The document introduces polynomials using the Indian place value decimal system, explaining how a number is represented and manipulated as a polynomial in base 10. It discusses polynomial arithmetic operations, the meaning of polynomial terms and coefficients, and outlines the division algorithm, remainder theorem, and factor theorem, which relate to polynomial functions. Additionally, it mentions the fundamental theorem of algebra, asserting that any polynomial can be expressed as a product of linear factors.

1. Introduction to polynomials
Let us write 63754 in Indian place value decimal system. It would be
6 × 10000 + 3 × 1000 + 7 × 100 + 5 × 10 + 4 × 1
Or, 63754 = 6 × 104
+ 3 × 103
+ 7 × 102
+ 5 × 101
+ 4 × 100
…(1)
Yes, 100
= 1, in fact for any non-zero number 𝒙, 𝒙𝟎
= 𝟏.
Why?
We have the laws of indices as follows:
𝑎2
× 𝑎3
= (𝑎 × 𝑎) × (𝑎 × 𝑎 × 𝑎) = 𝑎 × 𝑎 × 𝑎 × 𝑎 × 𝑎 =
𝑎5
= 𝑎2+3
In brief, 𝒂𝟐
× 𝒂𝟑
= 𝒂𝟐+𝟑
………………………………………………………….(2)
Similarly 𝑎4
÷ 𝑎3
=
𝑎×𝑎×𝑎×𝑎
𝑎×𝑎×𝑎
= 𝑎 = 𝑎1
= 𝑎4−3
In brief, 𝒂𝟒
÷ 𝒂𝟑
= 𝒂𝟒−𝟑
………………………………………………..(3)
Thus 𝒂𝟎
= 𝒂𝟑−𝟑
=
𝒂𝟑
𝒂𝟑 = 𝟏, also 𝒂−𝟏
= 𝒂𝟑−𝟒
=
𝒂𝟑
𝒂𝟒 =
𝟏
𝒂
Again, (𝑎4)3
= 𝑎4
× 𝑎4
× 𝑎4
= 𝑎4+4+4
= 𝑎4×3
In brief, (𝒂𝟒)𝟑
= 𝒂𝟒×𝟑
…………………………………………………….(4)
The expression
63754 = 6 × 104
+ 3 × 103
+ 7 × 102
+ 5 × 101
+ 4 × 100
Is called a polynomial in base 10.
Our number system is called decimal system because we have 10
digits, 0,1,2,3,……….9 to express any number however big. This has

2. been possible because of old Indian Gurus, such as Arya Bhatt, Bhaskar
I, Bhaskar II, Nilakanth, Madhav, Lilavati and others.
Arya Bhatt is credited to have recognized 0 as a number, a name for
‘nothing’, so that we could assign a place value to any number ; such as
, 3 in 345 stands for 300, whereas 3 in 1436 stand for 30, et al. Maybe,
this system was used much before them, as it should be; we do not
know for certain. For, we can look back as far as history can, and
beyond that, there are folk lore, ancient scriptures like The Vedas,
Puranas, Upanishats etc. In the dim cosmic background of the colossus
of human knowledge, there is the domain of archeology, time travel
and anything that we can accept with an open mind.
There are some peoples, say for example, the old Chinese people who
used only three digits, such as 0,1,2 only. With place value system in
place, they could write four as 10, five as 11, 7 as 100, and so on. Today
we have the binary number system used by, (maybe), all computers.
This comprises of only two digits, 0, and 1, we write two as 10, three as
11, four as 110 and so on. We can carry out the arithmetic operations in
all such scales of notation as easily we do so in decimal system and
easily enough we can any number in the base 10 to its form in base 3 or
base 2 or any base; the total number if digits used in any scale of
notation I called its base.
Now return to our polynomial in equation (1)
63754 = 6 × 104
+ 3 × 103
+ 7 × 102
+ 5 × 101
+ 4 × 100
…..(a)
Also 375 = 3 × 102
+ 7 × 101
+ 5 × 100
…………………………………..(b)
We can add , subtract, multiply, divide, extract square root, in fact
perform all arithmetic operations on the numbers, such as in equation
(a) and in equation (b) easily, with the help of place value decimal
system of ours. Just to exhibit, take addition for example.

