=(-1+i)/2.

Q2. In many ways, this is easier. This one does not involve imaginary numbers.
Set this up as (5x^2+17x+15)/[(x+2)^2(x+1)] = A/(x+1) + B/(x+2) + C/[(x+2)^2].
Note: The RHS always takes this form when there is a repeated root like (x+2)^2.

Using the Heaviside method:
To find A, imagine that we multiply both sides by (x+1) and evaluate the LHS expression at x=-1 (again, the root of (x+1)). Proceeding, we evaluate
(5x^2+17x+15)/[(x+2)^2] = A + B(x+1)/(x+2) + C(x+1)/[(x+2)^2] at x=-1.

Observe: This has the net effect of nullifying every term on the RHS, except the coefficient A, and the LHS appears as though the factor of (x+1) has been cancelled in the denominator. For this reason, it's also called the "cover-up method".

This gives A=(5-17+15)/(1)^2=3.

Next, we find C. We multiply both sides by (x+2)^2 and evaluate the LHS expression at x=-2 (the root of (x+2)). The cover up technique also works here. We basically evaluate
(5x^2+17x+15)/(x+1) = A(x+2)^2 + B(x+1)(x+2) + C at x=-2. This gives C=(5*4-34+15)/(-1)=-1.

To find B, write everything on the RHS over a common denominator.
RHS = [A(x+2)^2 + B(x+1)(x+2) + C(x+1)]/[(x+1)(x+2)^2].
Equating numerators on both sides, 5x^2+17x+15 = A(x+2)^2 + B(x+1)(x+2) + C(x+1).
Since A=3 and C=-1, 5x^2+17x+15 = 3(x+2)^2 + B(x+1)(x+2) -(x+1).  call this [#1]
Provided we do not select x=-1, or x=-2, we can find B by substituting any other x value into [#1]. For instance, pick x=0. Then, 15 = 3*4 + 2B -1, this yields B=2.

Ans: (5x^2+17x+15)/[(x+2)^2(x+1)] = 3/(x+1) + 2/(x+2) + -1/[(x+2)^2].


Q3. You can decompose the denominator (x^2+x+2)(2x-1) into three linear factors. In particular, (x^2+x+2) has roots of x=[-1+sqrt(-7)]/2, x=[-1-sqrt(-7)]/2. If we let a=[-1+sqrt(7)i]/2, b=[-1-sqrt(7)i]/2, then the LHS denominator may be written as (x-a)(x-b)(2x-1). The partial fractions on the RHS may be written A/(x-a)+B(x-b)+c/(2x-1). The cover-up method (seen in Q1 and Q2) can again be used to find A, B and C; but A and B may involve complex numbers.

Alternative: set (x^2-3x-7)/(x^2+x+2)(2x-1) = (Ax+B)/(x^2+x+2) + C/(2x-1).
How did we come up with this on the RHS?
Notice that the numerator is a polynomial of degree 2, the denominator is a polynomial of degree 3 [when we multiply (x^2+...)(kx+m)], we cannot extract a constant term in any way. So, something like (Kx^2+Ax+B)/(x^2+x+2) is out of the question. For this reason, we select Ax+B as the numerator for the first fraction on the RHS.

Using the cover-up method (the rationale was explained in Q1), we evaluate (x^2-3x-7)/(x^2+x+2) at x=1/2 to get C=(0.25-1.5-7)/(0.25+2.5)=-3. Then, write RHS as [(Ax+B)(2x-1)+C(x^2+x+2)]/[(2x-1)]. Comparing with LHS, using C=-3,

(x^2-3x-7)=[(Ax+B)(2x-1)-3x^2-3x-6]
         =(2A-3)x^2 + (2B-A-3)x -(B+6)
Equating coefficients, we readily see B=1 [since -7=-(B+6)]. From the leading coefficient (for x^2), 1=2A-3 implies A=2.

Ans: (x^2-3x-7)/(x^2+x+2)(2x-1) = (2x+1)/(x^2+x+2) -3/(2x-1).

