This document discusses polynomials and their properties. It begins by defining a polynomial as an algebraic expression of the form anxn + an-1xn-1 + ... + a2x2 + a1x + a0, where an ≠ 0. It then defines terms, monomials, binomials, and trinomials. It discusses the degree of a polynomial and the concept of roots. It introduces the remainder theorem and uses examples to illustrate factoring polynomials using the factor theorem. It concludes by discussing some algebraic identities involving polynomials.

1. B.J.P.S Samiti’s
M.V.HERWADKAR ENGLISH MEDIUM HIGH SCHOOL
Polynomials
Program:
Semester:
Course: NAME OF THE COURSE
Sandhya S.Kulkarni 1

2. Polynomials
A polynomial p(x) in one variable x is an algebraic expression in x of the
form
p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0 ,
where a0 , a1 , a2 , . . ., an are constants and an ≠ 0.
a0 , a1 , a2 , . . ., an are respectively the coefficients of x0, x, x2, . . ., xn,
and n is called the degree of the polynomial. Each of anxn, an–1 xn–1, ...,
a0 , with an ≠ 0, is called a term of the polynomial p(x).
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 2

3. Polynomials
A polynomial of one term is called a monomial
Example : 2x, x2 , 5x3
A polynomial of one term is called a binomial
Example : x2 - 2x , 5x3- 3
A polynomial of one term is called a trinomial
Example : x2 - 2x-3 , 5x3- 3 x2- 4
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 3

4. Polynomials
In the polynomial
p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0
Highest power of x i.e. ‘n’ is the degree of the polynomial
If degree n =1 then it is linear polynomial
If degree n = 2 then it is Quadratic polynomial
If degree n = 3 then it is Cubic polynomial
For x=a, if p(x) = 0 i.e. p(a)=0
then ‘a’ called a root of the equation p(x) = 0.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 4

5. Polynomials
Every linear polynomial in one variable has a unique zero.Example:
x-2, where x=2 is the unique zero
Non-zero constant polynomial has no zero
Example: p(x)=2= 2x0 has no zero. Degree polynomial is zero.
Example :
Check whether –2 and 2 are zeroes of the polynomial x + 2.
Solution : Let p(x) = x + 2.
Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0
Therefore, –2 is a zero of the polynomial x + 2, but 2 is not.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 5

6. Polynomials
Remainder Theorem : If p(x) is any polynomial of
degree greater than or equal to 1 and p(x) is
divided by the linear polynomial x – a, then the
remainder is p(a).
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 6

7. Polynomials
Example : Find the remainder when x4 + x3 – 2x2 + x + 1 is
divided by x – 1
Solution : Here, p(x) = x4 + x3 – 2x2 + x + 1, and the zero of
x – 1 is 1
So, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1= 2
So, by the Remainder Theorem, 2 is the remainder when
x4 + x3 – 2x2 + x + 1 is divided by x – 1
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 7

8. Polynomials
Example : Check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is
a multiple of 2t + 1.
Solution : As you know, q(t) will be a multiple of 2t + 1 only, if 2t
+ 1 divides q(t) leaving remainder zero.
Now, taking 2t + 1 = 0, we have t = -1/2
q(-1/2)= 4(-1/2)3 + 4(-1/2)2 – (-1/2) – 1 = 0
So the remainder obtained on dividing q(t) by 2t + 1 is 0.
So, 2t + 1 is a factor of the given polynomial q(t), that is q(t) is a
multiple of 2t + 1.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 8

9. Factorisation of Polynomials
Let us now look at the situation of Example 10 above more
closely. It tells us that since the remainder, q(-1/2) = 0,
(2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t)for some
polynomial g(t). This is a particular case of the following
theorem.
Factor Theorem :If p(x) is a polynomial of degree n ≥ 1 and a is
any real number, then (i) x – a is a factor of p(x), if p(a) = 0,
and (ii) p(a) = 0, if x – a is a factor of p(x)
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 9
Polynomials

10. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 10
Example : Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6
Solution : The zero of x + 2 is –2. Let p(x) = x3 + 3x2 + 5x + 6 and
s(x) = 2x + 4 then,
p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6= –8 + 12 – 10 + 6= 0
Hence by the Factor Theorem,
x + 2 is a factor of x3 + 3x2 + 5x + 6.
Polynomials

11. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 11
HOME WORK
Ex.1 Find the value of the polynomial 5x2 – 4 x+ 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Ex.2 Find p(0), p(1) and p(2) for each of the following
polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t2 – t3
Polynomials

12. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 12
HOME WORK
Ex.Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1 (ii) x –1/2 (iii) x
Ex. Check whether (7 + 3x) is a factor of 3x3 + 7x.
Polynomials

13. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 13
Factorisation of Polynomials
Factor Theorem : If p(x) is a polynomial of degree n ≥ 1
and a is any real number, then (i) x – a is a factor of
p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of
p(x).
Polynomials

14. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 14
Factorisation of Polynomials
Example: Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6
Solution : The zero of (x + 2) is –2. Let p(x) = x3 + 3x2 + 5x + 6
and s(x) = 2x + 4 then,
p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = = –8 + 12 – 10 + 6 = 0
By the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6
Polynomials

15. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 15
Factorisation of Polynomials
Ex: Find the value of k, if x – 1 is a factor of 4x3 + 3x2 – 4x + k
Sol:As x – 1 is a factor of p(x)= 4x3 + 3x2 – 4x + k , p(1) = 0
Means p(1) = 4(1)3 + 3(1)2 – 4(1) + k = 4 + 3 – 4 + k=0
Hence k = - 3
Polynomials

16. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 16
Algebraic Identities
Identity I : (x + y)2 = x2 + 2xy + y2
Identity II : (x – y)2 = x2 – 2xy + y2
Identity III : x2 – y2 = (x + y) (x – y)
Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab
Polynomials

17. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 17
HOME WORK
1.Determine which of the following polynomials has (x + 1) as
a factor : (i) x3 + x2 + x + 1 (ii) x4 + x3 + x2 + x + 1
2. Using Factor theorem find whether g(x) is a factor of p(x) if
p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
3. Factorise :
(i) x3 – 2x2 – x + 2 (ii) x3 – 3x2 – 9x – 5
Polynomials