# Polynomials

Notes of chapter: Polynomials are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 9 and Class 10.

(1) Polynomial-

Polynomial is an algebraic sum that consists of terms [variables, constants], variables can be in positive integer exponentiation (raise to the power) only and variables can not appear in denominators and under radical sign.

(2) Terms of the polynomials-

Terms of the polynomials are the parts of the polynomials that are separated by addition or subtraction.

Eg:- 2x2 + x + 5

In above examples, 2x2, x and 5 are terms that make a polynomial.

(3)Coefficient-

The number that is multiply to a variable is called coefficient of each term.

Eg:-

(i) 2x2 + x + 5

Coefficient of x2 = 2 (2 is multiply to x2)

Coefficient of term x = 1(1 is multiply to x)

5 is a coefficient of x0 (x0 =1) and known as constant polynomial.

(ii) -2x2 – x – 5

Coefficient of x2 = -2 (-2 is multiply to x2)

Coefficient of term x = -1(-1 is multiply to x)

-5 is a coefficient of x0 (x0 =1) and known as constant polynomial.

(iii)Zero polynomial –

The constant polynomial 0 is called the zero polynomial.

(4) Types of polynomial on the basis of terms

(i)Monomial-

Polynomials having one term only are known as monomial.

Eg:- 2x, 4x2 or 2

(ii)Binomial-

Polynomials having two terms only are known as binomial.

Eg:-  4x2  + 2x (4x2  and 2x are two terms)

x + 2(x and 2 are two terms)

(iii) Trinomial-

Polynomials having three terms only are known as trinomial.

Eg:-  4x2  + 2x  -5(4x2, 2x and – 5  are three terms)

x2 –  2x – 7(x2, -2x and -7 are three terms)

(5)Degree of the polynomial-

The highest power of the variable of the polynomial is known as degree of the polynomial.

Eg:-

(i) 4x7 + 2x2 – 5x + 6

The degree of the polynomial is 7.

(ii) 4x11 + 2x10 – 5x9 + 7x – 2

The degree of the polynomial is 11.

(iii)The degree of a non – zero constant polynomial is zero.

The constant of a polynomial is a coefficient of x0.

The value of x0 =1

Therefore, the degree of a non zero constant polynomial is zero.

Eg:- Find the degree of constant polynomial in given polynomial.

4x7 + 2x2 – 5x + 6

Ans- The constant polynomial is 6.

6 is a coefficient of x0 [ x0 =1]

Therefore, we can write 6 in multiplication of variable with powers.

6 = 6 x0

Hence, the degree of 6 is zero.

(iv) Degree of zero polynomial-

Zero polynomial = 0

We can write it,

0 = 0.x0 = 0 [Multiply zero with any number, result will be zero]

or

0 = 0.x5 = 0 [Multiply zero with any number, result will be zero]

or

0 = 0.x100 = 0 [Multiply zero with any number, result will be zero]

Therefore, degree of zero polynomial is not defined.

(7) The value of the variable of the polynomial for which value of the polynomial become zero is known as Zeroes of the polynomials.

Eg:- Find zeroes and root of the polynomial  x – 2.

Ans-

x – 2

x -2 = 0

x = 2

Therefore, 2 is the zero of the polynomial x – 2 or root of the polynomial equation x- 2 = 0.

Zeroes of the Linear Polynomial-

Linear polynomial has one and only one zero.

Eg:- Find the zeroes of polynomial p(x) = x + 3

Ans –

We know, x + 3 is a linear polynomial (degree of polynomial is 1)

p(x) = 0

x + 3 = 0

x = -3

Therefore, -3 is only zeroes of the linear polynomial.

Eg:- Verify whether 2 and 0 are zeroes of the polynomial x2 – 2x.

Ans-

Let p(x) = x2 – 2x

Put x = 2

p(2) = 22 – 2 × 2

p(2) = 4 – 4 = 0

Now,

Put x = 0

p(0) = 02 – 2 × 0

p(0) = 0 – 0 = 0

Hence, 2 and 0 are both zeroes of the polynomial x2 – 2x.

Therefore, our observations are listed below –

(i) A zero of a polynomial need not be 0.

(ii) 0 may be a zero of a polynomial.

(iii) Every linear polynomial has one and only one zero.

(iv) A polynomial can have more than one zero.

(8) A non – zero constant polynomial has no zero.

Eg:- 5 is a constant non – zero polynomial.

p(x) =5

p(x) = 5x0

Put x = 1

p(1) = 5(1)0

p(1) = 5 [ x0=1]

If we put x = 2

p(2) = 5(2)0

p(2) = 5 [ x0=1]

Hence, a non – zero constant polynomial has no zero.