These are good representative questions. Knowing how to use the cover-up method and understanding why it is used (see Q2) will help you a long way.

---------- FOLLOW-UP ----------

QUESTION: Hi Josh,

referring back to question 1

lhs numerator.   write LHS numerator as (x^3+x-x+1) = x(x^2+1) -x+1. where does -x+1 come from?  
this makes no sense to me at all,I have reread so many times,,,im sure i can get the rest of the question if i know this by following your instructions.

Q2

i seem to be missing something as i don't still understand how to get A=2

i get -1=B  and 3=C  but can not solve for A,,i understand everything until when it comes to solving A. I done a further question similar to this one,and again i only got 2 out of the 3 answers. Do i take C 3 and subtract from B -1  to get 2? I don't know if im confused because my answers for C&B are different to your A&C when i'm reading it.I set it out different (following a book also) yet i end up with the same answers apart from A.Does this make any sense?

sorry about this.

ps thanks for sending a edited answer to complete square solution.

Answer
Hi Richard,

Once again, I'm glad you keep asking questions. Sometimes, little things stand in the way of understanding the big picture.

Re: (x^3+x-x+1) = x(x^2+1) -x+1.
Ans: Remember that addition is the opposite of subtraction. If I add x to something, then take x away from it, I end up NOT changing anything. i.e., If I add x to (x^3+1), then subtract x from it, I still have (x^3+1).

From x^3+1, I proposed adding x and subtracting x from it. Since this changes nothing overall, we have x^3+1 = x^3+1 +x -x. Now, because additions and subtractions can be done in any order we like (without affecting things), I decided to rewrite x^3+1+x-x as x^3+x -x+1. Nothing new has been introduced, except the components have been reordered. Finally, I factorized the first two terms because they shared "x" as a common factor. It all follows from one line to the next.

x^3+1
= x^3+1 +x-x     [no net changes when we add x, then subtract x from the expression]
= x^3+x -x+1     [nothing new is added, just ordering the operations differently]
= x(x^2+1) -x+1  [pulled "x" out as common factor of first two terms]

It's little things like these that can get us frustrated sometimes. Try and not let this upset you too much.

Q2. You haven't told me what each coefficient sits on top of. I can't see what is written on the RHS of the equation in your book. I have a feeling that you have labelled A, B and C differently in your partial fraction decomposition. This may explain why you have obtained "B=-1, C=3", which would have been wrong given the way we formulated the problem here. If you have chosen to find B and C first for some reason, and a mistake has somehow crept in, then there would be little hope of getting A right. Just so that we are on the same page, can we use the following assumption instead? For the purpose of our discussion, why don't we stick with the following representation.

(5x^2+17x+15)/[(x+2)^2(x+1)] = A/(x+1) + B/(x+2) + C/[(x+2)^2]

If your textbook uses "c" for A, and "b" for C, we can easily change the labels when everything is done. If we can't agree on the same representation, it would be difficult for me to check what you have done, or where you are stuck.

As the original expression contains a repeated factor in the denominator, this is the correct way to set up the problem.

The approach I have shown you last time is the standard technique for finding unknown coefficients in partial fraction decomposition. I recommend that you follow that post closely, go over the comments and arithmetic step-by-step, to make sure that you have understood and executed everything properly. If anything becomes confusing at any point, I would be more than happy to provide you with further explanation. This is really worth persisting because learning this will serve you well in the long run. Referring to my last post, please pay special attention to the "Heaviside method" and "observation". It reveals the magic behind the "cover-up" method.

Given (5x^2+17x+15)/[(x+2)^2(x+1)] = A/(x+1) + B/(x+2) + C/[(x+2)^2],
to find A, we multiply both sides by (x+1). Straight away, we get
(5x^2+17x+15)/[(x+2)^2] = A + B(x+1)/(x+2) + C(x+1)/[(x+2)^2].

Putting x=-1 (the root of (x+1)=0) into this equation, we get
A=(5-17+15)/(1)^2=3.

Comment: I think you have also managed to find C=-1. Refer to my previous post on how to find B. I think you are very close to getting this now.

Cheers.