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In the previous lesson, we introduced the concept of polynomial division and how it relates to arithmetic division. We also saw how we can use the Madhyamadhyena Adhyamanthyena sutra to rewrite the dividend so that we can factor out the divisor and get a quotient and remainder. In this lesson, we will extend the method we developed, to apply it to divisors that are not linear.

How do we use the method developed in the previous lesson to perform division when the divisor contains more than 2 terms, and the unknown term is raised to powers higher than 1? One obvious way to do this would be to factorize the divisor into as many linear factors as necessary and then divide by each sequentially. This method requires no extension of the methodology developed in the previous lesson.

However, this may not be possible for several reasons. First of all, it may not be possible to factorize the divisor into linear factors easily. Moreover, even if that were possible or easy, it becomes messy to deal with multiple remainders from multiple steps of the operation. So, how do we perform the division directly in one step? To answer this question, it is better to work out an example rather than trying to explain the procedure in abstract mathematical terms.

Therefore, let us try the division of (3x^4 + 4x^3 + 7x^2 + 5x + 4) by (x^2 + x + 1). Notice that the divisor is a quadratic expression (polynomial of degree 2). Notice also that it has three terms, but in general quadratic expressions need not have 3 terms. But a full quadratic expression will have 3 terms. The three is important because it tells us that the dividend has to be arranged in sets of three terms for the division. Lastly, notice that the divisor really has no linear factors that use only real numbers (the quadratic expression itself has only complex roots).

Notice that the coefficients of the three terms in the divisor are in the ratio 1:1:1. This tells us that the sets we derive for division of the dividend have to be in that ratio just as in linear division. But, what do we mean by sets of three terms? What we mean is that the dividend has to be arranged as below:

3x^4 + ax^3 + bx^2 + cx^3 + dx^2 + ex + fx^2 + gx + h + jx + k

Our first set of three terms starts with the x^4 term, and contains the next two lower powers of x (x^3 and x^2). Our next set of 3 terms starts with an x^3 term, and similarly contains the next two lower powers of x (x^2 and x). Our next set of 3 terms starts with an x^2 term, and contains an x term and a constant. We stop making sets when the last set contains a constant term. We will notice that this last set will also start with a term that is of the same degree as the divisor (in this case, our divisor is of degree 2, and our last set starts with an x^2 term). The sets are shown in separate colors above.

After the last set, we write our reminder terms. Note that in general, the remainder will always be a polynomial of degree 1 lower than the divisor. Thus, in this case, jx + k is the remainder and it is of degree 1 whereas our divisor is of degree 2. One can consider the remainder term to be an incomplete set of 3 terms starting with an x term (the set contains only 2 terms, and hence is incomplete).

Notice that this method is what was used in the case of division by linear divisors also in the previous lesson. In that case, each set contained 2 terms, with the second term of each set being the same degree as the first term of the next set. This positioning of terms of similar degree next to each other masked the fact that we were creating sets of 2 terms, just as we are creating sets of 3 terms here.

Now that we have decided what the sets should be, what should the coefficients a, b, etc., be? This is where the ratio of coefficients in the divisor comes into play. The coefficients in each set should be in the same ratio as they are in the divisor. Thus, in the expression we wrote above, we can actually impose the following conditions:

3/a = 1/1
a/b = 1/1
c/d = 1/1
d/e = 1/1
f/g = 1/1
g/h = 1/1

These establish the ratios in the two sets of terms we created. In addition, we also have to make sure that the like terms can be added to give us the original dividend. So, we write the following equations also:

a + c = 4
b + d + f = 7
e + g + j = 5
h + k = 4

Working from the very first ratio, we can mechanically satisfy the ratios and sums by rewriting the dividend as below:

3x^4 + 3x^3 + 3x^2 + x^3 + x^2 + x + 3x^2 + 3x + 3 + x + 1

We can now factorize it as below:

3x^2(x^2 + x + 1) + x(x^2 + x + 1) + 3(x^2 + x + 1) + x + 1

This then tells us that our quotient is 3x^2 + x + 1, and our remainder is 3x + 3. One can indeed verify that (3x^2 + x + 3)(x^2 + x + 1) + x + 1 = 3x^4 + 4x^3 + 7x^2 + 5x + 4. Thus, our division is correct.

Just as in the case of division by a linear polynomial, note that it is possible for any of the coefficients of the quotient or remainder to be fractional or zero. In addition, some of the coefficients, including the remainder terms may be negative too. We will encounter some such cases in the next couple of examples.

Next let us try 5x^5 + 4x^3 + 3x^2 + 6x + 3 divided by 2x^2 + x + 2. Here, we note that the dividend is missing the x^4 term. This does not mean that we leave it out of the factorization entirely. It just means that the sum of all x^4 terms in the rewritten dividend expression should be 0.

