 # Polynomial Division 3: Applications To Arithmetic Division

This post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! Thank you!

In the previous lesson, we saw how we can extend the method we developed for polynomial division by linear divisors in this earlier lesson to divisors of higher degree. In this lesson, we will now look at some applications of this methodology for arithmetic division.

Before we proceed, we need to be clear that arithmetic division is a messy problem. We have dealt with arithmetic division using different methods in some previous lessons (here, here and here). Using polynomial division techniques for arithmetic division is just one more tool to tackle division that we can add to our arsenal. It is not a magic bullet that is going to make division trivial. But, the more tools we have in our arsenal, the better off we are.

As in this earlier lesson, using polynomial division techniques for arithmetic division relies on our ability to express numbers as polynomials. There is not a single way to express numbers as polynomials, but a multitude of them. In fact, each numeric base will result in a different polynomial expression of a number. The most common numeric base, of course, is 10. Most of this lesson will therefore rely on using 10 as a base in converting numbers to polynomials, but there is no rule that says the base has to be 10. We can use any other power of 10 or any random number such as 2, 3, 14 or whatever else we think will make the problem easier to tackle.

Let us first tackle the simple case of dividing 32895 by 123. It is easy to see that this can be rewritten as dividing 3x^4 + 2x^3 + 8x^2 + 9x + 5 by x^2 + 2x + 3, where x = 10 (thus we have used a base of 10 to convert the numbers into polynomials). Since the divisor contains 3 terms, we need to arrange our dividend into groups of three terms. We also notice that the ratio of coefficients in the divisor is 1:2:3. So, the ratio of coefficients in each of the groups of three terms in the dividend has to 1:2:3 also. Based on this, we see that the following expression of the dividend will work:

3x^4 + 6x^3 + 9x^2
– 4x^3 – 8x^2 – 12x
+ 7x^2 + 14x + 21
+ 7x – 16

This will then lead to the following factorization:

3x^2(x^2 + 2x + 3) – 4x(x^2 + 2x + 3) + 7(x^2 + 2x + 3) + 7x – 16

We can then collect the terms and conclude that the quotient is 3x^2 – 4x + 7, which is simply 300 – 40 + 7 = 267. Similarly the remainder is 7x – 16, which is simply 70 – 16 = 54. We can easily check to make sure that 267*123 + 54 = 32895, so we have performed the division correctly.

Consider another problem, 38979 divided by 132. Using x = 10 as our base, we can express this as the polynomial division of 3x^4 + 8x^3 + 9x^2 + 7x + 9 by x^2 + 3x + 2. We need to express the dividend as groups of three terms, with the coefficients within each group being in the ration 1:3:2. We then write the dividend as below for factorization:

3x^4 + 9x^3 + 6x^2
– x^3 – 3x^2 – 2x
+ 6x^2 + 18x + 12
– 9x – 3

Factorization of the dividend gives us:

3x^2(x^2 + 3x + 2) -x(x^2 + 3x + 2) + 6(x^2 + 3x + 2) – 9x – 3

This will give us a quotient of 3x^2 – x + 6, which is 300 – 10 + 6 = 296. We get a remainder of -9x – 3 = -93. We see that the remainder is negative. This is not the standard way in which the results of a division problem are presented. But the way around it is quite simple. We simply have subtract one from the quotient, and add the divisor to the remainder. Repeat this process until the remainder becomes positive. This procedure then gives us a final quotient of 295, and a final remainder of 39. I encourage you to verify that the answer derived here is indeed correct.

Now consider the division of 56424 by 1421. We can express this as the division of 5x^4 + 6x^3 + 4x^2 + 2x + 4 by x^3 + 4x^2 + 2x + 1. This will require the expression of the dividend in the form of groups of 4 terms, whose coefficients are in the ratio 1:4:2:1. This can be accomplished by expressing the dividend as below:

5x^4 + 20x^3 + 10x^2 + 5x
– 14x^3 – 56x^2 – 28x – 14
+ 50x^2 + 25x + 18

This can be factorized as:

5x(x^3 + 4x^2 + 2x + 1) -14(x^3 + 4x^2 + 2x + 1) + 50x^2 + 25x + 18. This yields us a quotient of 5x – 14, which is 36 when x = 10. We also get a remainder of 50x^2 + 25x + 18, which is 5268 when x = 10. Obviously, this is not the answer expressed in standard form since the remainder is much higher than the divisor. However, the situation is easy to correct. All we have to do is subtract the divisor from the remainder and add 1 to the quotient. Repeat until the remainder becomes smaller than the divisor. Following that procedure, we get a final quotient of 39, and a final remainder of 1005. You can verify that this indeed the correct answer.

