Class 10 Maths Chapter 2 Polynomials Notes | Free PDF Download

Class 10 Maths Chapter 2 Polynomials Notes Free PDF Download is prepared based on the latest CBSE and NCERT syllabus. These notes will help in school exams, board exams and quick revision. They help students to understand the chapter clearly, revise faster and prepare for exams with confidence.

Table of Contents

• Algebraic Expressions
• Important Terms in Polynomials
• Zeroes of a Polynomial
• Relationship Between Zeroes and Coefficients
• Quadratic Polynomial Graph
• Formation of Polynomial from Zeroes
• Division Algorithm for Polynomials
• Steps for Polynomial Long Division
• Important Formulas Algebraic Identities Used in Polynomials
• Solved Examples on Class 10 Chapter 2 Polynomials
• NCERT Based Questions on Class 10 Chapter 2 Polynomials

Algebraic Expressions

A polynomial is an algebraic expression made up of variables and coefficients, combined using addition, subtraction, and multiplication. The powers of variables must always be whole numbers (0, 1, 2, 3...).

A polynomial is any expression that looks like this:

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀

Where:

• x = variable

• aₙ, aₙ₋₁... = coefficients (numbers in front of variable)

• n = whole number (power/degree)

In a polynomial:

• Powers of variables must be whole numbers (0, 1, 2, 3...)

• No variable in the denominator

• No variable under a square root

Important Terms in Polynomials

1. Terms

Each part of a polynomial separated by + or - is called a term.

3x² + 2x - 5

Terms:

• First term: 3x²
• Second term: 2x
• Third term: -5

2. Coefficients

The number multiplied with the variable in each term.

In 3x²:

Coefficient = 3

Variable = x

Power = 2

In -7x:

Coefficient = -7

In 5 (constant term):

Coefficient = 5

3. Degree of a Polynomial

The highest power of the variable in a polynomial is called its degree.

3x² + 2x - 5    → Degree = 2 (highest power is 2)

4x³ - x + 1     → Degree = 3

7x⁵ - 2x² + x  → Degree = 5

6               → Degree = 0 (constant polynomial)

3x              → Degree = 1

4. Types Based on Degree

5. Types Based on Number of Terms

6. Standard Form of Polynomial

Writing polynomial with powers in descending order (highest to lowest):

Standard form of a quadratic polynomial:

ax² + bx + c

Where:

a ≠ 0

a, b, c are real numbers

a = coefficient of x²

b = coefficient of x

c = constant term

Zeroes of a Polynomial

The zeroes of a polynomial are the values of the variable that make the polynomial equal to zero. If you substitute a value of x in the polynomial and the answer comes out to be 0, then that value is a zero of the polynomial.

Formula:

If p(x) is a polynomial and p(a) = 0, then a is a zero of the polynomial p(x).

Example 1: Finding Zeroes of Linear Polynomial

Find the zero of p(x) = 2x + 4

Put p(x) = 0

2x + 4 = 0

2x = -4

x = -2

p(-2) = 2(-2) + 4 = -4 + 4 = 0

Zero of p(x) = 2x + 4 is x = -2

Example 2: Checking if a Value is a Zero

Check if x = 2 is a zero of p(x) = x² - 3x + 2

Put x = 2:

p(2) = (2)² - 3(2) + 2

p(2) = 4 - 6 + 2

p(2) = 0

Yes x = 2 is a zero

Check if x = 3 is a zero:

p(3) = (3)² - 3(3) + 2

p(3) = 9 - 9 + 2

p(3) = 2 ≠ 0

No x = 3 is NOT a zero.

Number of Zeroes

Geometrical Meaning of Zeroes

The zeroes of a polynomial are the x-coordinates of the points where the graph crosses or touches the x-axis.

Where the curve of y = p(x) meets the x-axis, those x values are the zeroes!