Once again, we notice that the divisor is of degree 2, and has 3 terms. So, we need to make sets of 3, and the remainder will contain an x term and a constant. The dividend is rewritten as below:

5x^5 + ax^4 + bx^3 + cx^4 + dx^3 + ex^2 + fx^3 + gx^2 + hx + jx^2 + kx + m + nx + p

The different sets of 3 terms have been colored differently to make them easier to identify. Since the terms in the divisor are in the ration 1:2:1, we need to make the terms in each set the same ratio too. In addition, we impose the following conditions:

a + c = 0
b + d + f = 4
e + g + j = 3
h + k + n = 6
m + p = 3

This then gives us the following rewritten equation and its factorization:

5x^5 + 10x^4 + 5x^3 -10x^4 – 20x^3 – 10x^2 + 19x^3 + 38x^2 + 19x – 25x^2 – 50x – 25 + 37x + 28, which can be factorized as
5x^3(x^2 + 2x + 1) – 10x^2(x^2 + 2x + 1) + 19x(x^2 + 2x + 1) – 25(x^2 + 2x + 1) + 37x + 28

Thus, the quotient is (5x^3 – 10x^2 + 19x – 25), and the remainder is 37x + 28. It is easy to verify that (5x^3 – 10x^2 + 19x – 25)(x^2 + 2x + 1) + 37x + 28 = 5x^5 + 4x^3 + 3x^2 + 6x + 3. Notice that the quotient is of degree 3, which is what we expect given that the dividend is of degree 5 and the divisor is of degree 2 (5 – 2 = 3).

Now let us divide 4x^3 – 7x^2 + 15x – 12 by 2x^2 – 3. In this case, notice that the divisor is of degree 2, but has only 2 terms. It is missing the x term (therefore the coefficient of the x term in the divisor is zero). Note that if the divisor were missing the constant, then x would be a common factor in the divisor and we could divide by x separately before dividing by the rest of the divisor if we chose. But it is not necessary to factor out the divisor in any case. Simply replace the missing term by 0, and proceed as usual.

Thus, the above division is actually the division of 4x^3 – 7x^2 + 15x – 12 by 2x^2 + 0x – 3. Thus, we see that we have to create sets of 3 in the dividend, and the coefficients have to be in the ratio 2:0:-3. This is shown below:

4x^3 + 0x^2 – 6x – 7x^2 – 0x + 10.5 + 21x – 22.5, which can be factorized as
2x(2x^2 – 3) – 3.5(2x^2 – 3) + 24x – 22.5

Thus, the quotient is (2x – 3.5), and the remainder is 21x – 22.5. We see that both the quotient and remainder can have fractional terms, as mentioned before.

Note that if the dividend expression is of lower degree than the divisor, the quotient is zero and the dividend becomes the remainder. For instance, if we attempt division of 3x^2 + 4x – 8 by x^3 + 4x^2 – 6x + 4, then the quotient is zero, and the remainder would be the dividend itself, which is 3x^2 + 4x – 8.

Note also that the same rules as we derived in the previous lesson apply regarding divisions in which the divisor has a constant common factor. Consider, for example, the division of 5x^4 + 7x^3 – 6x^2 + 3x – 5 by 2x^2 – 4x + 2. In this case, the common factor in the divisor is 2. There are two alternative approaches to performing this division. We can divide each term in the dividend by the common factor, perform the division as usual, and then multiply the remainder by the common factor at the end. Or, we can divide first, then divide the quotient by 2 while leaving the remainder as is. Let us consider each of these methods in turn below.

Using the first method, we find the ratio of terms in the divisor is 1:-2:1. We change the dividend as below, by dividing by the common factor in the divisor, 2:

2.5x^4 + 3.5x^3 – 3x^2 + 1.5x – 2.5

We then create sets of 3 terms as below for factorization:

2.5x^4 – 5x^3 + 2.5x^2 + 8.5x^3 – 17x^2 + 8.5x + 11.5x^2 – 23x + 11.5 + 16x -14, which can be factorized as
2.5x^2(x^2 – 2x + 1) + 8.5x(x^2 – 2x + 1) + 11.5(x^2 -2x + 1) + 16x – 14

This tells us that the quotient is 2.5x^2 + 8.5x + 11.5. The remainder is twice the remainder in the above line since we have to multiply it by the common factor in the divisor. So, we get a remainder of 32x – 28.

Now, let us perform the division the other way. We retain the dividend as it is, but rewrite it for factorization as below:

5x^4 – 10x^3 + 5x^2 + 17x^3 – 34x^2 + 17x + 23x^2 – 46x + 23 + 32x – 28 which can be factorized as
5x^2(x^2 – 2x + 1) + 17x(x^2 – 2x + 1) + 23(x^2 – 2x + 1) + 32x – 28

We can then conclude that the quotient is one half of 5x^2 + 17x + 23, which is 2.5x^2 + 8.5x + 11.5, and the remainder is 32x – 28. Notice that the two solutions are identical in the end.

For an example of dividing by a divisor of higher degree than 2, please read the full lesson here. The methodology is the same, but there are obviously more terms involved, so it is important to be organized about it when doing it to avoid errors.

Hopefully, these examples have provided a good insight into how these problems need to be approached and tackled. Obviously, things are relatively easy when the ratios in the divisors are easier to deal with rather than being a wild mix of numbers. The principle still does not change, but the difficulty of performing the arithmetic operations required to implement the method will obviously increase as the ratios become more and more unwieldy. Unfortunately, that is something no system of mathematics can alleviate!

We have not dealt with the extension of this methodology to deal with arithmetic division in this lesson. We will do this in the next lesson. In the meantime, hope you will find the time to practice the kinds of divisions illustrated here so that its application to arithmetic division will be much easier. Good luck, and happy computing!

– The Vedic Maths Forum India