Issue 137 - Polynomial Osculation

A warm welcome to our new subscribers.

This issue’s article is titled POLYNOMIAL OSCULATION and describes how Tirthaji’s osculation technique can be applied to polynomials.
“When applied to polynomials however it not only detects factors but gives the quotient as well. It can therefore give two factors simultaneously”.





This was held last Oct 7 and was a great success.

“We are very happy to share the news with you that we had over 4,800 participants. This just goes to show that VM’s popularity is fast spreading here.”


on Saturday, 25th November, 2023.

International Vedic Mathematics Olympiad 2023 registrations are open.

Teachers and students are invited to participate...

An international competition in speed, problem-solving, mathematical acumen and Vedic Maths techniques.

IVMO consists of one hour exam-like papers that test ability and speed in using Vedic Maths techniques and their applications to problem-solving. Students who do not know the Vedic techniques may find difficulty in completing all the questions in the time allocated but are welcome to have a go.
Any queries , please write to us :


There are two new articles are accessible here.
Item 61: "Polynomial Osculation"
by K. Williams

Two interlinked methods of factorising polynomials are shown. How Tirthaji’s osculation technique can be applied to the factorisation of polynomials, and how if the result of the osculation is not zero that result can itself be used to factorise the polynomial.

Item 60: "Proof of Nikhilam Sutra by using modular arithmetic" by Ranjeet Kumar, K. Senthil, S. Mitra, Amitava Roy, R. I. Bakhtsingh, 2023.
The range of application of the Nikhilam Sutra is explored and proved for general modular number systems.



Dr Kapoor has prepared:

  1. Index of Lessons of my V M Courses 1 to 9.   

It runs over 30 pages.   

It is uploaded on website with location

  1. Index of all material of my website. 

It runs over 285 pages. (Part -1   61 pages and Part-2  224 pages).   

It is uploaded with location: 



Here is the item from Newsletter 136 (in red) followed by my solution.

I recently had a request to factor a 7th power polynomial equation, ‘Vedic-style’. I sent back my answer, and the next day, wondering how the method used would extend generally I looked online for some polynomials to solve, and came across the exact same septic I had been sent the day before. There were several solutions, all very long and cumbersome. The polynomial is x7 + x2 + 1. If you can factor this easily please send your solution. We can publish it, and mine, in the next newsletter.

My solution, in brief:

Putting x = 2 in the septic we obtain 27 + 22 + 1 = 133 = 7 × 19.
In base 2, 7 is written 111, which is x2 + x + 1.
The other factor is obtained by division or by expressing 19 in base 2.

We find x7 + x2 + 1 = (x2 + x + 1)(x5 – x4 + x2 – x + 1).
Since 7 and 19 are prime there can only be 2 factors in this case and so no further factorisation is possible.
The same result is obtained using base 10.

The rationale behind this is given by the Specific and General Sutra. The septic and its equivalent factorised form will be true in any base (i.e. for any value of x), so choosing base 2 and factorising the result, we then put it back into general form. This approach to factorising polynomials is developed more fully in my recent paper.




Muthuselvi Prabhu: “I conducted the first Math competition in my area yesterday in the name of Infinity 2023. More than 20 students participated from 4 schools, and most of the students are interested to learn Vedic Math and parents got an idea about Vedic Math.  I hope this event created an awareness about Vedic Math.”

-          Please send us accounts of any talks, courses etc. you give so that we can mention them in this newsletter.



Here is the item from Newsletter 134 (in blue) followed by my solution.

Chapter 37 Of Bharati Krishna Tirthaji’s book “Vedic Mathematics” shows five proofs of Pythagoras’ Theorem. As is normal with mathematical proofs each of these rests on some simpler known result.

For the fifth proof Tirthaji writes:

“This proof is from Co-ordinate Geometry. And, as modern Conics and Co-ordinate Geometry (and even Trigonometry) take their genesis from Pythagoras' Theorem, this process would be objectionable to the modern mathematician. But, as the Vedic Sutras establish their Conics and Co-ordinate Geometry (and even their Calculus), at a very early stage, on the basis of first principles and not from Pythagoras' Theorem (sic), no such objection can hold good in this case.

The proposition is the one which gives us the distance between two points whose co-ordinates have been given. Let the points be A and B and let their co-ordinates be (a, 0) and (0, b) respectively.

Then, BA = √[(a-0)2 + (0-b)2] = √(a2 + b2), therefore BA2 = a2  + b2  Q.E.D.”

Since the usual derivation for the formula for the distance between two points in coordinate geometry relies on Pythagoras’ Theorem, Tirthaji’s proof may appear at first sight to be contradictory.