Linear Polynomial Graph:

p(x) = ax + b (straight line)

Graph: y = 2x - 4

The straight line:

• Crosses the x-axis at one point
• That point: set y = 0, x = 2

Number of zeroes = 1

(A straight line cuts x-axis at exactly one point)

Quadratic Polynomial Graph

p(x) = ax² + bx + c (parabola)

A parabola can cross the x-axis in three ways:

Case 1: Two distinct zeroes

The parabola cuts x-axis at two different points.

Number of zeroes = 2

 

Case 2: One zero (repeated)

The parabola just touches the x-axis at one point.

Number of zeroes = 1

Case 3: No real zeroes

The parabola doesn't cross the x-axis at all.

Number of zeroes = 0

Cubic Polynomial Graph

p(x) = ax³ + bx² + cx + d

Can have 1, 2, or 3 zeroes

Relationship Between Zeroes and Coefficients

For Linear Polynomial

p(x) = ax + b

Zero = -b/a

If α is the zero:

α = -b/a

Example:

p(x) = 2x + 6

a = 2, b = 6

Zero = -b/a = -6/2 = -3

For Quadratic Polynomial

p(x) = ax² + bx + c

If α and β are the two zeroes:

Sum of zeroes:

α + β = -b/a

= -(coefficient of x) / (coefficient of x²)

Product of zeroes:

α × β = c/a

= (constant term) / (coefficient of x²)

Example: Find sum and product of zeroes of p(x) = 3x² - 5x + 2

a = 3, b = -5, c = 2

Sum of zeroes = -b/a = -(-5)/3 = 5/3

Product of zeroes = c/a = 2/3

First find actual zeroes:

3x² - 5x + 2 = 0

3x² - 3x - 2x + 2 = 0

3x(x - 1) - 2(x - 1) = 0

(3x - 2)(x - 1) = 0

x = 2/3 or x = 1

Sum = 2/3 + 1 = 5/3 

Product = 2/3 × 1 = 2/3

For Cubic Polynomial

p(x) = ax³ + bx² + cx + d

If α, β, γ are the three zeroes:

Sum of zeroes:

α + β + γ = -b/a

Sum of product of zeroes taken two at a time:

αβ + βγ + γα = c/a

Product of all three zeroes:

α × β × γ = -d/a

Formula Table:

Formation of Polynomial from Zeroes

Quadratic Polynomial from Zeroes

Formula:

p(x) = x² - (sum of zeroes)x + (product of zeroes)

p(x) = x² - (α + β)x + (αβ)Example 1: Find a quadratic polynomial with zeroes 3 and -2.

Sum of zeroes = 3 + (-2) = 1

Product of zeroes = 3 × (-2) = -6

p(x) = x² - (1)x + (-6)

p(x) = x² - x - 6

Verification:

x² - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 or x = -2

Example 1: Find quadratic polynomial with zeroes 1/2 and -3.

Sum = 1/2 + (-3) = 1/2 - 3 = -5/2

Product = 1/2 × (-3) = -3/2

p(x) = x² - (-5/2)x + (-3/2)

p(x) = x² + 5x/2 - 3/2

Multiply by 2:

p(x) = 2x² + 5x - 3

 Cubic Polynomial from Zeroes

Formula:

p(x) = x³ - (α+β+γ)x² + (αβ+βγ+γα)x - αβγ

Example 2: Find cubic polynomial with zeroes 1, 2, and 3.

Sum = 1 + 2 + 3 = 6

Sum of products = (1×2) + (2×3) + (3×1) = 2 + 6 + 3 = 11

Product = 1 × 2 × 3 = 6

 p(x) = x³ - 6x² + 11x - 6

Division Algorithm for Polynomials

The Division Algorithm

When polynomial p(x) is divided by g(x):

p(x) = g(x) × q(x) + r(x)

Where:

p(x) = dividend

g(x) = divisor

q(x) = quotient

r(x) = remainder

Conditions:

Degree of r(x) < Degree of g(x)

OR r(x) = 0 (exact division)

This is same as:

