Expert: Josh - 2/8/2005

Question
(x^3-1)/(x-1)
I know the answer is x^2 +x +1,however I don't understand how to get that.
One reason is:
when my teacher did this problem on the board, he told us to go like regular long-division and he said to do x into x^3     (ie. x^3/x). My question is what happened to using the -1?That's one of my many problems on this.
I know somewhere along the line I'm missing a point.
Can you please help me.Thanks

Answer
Hi Jeff,

I guess polynomial division is probably a good starting point for understanding this problem. Let me just say right now that the intuition behind polynomial division is pretty much the same as long division with numbers. The only difference is that the "dividend", "divisor", "quotient" and "remainder" are  generally polynomial expressions. So, we need to understand how to measure the "size" of a polynomial.

I'll start by showing you that everyday division with numbers is just a special case of polynomial division.

Consider the following:
33 divided by 4.
Of course, we all know that the answer is 8 and the remainder is 1.
Here, the dividend (the quantity that we try to split) is 32. The divisor (largest unit that we are limiting to) is 4. The quotient is simply the number of whole units that we obtain in the process. Finally, the residue is 33-4*8=1.

In this problem, by construction, I have chosen the numbers carefully, such that 33=32+1=2^5+1, 4=2^2 and 8=2^3. So, I could re-phrase the question and ask you what is the answer of "2^5+1 divided by 2^2" and the answer is "2^3 with a remainder of 1".

To make the polynomial connection really obvious, let x=2.
What is the answer of "x^5-1 divided by x^2, where x=2?"
The answer is x^3, with a remainder of x^0 (recall that anything raised to the power of zero is one).

Lesson 1: In polynomial division, we don't really talk about the size of a polynomial. Instead, we talk about the "degree" of the polynomial. For instance, x^5 is a quintic polynomial (it has degree 5). x^2 is a quadratic polynomial, since it is raised to the power of 2. The quotient, x^3, is a cubic polynomial (with degree 3). Finally, the remainder is a constant in this case, since it is some multiple of x^0.

Lesson 2: To establish the rules of polynomial division, we really don't care if a polynomial contains smaller terms. What I mean is that for instance, x^2+2x+1, x^2, x^2-5x are all considered to be quadratics and therefore have roughly the same magnitude (to a first approximation).

Lesson 3: In general, if asked what polynomial (instead of how many times) a divisor "x^2" will go into "x^5", you would expect the quotient to be of the order "x^3". We deduce this by looking at the degree of the relevant polynomials.
===========================================================
Rule regarding the balance of power:
Degree of dividend = degree of divisor + degree of quotient.
===========================================================
In the example above, 5=2+3. This is the point that your teacher was trying to get across.

Now, in reality, the polynomial expressions are generally more complicated. Let us consider A=D*Q+R, where
A is the dividend,
D is the divisor,
Q is the quotient,
R is the remainder.

Let A=x^3-1, D=x-1 as given in your question.
First consideration - applying the rule on balance of power, we care about nothing else, but the leading term in both the divisor, D (which is "x" here) and the dividend, A (which is "x^3" here).
How many times will "x" go into "x^3".
Of course, the answer is x^2.
Observe that the rule of power balance holds, since 3=1+2, where 3 is the degree of A, 1 is the degree of D and 2 is the degree of the leading term in quotient, Q.

So, the first part of the division yields "x^2".
Comparing with A=D*Q+R, we now know that
(x^3-1)=(x-1)*(x^2+....)+R

The next step is to figure out what the missing bit .... represents. Observe that (x-1)*(x^2)=x^3-x^2. Whereas, originally, A=x^3-1. We have a small problem.
i.e., if we are to enforce that
"x^3-1 equals (x-1)*(x^2+...) + some remainder"
we have got to eliminate the -x^2 term that we have introduced.

Look at our situation,
(ignore the dots, I use them to align things properly)

... x^2
  ___________
x-1)x^3     -1
... x^3-x^2    subtract
  ___________
   A'

we do what we normally do with numbers,
subtract x^3-x^2 from x^3-1, we get an intermediate term
A'=x^2-1

We treat A'=x^2-1 as if it is the dividend and repeat what we have done.

Applying the rule on balance of power, we care about nothing else, but the leading term in both the divisor, D (which is still "x") and the new dividend, A' (which is "x^2" here).
How many times will "x" go into "x^2".
Of course, the answer is x.

Observe that the rule of power balance holds, since 2=1+1, where 2 is the degree of A', 1 is the degree of D and 1 is the degree of the second term in quotient, Q.

So, the second part of the division yields "x".
[more generally, we can expect multiples of x, but we need not worry about that here, as things balance out]

Comparing with A=D*Q+R, we now know that
(x^3-1)=(x-1)*(x^2+x+...)+R

The last step is to figure out what the missing bit ... represents. Observe that (x-1)*(x^2+x) = x^3-x^2 +(x^2-x)=x^3-x . Whereas, originally, A=x^3-1. We have a small problem.
i.e., if we are to enforce that
"x^3-1 equals (x-1)*(x^2+x+...) + some remainder"
we have got to eliminate the -x term that we have introduced.

Look at the latest situation,
(again ignore the dots)

... x^2+x
  _____________
x-1)x^3       -1
... x^3-x^2
..._____________
........x^2    -1
........x^2 -x    subtract from prev. row
..._____________
........     A"

we do what we normally do with numbers,
subtract x^2-x from A'=x^2-1, we get an intermediate term
A"=x-1

We treat A"=x-1 as if it is the dividend and repeat what we have done.

This time, quite clearly, x-1 goes into x-1 once. I won't bother with the balance of power, since it is immediately obvious.

So, the third term in the quotient is "1".

Summing up, we have

... x^2 + x + 1
  _____________
x-1)x^3       -1
... x^3-x^2
..._____________
........x^2    -1
........x^2 -x    subtract from prev. row
..._____________
........     x-1
........     x-1
..._____________
              0

The remainder R is zero, so, (x-1) divides (x^3-1) perfectly. A=D*Q+R is in fact,
x^3-1=(x-1)(x^2+x+1).

Useful result to remember,
x^3+1=(x+1)(x^2-x+1)

I hope this clears things up a bit.

CHeers