The explanation, implied by the text, is that the formula for the distance between two points can be derived independently of Pythagoras’ Theorem.

It would be interesting to find a derivation of the formula Tirthaji quotes using coordinate geometry and without invoking Pythagoras’ Theorem.

This is an interesting challenge that you may wish to consider. Let us know if you have a solution.


Suppose we want the distance D between (A,B) and (a,b).

We need to prove that D2 = (A – a)2 + (B – b)2 =  p2  – q2 say,  using coordinate geometry and without using Pythagoras’ Theorem.

Suppose a |  p    q    D   is a triple with p as base, q as height and D as hypotenuse.

It is assumed in the following that the reader is familiar with addition and subtraction of triples. Those operations do not depend on Pythagoras’ Theorem.

Then, using triple notation:            a |    p        q     D
                                               2a | p2-q2   2pq    D2


And using triple subtraction:       2a | p2-q2                       2pq                  D2                                  [1]
                                              a |    p                             q                   D   –                               [2]
                                              a |    p3-pq2 + 2pq2   2p2q-p2q + q3          D3                                  [3]
                                                    =  p3+pq2                p2q + q3             D3                                  [4]

We equate elements in [2] and [4] above by multiplying [2] by D2 so that they have the same hypotenuse.

Then pD2 = p3+pq2  and  qD2 = p2q + q3

Both of which lead to D2 = p2 + q2.

Hence D2 = (A – a)2 + (B – b)2, as required.

Maybe there is a better proof?






Tirthaji’s osculation technique is used for testing for divisibility of a number by another number. It involves multiplying the right-most digit of a number by an ‘osculator’, adding on the next digit and repeating that process until the left-hand digit is reached.

When applied to polynomials however it not only detects factors but gives the quotient as well. It can therefore give two factors simultaneously.

So, the leftmost coefficient is multiplied by the osculator and the next coefficient is added. This is continued until the right-hand end of the polynomial is reached. In the example that follows the osculator is 2.


EXAMPLE 1  Find the value of 2x3 + 7x2 + 9x + 3 when x = 2.

The coefficients are 2, 7, 9, 3.

We start with the initial 2.

We multiply 2 by 2 and add the 7: 2×2 + 7 = 11.

Then 11×2 + 9 = 31.

Then 31×2 + 3 = 65.

65 is the value required.

This is a rapid and easy, one-line, cyclic process.

And it tells us the following:

  1. a) The value of the cubic at x = 2 is 65,
  2. b) x – 2 is not a factor of the cubic,
  3. c) The result of dividing the cubic by x – 2 is 2x2 + 11x + 31 remainder 65 (note the bold numbers in the example above),
  4. d) The value of 2793 in base 2, is 65.

The first two of these are probably obvious to the reader.

The 3rd shows that the steps in this Vedic method give the quotient as well as the remainder. If the remainder had in fact been zero instead of 65 we would not only have found a factor of the polynomial but we would have the quotient too, which gives the other factor.

The 4th point is significant because, though the osculator does not yield a zero, the 65 can itself be used to factorise the polynomial. See the full article for this.


EXAMPLE 2  Find the value of 2x3 + 3x2 – 5x + 12 when x = -3, and factorise the cubic.

The coefficients are 2, 3, -5, 12; the osculator is -3.

We start with the initial 2.

We multiply 2 by -3 and add the 3: 2×-3 + 3 = -3.

Then -3×-3 + -5 = 4.

Then 4×-3+ 12 = 0.

We find that x = -3 brings the cubic to zero and so (x + 3) is a factor.

But the bold values above (2, -3 and 4) give us the quotient, so we can write:

2x3 + 3x2 – 5x + 12 = (x + 3)(2x2 – 3x + 4).

Thus, this one operation gives both factors.


EXAMPLE 3   Factorise 2x3 – 5x2 + 8x – 3.

Suppose we have osculated here with 1, -1, 3, -3 (all the factors of the last coefficient) and found no zeros.

We can test if (2x – 1) is a factor by osculating with ½.

We start with the initial 2.

We multiply 2 by ½ and add the -5: 2×½ – 5 = -4.

Then -4×½ + 8 = 6.

Then 6×½ – 3 = 0.

We find (2x – 1) is a factor. But here we need to halve the bold figures to get the other factor.

So, 2x3 – 5x2 + 8x – 3 = (2x – 1)(x2 – 2x + 3).

(We can also osculate from right to left with x = 2 to obtain the same result.)

There are a great many powerful applications of osculation, and these are explored in my new book which is under preparation.

Kenneth Williams

End of article.


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Editor: Kenneth Williams

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3rd November 2023


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