Now, let us consider 435323 divided by 212. Once again, we can express this as 4x^5 + 3x^4 + 5x^3 + 3x^2 + 2x + 3 divided by 2x^2 + x + 2, where x = 10. We once again have to arrange the dividend in groups of 3, and the coefficients within each group have to be in the ratio 2:1:2. Based on this, we rewrite the dividend as below:

4x^5 + 2x^4 + 4x^3
+ x^4 + 0.5x^3 + x^2
+ 0.5x^3 + 0.25x^2 + 0.5x
+ 1.75x^2 + 0.875x + 1.75
+ 0.625x + 1.25

As you can see this is getting a little messy, but let us proceed with the factorization as below:

2x^3(2x^2 + x + 2) + 0.5x^2(2x^2 + x + 2) + 0.25x(2x^2 + x + 2) + 0.875(2x^2 + x + 2) + 0.625x + 1.25

This then gives us a quotient of 2x^3 + 0.5x^2 + 0.25x + 0.875. Substituting x = 10 in this expression gives us 2000 + 50 + 2.5 + 0.875 = 2053.375. The remainder is 0.625x + 1.25, which is 6.25 + 1.25 = 7.5, when x = 10. This answer is correct, but it is obviously not in standard form, where we expect our quotients and remainders to be whole numbers. To get rid of the fractional part from the quotient, let us take the fractional part, 0.375 (which is 3/8), multiply it by our divisor, 212, and add it to our remainder. This then gives us a final quotient of 2053, and final remainder of 7.5 + 212*3/8 = 87. We can verify that 2053*212 + 87 = 435323. So, not only was the division performed correctly, but we also managed to get the results into standard form without much difficulty.

Now, let us see how we can use polynomial division techniques to perform the division of 369670 by 1212. There are actually two ways to tackle this problem. The more traditional way is to consider this to be the division of 3x^5 + 6x^4 + 9x^3 + 6x^2 + 7x by x^3 + 2x^2 + x + 2. This would necessitate the expression of the dividend as groups of 4 terms, whose coefficients are in the ratio 1:2:1:2. This is accomplished by writing the dividend as below:

3x^5 + 6x^4 + 3x^3 + 6x^2
+ 0x^4 + 0x^3 + 0x^2 + 0x (note that this line can be skipped entirely)
+ 6x^3 + 12x^2 + 6x + 12
– 12x^2 + x – 12

This can be factorized as:

3x^2(x^3 + 2x^2 + x + 2) + 6(x^3 + 2x^2 + x + 2) – 12x^2 + x – 12

This gives us a quotient of 3x^2 + 6, which is 300 + 6 = 306, and a remainder of -12x^2 + x – 12, which is -1200 + 10 – 12 = -1202. Since the remainder is negative, we add the divisor to it and subtract 1 from the quotient. This gives us a quotient of 305, and a remainder of 10. One can verify that this is the correct answer.

But there is another way to approach this problem using the principle of polynomial division. We can also express it as the division of 36x^2 + 96x + 70 by 12x + 12, where x = 100. Now, we see that the divisor terms are in the ratio 1:1, and the divisor terms have a common factor of 12. We can choose to take care of the common factor in two different ways.

In the first case, let us divide by the dividend terms by the common factor before we factorize. This gives us:

3x^2 + 8x + 70/12

We then express the dividend in groups of two terms, with the coefficients in each group being the ratio 1:1. This gives us:

3x^2 + 3x
+ 5x + 5
+ 10/12

This can be factorized as:

3x(x + 1) + 5(x + 1) + 10/12

This gives us a quotient of 3x + 5, and remainder of (10/12)*12 = 10 (remember that in this method, the remainder has to be multiplied by the common factor to get the final remainder as explained this earlier lesson). Substituting x = 100 in the quotient expression, we see that we once again get a quotient of 305 and a remainder of 10.

The other way to handle the common factor is to ignore it initially, and divide the quotient by 12 after the factorization (remember not to divide the remainder by the common factor if you do it this way, again as explained in this earlier lesson). Using this technique, we can rewrite the dividend as below:

36x^2 + 36x
+ 60x + 60
+ 10

This then factorizes out as below:

36x(x + 1) + 60(x + 1) + 10

This then gives us a quotient of 36x + 60 and a remainder of 10. We divide this quotient by the common factor, 12, to get a final quotient of 3x + 5, which is equal to 305 when x = 100. The final remainder is 10, as before.

The example above is a demonstration of how changing the base in which we do our calculations can sometimes lead to shorter, easier and quicker calculations.

For an example of how to use vinculums (dealt with in this earlier lesson) to make some division problems easier than they would be otherwise, please read my full lesson here. The use of such creative techniques is sometimes necessary to solve division problems if we don’t want to use brute force. Keeping these various techniques in mind and applying them at the correct time under the right conditions is essential when you want to solve problems mentally, and the first technique that comes to mind is not the ideal method to tackle the problem with.

Hope this lesson has provided you with a wide variety of examples of solving arithmetic problems using polynomial division techniques. As mentioned at the beginning of this lesson, division is never a trivial problem, and it is impossible to make it universally easy in any system of mathematics. However, having a large arsenal of tools to rely on and using the right tool at the right time is the secret to tackling them effectively. Good luck, and happy computing!