3. 63754
+ 375
64129
( carry over process when any place exceeds 10).
In terms of polynomials,
63754 + 375
= (6 × 104
+ 3 × 103
+ 7 × 102
+ 5 × 101
+ 4
× 100) + (3 × 102
+ 7 × 101
+ 5 × 100)
= 6 × 104
+ 3 × 103
+ 7 × 102
+ 3 × 102
+ 5 × 101
+ 7
× 101
+ 4 × 100
+ 5 × 100
(we have just put the corresponding powers of 10 together, just
as we placed corresponding places one below the other )
= 6 × 104
+ 3 × 103
+ (7 + 3) × 102
+ (5 + 7) × 101
+ (4 + 5) × 100
= 6 × 104
+ 3 × 103
+ 10 × 102
+ 12 × 101
+ 9 × 100
(we chose not to perform carry over process with an intention to
be clarified afterwards)
Now let us take any number 𝑥 in place of base 10 and rewrite the
polynomials as under
𝑆(𝑥) = 6 × 𝑥4
+ 3 × 𝑥3
+ 7 × 𝑥2
+ 5 × 𝑥1
+ 4…………....(A)
(We have written 𝑥0
= 1, for brevity)
And T(𝑥) = 3 × 𝑥2
+ 7 × 𝑥1
+ 5…………………………….….....(B)
We can easily write
𝑆(𝑥) + 𝑇(𝑥) = 6 × 𝑥4
+ 3 × 𝑥3
+ 10 × 𝑥2
+ 12 × 𝑥1
+ 9…....(C)
Now understand that we have sacrificed the precious carry over
operation for working with polynomials (A), (B) and (C) in single

4. variable x. The numbers such as 6 in the term 6 × 𝑥4
, or 3 in the
term 3 × 𝑥3
are called coefficients of the terms. Now we
understand what is a polynomial. It must not involve negative or
fractional powers of 𝑥. The degree of the polynomial is the
highest degree of the variable(s). If we call 6 × 𝑥4
a polynomial, it
has one term only, it is called a monomial. The expression having
two terms like 7 × 𝑥1
+ 5 is called a binomial. And so on.
Consider 𝑓(𝑥) = 𝑥 −
𝑥3
1.2.3
+
𝑥5
1.2.3.4.5
−
𝑥7
1.2.3.4.5.6.7
+ ⋯ … ..
Would it be a polynomial? Of course not; for , it contains infinite
number of terms, though 𝑓(𝑥) is finite. A polynomial must
consist of a finite number of terms.
Should we discuss subtraction of polynomials? No issue, it is just
like addition.
Try a multiplication,
𝑆(𝑥)𝑇(𝑥) = (6𝑥4
+ 3𝑥3
+ 10𝑥2
+ 12𝑥 + 9)(3𝑥2
+ 7𝑥 + 5)
We can write 3 × 𝑥3
= 3𝑥3
, 5 × 𝑥1
= 5𝑥 etc. and multiply every
term of 𝑆(𝑥) with every term of 𝑇(𝑥) and add the results at each
step just as in the manner we multiplied the two numbers 63754
and 375 writing the 2nd number below the 1st one. No big deal
again. See details in one line, not one on the top of the other.
𝑃(𝑥) = 𝑆(𝑥)𝑇(𝑥)
= 18𝑥6
+ 42𝑥5
+ 30𝑥4
+ 9𝑥5
+ 21𝑥4
+ 15𝑥3
+ 30𝑥4
+ 70𝑥3
+ 50𝑥2
+ 36𝑥3
+ 84𝑥2
+ 60𝑥
+ 27𝑥2
+ 63𝑥 + 45
= 18𝑥6
+ 51𝑥5
+ 81𝑥4
+ 121𝑥3
+ 161𝑥2
+ 123𝑥 + 45

5. To check the result, open ‘algebra solver’ (MathPapa) by
searching in Google and paste the expressions on your browser!
Had been x=10, we could have made carry overs, or for x being
any fixed number for that matter. But we chose not to carry over
for generalization. For, mathematics is generalizations, i.e.,
viewing patterns, and patterns of patterns. We are not
undermining the carry over process in any way except as per
requirement as of now. It is because of the carry over process in
the place value system for numbers by Indians, we could write
numbers with a limited alphabet of 10 and do the arithmetic
processes with ease.
Just a quick note: First term of the product 𝑃(𝑥) is product of the
first terms of the factors 𝑆(𝑥) 𝑎𝑛𝑑 𝑇(𝑥) , (6𝑥4)(3𝑥2), and last
term of the product is the product of last terms of the factors 9×
5.
Euclid’s Division algorithm, Remainder theorem, Factor theorem
Let us we divide a number P by another number D and let Q be
the quotient and R be the remainder. Naturally we take for
granted that the divisor D is smaller than the dividend N and we
stop the process of division when the remainder R falls short of
the divisor D.
In formal language, 𝑃 = 𝑄. 𝐷 + 𝑅; 𝐷 < 𝑃, 𝑅 < 𝐷…………(I)
This is division Algorithm(Division process) of Euclid.
If 𝐷 > 𝑃, then 𝑃 is itself the remainder. If 𝑅 > 𝐷, more number
of 𝐷′𝑠 could be deducted from 𝑃 until the final remainder is less
than the divisor. For, by the division process, we are 𝑄 times
deducting 𝐷, from 𝑃. If 𝑅 = 0, we say 𝐷|𝑃 i.e. 𝐷 𝒅𝒊𝒗𝒊𝒅𝒆𝒔 𝑃.