Zeroes of zero polynomial-

Find zeroes of zero polynomial.

p(x) = 0 [Value of zero of polynomial is zero]

p(x) = 0.x0

Put x = 1

p(1) = 0. (1)0 = 0

If x = 20

p(20) = 0.(20)0 = 0

If x =√5

p(√5 ) = 0.(√5 )0 = 0

If x = 2/5

p(2/5 ) = 0.(2/5 )0 = 0

If x = π

p(π ) = 0.(π )0 = 0

As we now 1, 20 , 2/5 ,√5  and π are real numbers.

Therefore, all real numbers are zeroes of the zero polynomial.

(9)Remainder theorem-

We can also divide polynomials like numbers.

Let,

Dividend = p(x)

Divisor g(x)

Quotient = q(x)

Remainder = r(x)

We can write,

p(x) ÷ g(x) = q(x) with remainder r(x)

Therefore,

p(x) = (g(x) × (q(x)] + r(x)

Dividend = (Divisor × Quotient) + Remainder

Conditions:-

If Degree of p(x) ≥ Degree of g(x)

and g(x) ≠ 0

r(x) = 0 or degree of r(x) <degree of g(x)

Eg:- Divide x2 + 1 by x + 1

Let,

Dividend p(x) = x2 + 1

Divisor g(x) = x + 1

Division is showing below-

Now ,

Quotient q(x) = x – 1

Remainder r(x) = 2

We know that

Dividend =(Divisor × Quotient) + Remainder

= (x + 1) × (x -1) + 2

= x2 -1 + 2

= x2 + 1

It is proof of our division.

Finding link between the remainder and certain values of the dividend-

We know g(x) is a linear polynomial.

g(x) = x + 1

x + 1 = 0 (Zeros of the polynomial g(x))

x = -1

Therefore, we will put x = -1 in p(x)

p(x) = x2 + 1

= (-1)2 + 1

= 1+ 1

= 2

Hence, the remainder obtained on dividing p(x) by x + 1 is same as the value of the polynomial p(x) at the zero of the polynomial x + 1, ie, -1.

Generalisation of this fact is known as remainder theorem.

Remainder theorem is used to avoid long division in polynomials to get remainder.

Statement of Remainder Theorem –

Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number. If p(x) is divided by the linear polynomial x-a, then the remainder is p(a).

Proof-

Dividend p(x), degree of p(x) ≥ 1

Divisor g(x) = x – a, where a is real number

Let quotient = q(x) and remainder = r(x)

We know,

Dividend =(Divisor × Quotient) + Remainder

p(x) = [(x-a) × q(x)] + r(x)

We know that,

Degree of x –a = 1

and

Degree of r(x) < Degree of (x-a)

Therefore, r(x) is a constant, ie r.

So, for every value of x, r(x) = r

Therefore,

p(x) = [(x-a) × q(x)] + r

In particular,

x = a

Then,

p(x) = [(x-a) × q(x)] + r

p(a) = [(a-a) × q(a)] + r

= 0 + r

= r

Hence, proved.

Eg:- Find the remainder on dividing p(x) = x3 + 1 by x +1

Ans-

Finding remainder by remainder theorem

p(x) = x3 + 1

Root of x + 1 = 0

x = -1

Therefore,

p(-1) = -1 +1

=0

Hence, Value of remainder is zero.

(10)Factor theorem is a special case of the remainder theorem which links factors and zeroes of the polynomials.

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

(ii) p(a) = 0,

If x –a is a factor of p(x)

Eg:- Examine whether x + 1 is a factor of x3 + 2x2 +1

Ans-

The zero of x +1 is x = -1

Put value in polynomial p(x) = x3 + 2x2 +1

P(-1) = (-1)3 + 2(-1)2 + 1

p(-1) = -1 + 2 + 1

=2

Hence, x + 1 is not a factor of x3 + 2x2 +1.

(11) Factorisation of polynomials

(i) Factorisation of quadratic polynomial (ax2 + bx + c) can be done by splitting middle term (bx) of polynomial after finding two numbers p and q such that (px+ qx = b) so that product is (pq=ca) and get desired factors by grouping the terms.

where a  0 and a, b are coefficients

c is a constant

p and q are the factors of product

Eg:- Find the factors of polynomial 4x2 + 12x +5

Ans- We have to find two numbers p and q, such that

p + q = 12 and

pq = 4 × 5 = 20

Polynomial 4x2 + 12x +5

Sum of the two numbers should be any two factors of product of the coefficient of the variable of power 2 and constant term of the polynomial.

