**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:-** 2x^{2} + x + 5

In above examples, 2x^{2}, 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)** 2x^{2} + x + 5

Coefficient of x^{2} = 2 (2 is multiply to x^{2})

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

5 is a coefficient of x^{0} (x^{0} =1) and known as constant polynomial.

**(ii)** -2x^{2} – x – 5

Coefficient of x^{2} = -2 (-2 is multiply to x^{2})

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

-5 is a coefficient of x^{0} (x^{0} =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, 4x^{2} or 2

**(ii)Binomial-**

Polynomials having two terms only are known as **binomial.**

**Eg:-** 4x^{2} + 2x (4x^{2} 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:-** 4x^{2} + 2x -5(4x^{2}, 2x and – 5 are three terms)

x^{2} – 2x – 7(x^{2}, -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)** 4x^{7} + 2x^{2} – 5x + 6

The degree of the polynomial is 7.

**(ii)** 4x^{11} + 2x^{10} – 5x^{9} + 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 x^{0}.

The value of x^{0} =1

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

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

4x^{7} + 2x^{2} – 5x + 6

**Ans-** The constant polynomial is 6.

6 is a coefficient of x^{0} [ x^{0} =1]

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

6 = 6 x^{0}

Hence, the degree of 6 is zero.

**(iv)** **Degree of zero polynomial-**

Zero polynomial = 0

We can write it,

0 = 0.x^{0} = 0 [Multiply zero with any number, result will be zero]

or

0 = 0.x^{5} = 0 [Multiply zero with any number, result will be zero]

or

0 = 0.x^{100 }= 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 x^{2} – 2x.

**Ans-**

Let p(x) = x^{2} – 2x

Put x = 2

p(2) = 2^{2} – 2 × 2

p(2) = 4 – 4 = 0

Now,

Put x = 0

p(0) = 0^{2} – 2 × 0

p(0) = 0 – 0 = 0

Hence, 2 and 0 are both zeroes of the polynomial x^{2} – 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) = 5x^{0}

Put x = 1

p(1) = 5(1)^{0}

p(1) = 5 [ x^{0}=1]

If we put x = 2

p(2) = 5(2)^{0}

p(2) = 5 [ x^{0}=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.x^{0}

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 x^{2} + 1 by x + 1

Let,

Dividend p(x) = x^{2} + 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

= x^{2} -1 + 2

= x^{2} + 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) = x^{2} + 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) = x^{3} + 1 by x +1

**Ans-**

Finding remainder by **remainder theorem**

p(x) = x^{3} + 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 x^{3} + 2x^{2} +1

**Ans-**

The zero of x +1 is x = -1

Put value in polynomial p(x) = x^{3} + 2x^{2} +1

P(-1) = (-1)^{3 }+ 2(-1)^{2} + 1

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

=2

Hence, x + 1 is not a factor of x^{3} + 2x^{2} +1.

**(11)** **Factorisation of polynomials**

**(i) **Factorisation of **quadratic polynomial** (ax^{2 }+ 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 4x^{2} + 12x +5

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

p + q = 12 and

pq = 4 × 5 = 20

Polynomial 4x^{2} + 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,

4x^{2} + 12x +5

=4x^{2} + 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 4x^{2} + 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.

**Zeroes of quadratic polynomial**

For any quadratic polynomial, the zeroes of a quadratic polynomial ax^{2} + bx + c, a ≠ 0 are the x – coordinates of the points where the parabola representing y = ax^{2} + bx + c intersects the x – axis.

**Eg:- **Quadratic polynomial y = x^{2} – 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 = x^{2} – 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 = x^{2} – 3x – 4 intersects the x – axis.

Therefore, zeroes of the quadratic polynomial x^{2} – 3x – 4 are x – coordinates of the points where the graph of y = x^{2} – 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 ax^{2} + 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 ax^{2} + bx + c.

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

Therefore, x – coordinates is the only zero for the quadratic polynomial ax^{2} + bx + c.

**Relationship between zeroes and Coefficients of a Polynomial**

If, quadratic polynomial(x) = ax^{2} + 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,

ax^{2} + bx + c = k(x – α)( x – β), Where, k is a constant.

= k[x^{2} –(α +β)x +αβ]

= kx^{2} – k(α+β)x + kαβ

= kx^{2} – 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 x^{2} + 6x + 8, and verify the relationship between the zeroes and the coefficients.

**Ans –**

x^{2} + 6x + 8

= x^{2} + 4x + 2x + 8

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

= (x + 4) (x + 2)

Therefore, the value of zeroes of polynomial x^{2} + 6x + 8 is

x + 4 = 0, ie, x = – 4

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

The zeroes of polynomial x^{2} + 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 3x^{2} –x^{3} -3x + 5 by x – 1 – x^{2}, and verify the division algorithm.

Ans- Write in decreasing order of their degree.

-x^{3} + 3x^{2} -3x + 5 is dividend

– x^{2} + x – 1 is divisor

Degree (3) = 0 < 2 = degree (– x^{2} + x – 1).

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

Divisor Quotient + Remainder

= (– x^{2} + x – 1) (x- 2) + 3

= -x^{3} + x^{2} – x + 2x^{2} – 2x + 2 + 3

= -x^{3} + 3x^{2} -3x + 5

= Dividend

Hence, the division algorithm is verified.

**Helping Topics**