This is our first post on Upavidhi Blog, and we decided to discuss polynomials  especially because polynomials are an intrinsic part of mathematics. From the first time that we learn what an equation is, to calculus, to numerical analysis to advanced mathematics  there is no other concept that finds such abundant usage in mathematics.
This is one of the many areas where knowledge of Vedic Mathematics can be effectively applied. Moreso because, it simplifies most of its basic operations  allowing oneself to concentrate on the overall mathematical problem, rather than getting entangled in the cumbrous polyoperations.
Let us refresh our memories, and start with the definition of a polynomial:
The word 'Polynomial' is derived from the Greek word '
poly', which means 'many', and '
nomial, which means 'term'. So, a polynomial is something that has 'many terms' like:
4xy
^{2} + 3x  5
In the example above, the polynomial consists of 3 terms: '4xy
^{2}', '3x', and '5'.
By definition, a polynomial is an expression consisting variables and coefficients, that are interralated only by the operations of addition, subtraction, multiplication, division and exponentiation to a nonnegative number.

A polynomial, P(x) for a single indeterminate (variable) x is:
a_{n}x^{n} + a_{n1}x^{n1} + a_{n2}x^{2} + ... + a_{2}x^{2} + a_{1}x + a_{0}
And, further expressed in 'summation notation' as:
_{n}
∑ a_{i}x^{i}
^{i=0}

It should be noted that the exponent, also referred as 'index' or 'power', is the 'n' in x
^{n}, and the coefficient is a
_{n} in a
_{n}x
^{n}
The 'degree' of a polynomial is the highest degree of its terms. In other words, it is the highest sum of the exponents of any term. For example:
3x
^{3} + 5x
^{2} + 9, is a polynomial of degree 3
And, 3x
^{2}y
^{3} + xy, is a polynomial of degree 5 because the sum of exponents, 2 + 3 = 5, in 3x
^{2}y
^{3}
A polynomial is said to be expressed in 'standard' form, when the terms are organized according to the decreasing degrees of the individual terms. For example:
The standard form of 3x
^{2}  7 + 4x
^{3} + x
^{6} is:
x
^{6} + 4x
^{3} + 3x
^{2}  7
It is interesting to note that any real number may be represented as a polynomial. For example:

123 = (1 × 100) + (2 × 10) + (3 × 1)
= (1 × 10^{2}) + (2 × 10^{1}) + (3 × 10^{0})
= x^{2} + 2x + 3, where x = 10
And, 123 = (1 × 100) + (2 × 10) + (3 × 1)
= (1 × 10^{2}) + (2 × 10^{1}) + (3 × 10^{0})
= x^{2}  2x  3, where x = 10

Additionally, in Vedic Mathematics, we may also have something like: 1
23. Here,
2 is a
Rekhank, that signifies 2. Thus,

123 = (1 × 100) + (2 × 10) + (3 × 1)
= (1 × 10^{2}) + (2 × 10^{1}) + (3 × 10^{0})
= x^{2}  2x + 3, where x = 10
(More information on Rekhank and Vinculum Numbers, is available here »)

However, note that real numbers are not polynomials, because x = 10, and is not indeterminate.
ADDING & SUBTRACTING POLYNOMIALS:
The addition of two polynomials is achieved by adding the coefficients of the identical terms. For example:

When 2x^{2} + 6x + 5 is added with 3x^{2}  2x  1
x^{2}(2 + 3) + x(6  2) + (5  1)
= 5x^{2} + 4x + 4
Similarly, when x^{3} + 6x^{2} + 10x  7 is added with 2x^{2} + 4x
x^{3}(1 + 0) + x^{2}(6 + 2) + x(10 + 4) + (7 + 0)
= x^{3} + 8x^{2} + 14x  7

Subtraction of polynomials is achieved by changing the signs of the coefficients of all the terms of the subtrahend, and adding it with the minuend. For example:

(5y^{2} + 2xy  9)  (2y^{2} + 2xy  3)
= (5y^{2} + 2xy  9) + (2y^{2}  2xy + 3)
= y^{2}(5  2) + xy(2  2) + (9 + 3)
= 3y^{2}  6
Note: Since, the resulting coefficient of xy is Zero, it does not need to be mentioned anymore.