Some factors of 20 = ±1, ±2, ±4, ±5, ±10 and ±20

Product of factors 1 and 20 = 1  20 = 20

But, sum of factors 1 and 20 = 1 + 20 = 21, which is not 12(p + q = 12)

Product of factors 5 and 4 = 5  4 = 20

But, sum of factors 5 and 4 = 5 + 4 = 9, which is not 12(p + q = 12)

Product of factors 10 and 2 = 10  2 = 20

And, sum of factors 10 and 2 = 10 + 2 = 12, which is 12(p + q = 12)

Therefore, 10 and 2 are the factors whose sum is 12.

So,

4x2 + 12x +5

=4x2 + 2x + 10 x + 5

= 2x (2x + 1) + 5(2x +1) [Taking common]

=(2x + 1) (2x + 5)

Hence, (2x + 1) (2x + 5) are the factors of polynomial 4x2 + 12x +5.

(13) Geometrical Meaning of the Zeroes of a Polynomial

Zeroes of linear polynomial

Therefore, graph of linear polynomial has only one zero which is the x –coordinate of the point where graph intersects the x – axis.

For any quadratic polynomial, the zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0 are the x – coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x – axis.

Eg:- Quadratic polynomial y = x2 – 3x – 4

To draw graph find values of the x and y for various different values.

 x -2 -1 0 1 2 3 4 5 y = x2 – 3x – 4 6 0 -4 -6 -6 -4 0 6

Graph is plotted below –

It is clear from the table that -1 and 4 are two zeroes of the quadratic polynomial. It is also clear from the graph that -1 and 4 are the x -coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x – axis.

Therefore, zeroes of the quadratic polynomial x2 – 3x – 4 are x – coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x – axis.

The graph of the quadratic polynomial has one of the two shapes

(i) Open upwards parabola-

(ii) Open downwards parabola-

These curves are called parabola.

Different cases of parabola-

(i) Case 1 – The graph cuts x – axis at two points A and A’.

The x – coordinates of A and A’ are the two zeroes of the quadratic polynomial ax2 + bx + c.

(ii) Case 2 –The graph cuts x – axis at exactly one point A (The two points A and A’ coincident).

The x – coordinates of A is the only zero for the quadratic polynomial ax2 + bx + c.

(iii) Case 3 -The graph do not cuts x – axis.

Therefore, x – coordinates is the only zero for the quadratic polynomial ax2 + bx + c.

Relationship between zeroes and Coefficients of a Polynomial

If, quadratic polynomial(x) = ax2 + bx + c, a  0

And, Zeroes of the quadratic polynomial are α and β .

Then factors of p(x) = x – α and x –β

Therefore,

ax2 + bx + c = k(x – α)( x – β), Where, k is a constant.

= k[x2 –(α +β)x +αβ]

= kx2 – k(α+β)x + kαβ

= kx2 – k(  + )x + k

Comparing coefficients of both sides.

a = k ….(1)

b = – k( α + β)….(2)

c = kαβ …..(3)

We get

Eg:-  Find the zeroes of the quadratic polynomial x2 + 6x + 8, and verify the relationship between the zeroes and the coefficients.

Ans –

x2 + 6x + 8

=  x2 + 4x + 2x + 8

= x( x + 4) + 2(x + 4)

= (x + 4) (x + 2)

Therefore, the value of zeroes of polynomial x2 + 6x + 8 is

x + 4 = 0, ie, x = – 4

x + 2 + 0, ie, x = -2

The zeroes of polynomial x2 + 6x + 8 are -4 and – 2.

Now,

Sum of the zeroes = -4 + (-2) = – 6

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (-4)(-2) = 8

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

Division Algorithm for Polynomials

If p(x) and g(x) are any to polynomials with g(x)  0, then we can find poolynomials q(x) and r(x) such that

p(x) = g(x)  q(x) + r(x),

Where r(x) = 0 or degree of r(x) < degree of g(x).

This result is known as the Division Algorithm for polynomials.

Eg:- Divide 3x2 –x3 -3x + 5 by x – 1 – x2, and verify the division algorithm.

Ans- Write in decreasing order of their degree.

-x3 + 3x2 -3x + 5 is dividend

– x2 + x – 1 is divisor

Degree (3) = 0 < 2 = degree (– x2 + x – 1).

So, quotient = x-2, remainder = 3.

Divisor  Quotient + Remainder

= (– x2 + x – 1) (x- 2) + 3

= -x3 + x2 – x + 2x2 – 2x + 2 + 3

= -x3 + 3x2 -3x + 5

= Dividend

Hence, the division algorithm is verified.

Helping Topics

NCERT Solutions Class 9

Worksheet Class 9

Algebraic expressions

NCERT Solutions Class 10

Worksheet Class 10