Expert: Josh - 12/1/2009

Question
"Hi Josh,



problems with polynomial division, can you help me?



1.      x^3-4x^2+9

       -----------

        x-3



answer x^2-x-3        i get the x^2-x    where does the -3 come from?







2.     x^3-2x^2-4

      -----------

       x-2



answer x^2+2x+2           i get x^2  but don't know how to get

                         +2x+2









and finally



3.      7x^4-10x^3+3x^2+3x-3

       --------------------

              x-1







answer   7x^3-3x^2+3                 how do i get +3, i got the

                                    rest of answer.





any help is greatly appreciated, thanks  "  

Answer
Hi Richard,

To be honest, long division is not easy to do in this format because things don't line up properly. Using pencil and paper, these concepts can be explained a lot quicker. I'll have a go at explaining the procedure. For a more visual approach, you can always look up a "polynomial division" tutorial on the web for more examples on this topic. I hope you can follow what I'm about to do.

Q1. Let me write the question as (x^3-4x^2+9)/(x-3) instead. Recall that (x^3-4x^2+9) is called the dividend, (x-3) is the divisor and the polynomial that we are trying to find which sits on top of )------ is called the quotient. Just repeating, the answer we seek is called the QUOTIENT.

First, identify the dominant term in the divisor. For example, if a polynomial is divided by 2x^3+x-1, then the dominant term is "2x^3", because it corresponds to the highest power. In this question, the dominant term is simply "x". We pay attention to the dominant term, because in working out the quotient, nothing else really matters.

Second, you need to learn how to use subtraction to find the residue in each step. 90% of the time, this part is where people make mistakes.

Notations
=========
Let Q(x) be the quotient (what we seek to find).
Let D(x) be the divisor  (this refers to x-3).
Let P(x) be the dividend (initially same as x^3-4x^2+9).
Let R(x) be the remainder at an intermediate stage.

We begin with P(x)=x^3-4x^2+9.

Step 1:
The initial portion of the quotient Q(x) reads x^2, as x goes into x^3 only x^2 times.
| We calculate x^2*D(x):
| i.e., x^2(x-3)=x^3-3x^2
| Then, subtract this from P(x) to find the remainder
| R(x)=P(x)-(x^3-3x^2)
|     =x^3-4x^2+9 -x^3+3x^2 ......watch out for the signs
|     =(-4+3)x^2+9          ......terms with highest power always cancel
|     =-x^2+9

Step 2:
Let the reduced polynomial P(x)=R(x)=-x^2+9 from the previous step.
Continue to divide P(x) by D(x).
We are only interested in the leading coefficients in P(x) and D(x).
Since -x^2/x gives -x, we update the quotient as Q(x)=x^2-x.
| We calculate x*D(x):
| i.e., -x*(x-3)=-x^2+3x ....again, be careful with minus signs
| Then, subtract this from P(x) to find the remainder
| R(x)=-x^2+9 -(-x^2+3x) ....watch out for the signs
|     =9-3x ....as usual, terms with the highest power cancel out.

Step 3:
Let the reduced polynomial P(x)=R(x) from the previous step.
Following convention, we will write P(x)=-3x+9 (instead of 9-3x).
Continue to divide P(x) by D(x) until it becomes impossible (irreducible).
Again, we only focus on the leading coefficients of P(x) and D(x).
Since -3x/x gives -3, we update the quotient as Q(x)=x^2-x-3.We calculate x*D(x):
| i.e., -3*(x-3)=-3x+9
| Then, subtract this from P(x) to find the remainder
| R(x)=-3x+9-(-3x+9)=0
| There is no remainder. So, we stop here.

Conclusion: (x^3-4x^2+9)/(x-3)=x^2-x-3.

Q2. There's something wrong with the question. I think the dividend is meant to be (x^3-4), NOT x^3-2x^2-4.

Q3. For (7x^4-10x^3+3x^2+3x-3)/(x-1),
let P(x)=7x^4-10x^3+3x^2+3x-3 and D(x)=x-1.

Step 1:
Initial portion of quotient Q(x) reads 7x^3, as x goes into 7x^4 only 7x^3 times.
| We calculate 7x^3*D(x):
| i.e., 7x^3(x-1)=7x^4-7x^3
| Then, subtract this from P(x) to find the remainder
| R(x)=P(x)-(7x^4-7x^3)
|     =7x^4-10x^3+3x^2+3x-3 -7x^4+7x^3
|     =-3x^3+3x^2+3x-3  ......terms with highest power always cancel

Step 2:
Let P(x)=R(x)=-3x^3+3x^2+3x-3 from the previous step.
Continue to divide P(x) by D(x).
We are only interested in the leading coefficients in P(x) and D(x).
Since -3x^3/x gives -3x^2, we update the quotient as Q(x)=7x^3-3x^2.
| We calculate x*D(x):
| i.e., -3x^2*(x-1)=-3x^3+3x^2
| Then, subtract this from P(x) to find the remainder
| R(x)=-3x^3+3x^2+3x-3 -(-3x^3+3x^2)
|     =3x-3 ....[x^3 terms cancel as expected, but x^2 terms also cancel out]

Step 3:
Let the reduced polynomial P(x)=R(x)=3x-3.
Continue to divide P(x) by D(x) until it becomes impossible (irreducible).
Clearly, 3x/x gives 3, we update the quotient as Q(x)=7x^3-3x^2+3.
We calculate x*D(x):
| i.e., 3*(x-1)=3x-3
| R(x)=3x-3-(3x-3)=0
| There is no remainder.

Conclusion: (7x^4-10x^3+3x^2+3x-3)/(x-1)=7x^3-3x^2+3

Note: Division does not always work out perfectly. Sometimes the reminder is non-zero. The procedure stops as long as the degree of the remainder is smaller than the degree of the divisor, D(x).

e.g., consider (x^2+3x+1)/(x+1)

Let P(x)=x^2+3x+1 and D(x)=x+1.
Initial portion of quotient Q(x)=x, since x^2/x=x as far as the leading coefficients are concerned. x(x+1)=x^2+x is subtracted from P(x). This gives a remainder of R(x)=2x+1.
Next, continue to divide 2x+1 by x+1, (think: 2x/x=2) quotient is updated as Q(x)=x+2. We compute 2*P(x), i.e., 2(x+1)=2x+2 is subtracted from 2x+1. This gives a non-zero remainder of 2x+1-(2x+2)=-1. We stop here as the number "-1" or -1*(x^0) represents a constant, and it has a lower degree compared to the divisor D(x)=x+1. [x+1 has degree 1].