Dividend = Divisor × Quotient + Remainder

Steps for Polynomial Long Division

Divide p(x) = x³ - 3x + 2 by g(x) = x - 1

Step 1: Divide first term of dividend by first term of divisor

x³ ÷ x = x²  

Step 2: Multiply divisor by this term

x² × (x - 1) = x³ - x²

Step 3: Subtract from dividend

(x³ + 0x² - 3x + 2) - (x³ - x²)

= x² - 3x + 2

Step 4: Repeat with new expression

x² ÷ x = x  ← next term of quotient

x × (x - 1) = x² - x

(x² - 3x + 2) - (x² - x)

= -2x + 2

 Step 5: Repeat again

-2x ÷ x = -2

-2 × (x - 1) = -2x + 2

(-2x + 2) - (-2x + 2)

= 0 remainder.

 Result:

Quotient = x² + x - 2

Remainder = 0

Verification:

(x - 1)(x² + x - 2)

= x³ + x² - 2x - x² - x + 2

= x³ - 3x + 2Remainder Theorem

If polynomial p(x) is divided by (x - a), the remainder equals p(a).

Remainder = p(a)

Example:

Find remainder when p(x) = x³ - 2x + 1 is divided by (x - 2)

Remainder = p(2) = (2)³ - 2(2) + 1

          = 8 - 4 + 1

          = 5

 Factor Theorem

(x - a) is a factor of polynomial p(x) if and only if p(a) = 0.

If p(a) = 0. (x - a) is a factor of p(x)

If (x - a) is a factor p(a) = 0

Example:

Is (x - 2) a factor of p(x) = x² - 5x + 6?

p(2) = (2)² - 5(2) + 6

     = 4 - 10 + 6

     = 0

 Since p(2) = 0, (x - 2) is a factor.

Important Formulas Algebraic Identities Used in Polynomials

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

Quadratic Formula:

For ax² + bx + c = 0:

x = [-b ± √(b² - 4ac)] / 2a

 Discriminant (D) = b² - 4ac

If D > 0 ,Two distinct real zeroes

If D = 0 ,One repeated zero

If D < 0 ,No real zeroes

Formula Tables:

Solved Examples on Class 10 Chapter 2 Polynomials

Example 1: Finding Zeroes of Quadratic

Question: Find the zeroes of p(x) = x² - 5x + 6 and verify the relationship between zeroes and coefficients.

Solution:

Step 1: Factorise the polynomial

x² - 5x + 6 = 0

x² - 3x - 2x + 6 = 0

x(x - 3) - 2(x - 3) = 0

(x - 2)(x - 3) = 0

x = 2 or x = 3

Step 2: Identify coefficients

a = 1, b = -5, c = 6

α = 2, β = 3

Step 3: Verify sum of zeroes

α + β = 2 + 3 = 5

-b/a = -(-5)/1 = 5

Step 4: Verify product of zeroes

αβ = 2 × 3 = 6

c/a = 6/1 = 6

Answer: Zeroes are 2 and 3. Relationship verified

Example 2: Finding Quadratic Polynomial from Zeroes

Question: Find a quadratic polynomial whose zeroes are 5 and -3.

Solution:

Given:

α = 5, β = -3

Step 1: Calculate sum of zeroes

α + β = 5 + (-3) = 2

Step 2: Calculate product of zeroes

αβ = 5 × (-3) = -15

Step 3: Form polynomial

p(x) = x² - (α + β)x + (αβ)

p(x) = x² - 2x + (-15)

p(x) = x² - 2x - 15

Verification:

x² - 2x - 15

= x² - 5x + 3x - 15

= x(x - 5) + 3(x - 5)

= (x + 3)(x - 5)

Zeroes: x = -3 and x = 5

Answer: Required polynomial is x² - 2x - 15.

Example 3: Cubic Polynomial Zeroes

Question: Verify that 2, -1, and -3 are zeroes of p(x) = x³ + 2x² - 5x - 6. Also verify the relationship between zeroes and coefficients.