Simplifying the operations using śūddha, of Vedic Mathematics:
(Note that Upasūtra: śūddha is
discussed here »)
The operations of addition and subtraction of polynomials may be further simplified, especially when more than two polynomials are involved, with the aid of Vedic Mathematical Upasūtra: śūddha. For example:

(5y^{2} + 2xy  9)  (2y^{2} + 2xy  3) + (7xy + 4)
y^{2}  xy  c

5  2  9
2  2  3
+ 0  7  4

3  7  2
Thus, the operation gives us: 3y^{2} + 7xy  2
Note that, there will be no carry over from one section to another.

Using this method gives a marked advantage by reducing the chances of mistakes, especially with bigger coefficients and multiple polynomials.
MULTIPLYING POLYNOMIALS:
Multiplication of polynomials is achieved by multiplying every term of one of the polynomials, with every term of the other. For example:

(3x^{2} + 2x + 3) × (x + 5)
= x(3x^{2} + 2x + 3) + 5 (3x^{2} + 2x + 3)
= (3x^{3} + 2x^{2} + 3x) + (15x^{2} + 10x + 15)
Clubbing the identical terms together, we get:
= 3x^{3} + (2 + 15)x^{2} + (3 + 10)x + 15
= 3x^{3} + 17x^{2} + 13x + 15

Simplifying the operation using ūrdhva tiryagbhyāṃ, of Vedic Mathematics:
(Note that Sūtra: ūrdhva tiryagbhyāṃ is
discussed here »)
This operation may be drastically simplified with the aid of Vedic Mathematical Sūtra: ūrdhva tiryagbhyāṃ. For example:

(3x^{2} + 2x + 3) × (x + 5)
× 3 2 3
× 0 1 5

0,3,17,13,15
This merely implies: 0x^{4} + 3x^{3} + 17x^{2} + 13x + 15
Thus, (3x^{2} + 2x + 3) × (x + 5) = 3x^{3} + 17x^{2} + 13x + 15
Notes:
1. It is not required to convert the resultant to Base10, so there will be no carryover.
2. Converting the negative terms as Rekhanks would further help.
3. The order should be 'Standard Form', and the resultant derived is always in the increasing order of the power of x  from right to left.

So, for a practitioner of Vedic Mathematics, for something like:
(x
^{2}  2x + 3) × (5x
^{2} + 3x  1)

× 1 2 3
× 5 3 1

×5,7,8,11,3
This merely implies: 5x^{4}  7x^{3} + 8x^{2} + 11x  3
Thus, (x^{2}  2x + 3) × (5x^{2} + 3x  1) = 5x^{4}  7x^{3} + 8x^{2} + 11x  3

Apart from speed, use of this method simplifies the entire operation to an extent that it can also be executed mentally. Most importantly, the 'clubbing' of identical terms after the multiplication of individual terms  as in the conventional method, is not required in the Vedic Method. The simplification derived from using this method get exemplified with more terms, and bigger coefficients.
DIVIDING POLYNOMIALS:
Division pf polynomials gets a little tricky, and is achieved by the 'polynomial long division' as: Divide the first term of the Dividend by the first term of the Divisor, and create a new polynomial by multiplying this resultant quotient with the terms of the Divisor  subtracted from the original polynomial, and repeating for all terms. For example:

(x^{2}  3x  10) ÷ (x + 2)
x
x + 2 ) x^{2}  3x  10
 x^{2}  2x

 5x  10
After the next step we get:
x  5
x + 2 ) x^{2}  3x  10
 x^{2}  2x

 5x  10
+ 5x + 10

×
Thus, (x^{2}  3x  10) ÷ (x + 2) = (x  5)

An example with remainders, would be something like:
(2x
^{2}  5x  1) ÷ (x  3)

