Class 10 Maths Chapter 2 Polynomials Notes | Free PDF Download

Class 10 Maths Chapter 2 Polynomials Notes with Free PDF Download is a useful resource for quick revision and better exam preparation. Students can use them to understand the concepts clearly, revise faster, and prepare confidently for exams.

Table of Contents

• What is a Polynomial?
• 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
• Practice Questions on Class 10 Chapter 2 Polynomials

What is a Polynomial?

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

What are Zeroes?

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

This section explains what zeroes mean when you look at graphs!

Understanding Through Graphs

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

In other words:

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)

 Graph Shape:

     y

      |        /

      |       /

      |      /

      |     /

      |    /

------•---/------→ x

      |  / (2,0) ← Zero here

      | /

      |/

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

      y

      |

   \  |  /

    \ | /

     \|/

------×-×------→ x

     / \

    /   \

Case 2: One zero (repeated)

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

Number of zeroes = 1

 

      y

      |

      |

      |

------•------→ x

     / \

    /   \

   /     \ 

Case 3: No real zeroes

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

Number of zeroes = 0

 

      y

   \     /

    \   /

     \ /

      •  ← Doesn't touch x-axis

      |

------+------→ x

 

Cubic Polynomial Graph

p(x) = ax³ + bx² + cx + d

Can have 1, 2, or 3 zeroes

Three zeroes:          One zero:

      y                      y

      |                      |

      |    /                 |   /

  \   |   /                  |  /

   \  |  /                   | /

----×-+-×-×----→ x     ------×------→ x

     \  /                   /|

      \/                   / |

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

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 2: 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: 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

Table of Formulas:

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.

Practice 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.

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