Solution:

Verification of zeroes:

p(2) = (2)³ + 2(2)² - 5(2) - 6

     = 8 + 8 - 10 - 6 = 0 

p(-1) = (-1)³ + 2(-1)² - 5(-1) - 6

      = -1 + 2 + 5 - 6 = 0

p(-3) = (-3)³ + 2(-3)² - 5(-3) - 6

      = -27 + 18 + 15 - 6 = 0

All three are zeroes.

Verification of relationships:

α = 2, β = -1, γ = -3

a = 1, b = 2, c = -5, d = -6

Sum of zeroes:

α + β + γ = 2 + (-1) + (-3) = -2

-b/a = -2/1 = -2 

Sum of products (two at a time):

αβ + βγ + γα = (2×-1) + (-1×-3) + (-3×2)

= -2 + 3 + (-6) = -5

c/a = -5/1 = -5

Product of zeroes:

αβγ = 2 × (-1) × (-3) = 6

-d/a = -(-6)/1 = 6 

 Answer: All relationships verified.

Example 4: Division Algorithm

Question: Divide p(x) = 2x³ - 11x² + 17x - 6 by g(x) = 2x - 1. Find quotient and remainder.

Solution:

Step 1: Divide first terms

2x³ ÷ 2x = x²

x² × (2x - 1) = 2x³ - x²

(2x³ - 11x²) - (2x³ - x²) = -10x²

 Step 2: Bring down and repeat

-10x² ÷ 2x = -5x

-5x × (2x - 1) = -10x² + 5x

(-10x² + 17x) - (-10x² + 5x) = 12x

 Step 3: Continue

12x ÷ 2x = 6

6 × (2x - 1) = 12x - 6

(12x - 6) - (12x - 6) = 0

 Result:

Quotient = x² - 5x + 6

Remainder = 0

 Verification:

(2x - 1)(x² - 5x + 6) + 0

= 2x³ - 10x² + 12x - x² + 5x - 6

= 2x³ - 11x² + 17x - 6

Example 5: Using Remainder Theorem

Question: Find the remainder when p(x) = x³ - 6x² + 2x - 4 is divided by (x - 3).

Solution:

Using Remainder Theorem:

Remainder = p(3)

p(3) = (3)³ - 6(3)² + 2(3) - 4

     = 27 - 54 + 6 - 4

     = -25

 Answer: Remainder = -25.

Example 6: Discriminant Application

Question: Check the nature of zeroes of p(x) = 2x² - 3x + 5.

Solution:

a = 2, b = -3, c = 5

Discriminant = b² - 4ac

= (-3)² - 4(2)(5)

= 9 - 40

= -31

Since D = -31 < 0

No real zeroes exist

Answer: The polynomial has no real zeroes.

NCERT Based Questions on Class 10 Chapter 2 Polynomials

Section 1: Basic Concepts

Question 1: What is the degree of 5x⁴ - 3x² + 7x - 1?

Question 2: How many zeroes can a cubic polynomial have?

Question 3: Is x = -1 a zero of p(x) = x² + 2x + 1?

Question 4: Find the zero of p(x) = 3x - 9.

Question 5: Write the standard form of a cubic polynomial.

Section 2: Zeroes and Coefficients

Question 6: If the zeroes of x² + px + q are 3 and -4, find p and q.

Question 7: Find sum and product of zeroes of 5x² - 4x + 1.

Question 8: A quadratic polynomial has sum of zeroes = 6 and product = 8. Find the polynomial.

Question 9: If α and β are zeroes of 2x² - 5x + 3, find α² + β².

Question 10: Find the cubic polynomial with zeroes 1, -1, and 2.

Section 3: Division Algorithm

Question 11: Find remainder when x³ + 1 is divided by x + 1.

Question 12: Check if (x - 2) is a factor of x³ - 8.

Question 13: Divide 3x³ + x² + 2x + 5 by (x + 1).

Question 14: If p(x) = x³ - 4x² + x + 6, check if x = 3 is a zero.

Question 15: Find a and b if (x² - 4) is a factor of ax⁴ + 2x³ - 3x² + bx - 4.

Download PDF - Class 10 Maths Chapter 2 Polynomials Notes

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