2x + 1
x  3 ) 2x^{2}  5x  1
 2x^{2} + 6x

x  1
 x + 3

2
Thus, (2x^{2}  5x  1) ÷ (x  3) = (2x + 1), with Remainder: 2

And, an example of more than one variable, would be something like:
(x
^{2} + 2x
^{2}y  2xy + 2xy
^{2}  3y
^{2}) ÷ (x + y)

x + 2xy  3y
x + y ) x^{2} + 2x^{2}y  2xy + 2xy^{2}  3y^{2}
 x^{2}  xy

2x^{2}y  3xy + 2xy^{2}  3y^{2}
 2x^{2}y  2xy^{2}

 3xy  3y^{2}
+ 3xy + 3y^{2}

×
Thus, (x^{2} + 2x^{2}y  2xy + 2xy^{2}  3y^{2}) ÷ (x + y) = (x + 2xy  3y)

Simplifying the operation using parāvartya yojayet, of Vedic Mathematics:
(Note that Sūtra: parāvartya yojayet is
discussed here »)
This operation may be drastically simplified with the aid of Vedic Mathematical Sūtra: parāvartya yojayet. For example:

x^{2} x  c
^{x + 2} 2 ) 1 3  10
2  10

1, 5  0
The above gives us: 1,5 that merely implies: x  5
Thus, (x^{2}  3x  10) ÷ (x + 2) = (x  5)
Notes:
1. Note that the sign needs to be changed in the same way that Base  Divisor is implemented.
2. The 2nd part for Remainders should be adjusted according to anything below the power of the first term of the Divisor.
3. The order should be 'Standard Form', and the resultant derived is always in the increasing order of the power of x  from right to left.

Let us take another example, with higher power of x and missing terms, for something like:
(x
^{6} + 2x
^{4} + 6x  9) ÷ (x
^{3} + 3)

Since, the Divisor is of order x^{3}, which means x × x × x, the Transposition will be 003
x^{6} x^{5} x^{4} x^{3}  x^{2} x c
^{x3 + 3} 003 ) 1 0 2 0  0 6 9
0 0 3 
0 0  0
0  0 6
 0 0 9

1 0 2 3  0 0 0
The above gives us: 1x^{3} + 0x^{2} + 2x  3
Thus, (x^{6} + 2x^{4} + 6x  9) ÷ (x^{3} + 3) = (x^{3} + 2x  3)

So, for a practitioner of Vedic Mathematics, for the above examples used to explain conventional methods:

For, (2x^{2}  5x  1) ÷ (x  3):
x^{2} x  c
^{x  3} 3 ) 2 5  1
6 
 3

2 1  2
Thus, (2x^{2}  5x  1) ÷ (x  3) = (2x + 1), with Remainder: 2
And for, (x^{2} + 2x^{2}y  2xy + 2xy^{2}  3y^{2}) ÷ (x + y):
x^{2} x  c
^{x + y} y ) (1+2y) (2y+2y^{2})  3y^{2}
(1y+2y^{2})  3y^{2}

(1+2y), 3y  0
Thus, (x^{2} + 2x^{2}y  2xy + 2xy^{2}  3y^{2}) ÷ (x + y) = (x + 2xy  3y)

MISCELLANEOUS NOTES:
It should be noted that polynomials with more than one indeterminate (variable) tend to get tricky, and Vedic Mathematics should only be applied after much practice. For example:

(x + y + 3) × (2x + 3y + 4) Applying ūrdhva tiryagbhyāṃ would mean:
x c y
1 3 1
2 4 3

2,10,5+12,13,3
This would mean: 2x^{2} + 10x + 5xy + 12 + 13y + 3y^{2}
Thus, (x + y + 3) × (2x + 3y + 4) = 2x^{2} + 10x + 5xy + 13y + 3y^{2} + 12
Note that, the constant part is placed in the middle, and when it is multiplied with itself  it is kept apart.

To conclude, knowledge of Vedic Mathematics indeed works wonders  especially in speeding up the operations, eradicating chances of silly mistakes, and most importantly, in helping one to visualize the entire process in a simple, transparent manner. These advantages, in turn, allows oneself to concentrate on the overall mathematical problem.
However, in line with our stand at Upavidhi, it should only be viewed as an extension to the conventional methods, and applied only after expertise  that comes though practice, and a certain amount of mathematical skill.