6. For polynomials the same division algorithm may be restated like:
𝑷(𝒙) = 𝑸(𝒙). 𝑫(𝒙) + 𝑹(𝒙);……………………………………(I)
𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝐷(𝑥) < 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑃(𝑥), and
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑅(𝑥) < 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝐷(𝑥)
Since 𝑥 is a variable , for some 𝑥, 𝑅(𝑥) > 𝐷(𝑥),
or 𝐷(𝑥) > 𝑃(𝑥); but that does not affect the statement of the
division algorithm for polynomials. Only the degrees of the
polynomials are concerned. One example follows where we have
carried out the process of long division.
Suppose 𝑓(𝑥) = 𝑄(𝑥).(𝑥 − 𝑟) + 𝑅 . Since the degree of 𝑅 must
be less than the degree of the divisor 𝑥 − 𝑟, which is linear
(of degree one) then the degree of 𝑅 must be 0, i.e., 𝑅
must be a constant.
This is the reason why they call any constant as a
polynomial of degree 0.
𝑓(𝑥)
𝑥 − 𝑟
=
𝑎𝑥2
+ 𝑏𝑥 + 𝑐
𝑥 − 𝑟
=
𝑎𝑥2
− 𝑎𝑟𝑥 + 𝑎𝑟𝑥 + 𝑏𝑥 + 𝑐
𝑥 − 𝑟
=
𝑎𝑥(𝑥−𝑟)+𝑥(𝑏+𝑎𝑟)+𝑐
𝑥−𝑟
= 𝑎𝑥 +
(𝑥−𝑟)(𝑏+𝑎𝑟)+𝑟(𝑏+𝑎𝑟)+𝑐
𝑥−𝑟
=𝑎𝑥 + 𝑏 + 𝑎𝑟 +
𝑎𝑟2+𝑏𝑟+𝑐
𝑥−𝑟
= 𝑎𝑥 + 𝑏 + 𝑎𝑟 +
𝑓(𝑟)
𝑥−𝑟
This is the sketch of the long division process which you could do
just as when you divide two numbers. Thus we could write

7. 𝒇(𝒙) = (𝒂𝒙 + 𝒃 + 𝒂𝒓)(𝒙 − 𝒓) + 𝒇(𝒓)……………………….(II)
Or, 𝒇(𝒙) = 𝑸(𝒙)(𝒙 − 𝒓) + 𝒇(𝒓)………………………..…….(III)
The equations (II) and (III) are special cases of (I) where 𝑓(𝑥) is
divided by 𝑥 − 𝑟, may 𝑓(𝑥) be a quadratic function (of degree 2)
or a function of any higher degree. None the less, they are general
results to remember.
Equation (III) tells us that , to know the remainder when 𝑓(𝑥) is
divided by 𝑥 − 𝑟, one need not undergo the long division process
and need not get the quotient 𝑄(𝑥), but only replace 𝒙 by 𝒓
in 𝒇(𝒙); i.e.,
Remainder when , 𝒇(𝒙) is divided by 𝒙 − 𝒓 is simply 𝒇(𝒓).
This is the remainder theorem for polynomials.
Still there is a more special case. When the remainder𝑓(𝑟) = 0, as
evident from equation (III), 𝑓(𝑥) = 𝑄(𝑥)(𝑥 − 𝑟). In short,
When 𝒇(𝒓) = 𝟎, 𝒙 − 𝒓 is a factor of 𝒇(𝒙).
This is the factor theorem, a special case of the remainder
theorem, which is a special case of the Euclidean division
algorithm.
The Fundamental Theorem of Algebra
If we multiply two linear factors such as 𝑎𝑥 + 𝑏 and 𝑐𝑥 + 𝑑, we
get, (𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑) = 𝑎𝑐𝑥2
+ (𝑎𝑐 + 𝑏𝑑)𝑥 + 𝑏𝑑; which
is a quadratic polynomial. Similarly if we multiply three linear
factors or one quadratic factor we get a polynomial of degree 3.
And so on we can get a polynomial of degree n in this manner.

8. Now the question is, if we have a polynomial like 𝑝𝑥2
+ 𝑞𝑥 + 𝑟,
where we have chosen 𝑝, 𝑞, 𝑟, as per our sweet will, would it be
product of two linear factors.
Yes! Any quadratic expression whatsoever, is, the product of two
appropriate linear factors. Not only that , any polynomial of
degree n, has precisely n linear factors, unique, factors, may be,
multiplied by constants, that does not matter to us. They call this,
or any if its equivalent statement The Fundamental Theorem of
Algebra. For the proof, a towering mathematician called Carl
Fredrik Gauss has proved it thrice in his lifetime; once when he
was 22, again at 44 and yet again at 82, in different methods. We
are yet a little too far away from that.