Class 9 Mathematics Notes Chapter 2 (Chapter 2) – Examplar Problem (Englisha) Book

Alright class, let's focus on Chapter 2, Polynomials, from your NCERT Exemplar book. This chapter is crucial, not just for your class exams, but also forms the foundation for many topics in higher mathematics and competitive exams. Pay close attention.

Chapter 2: Polynomials - Detailed Notes for Government Exam Preparation

1. Introduction to Polynomials

• Definition: An algebraic expression p(x) of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where a_0, a_1, ..., a_n are real numbers (coefficients) and n is a non-negative integer (whole number), is called a polynomial in one variable x.
  • Key Point: The exponents of the variable (x) must be whole numbers (0, 1, 2, 3,...). Expressions like x + 1/x, √x + 3, or x⁻² + 5 are not polynomials.
• Terms: The parts of the polynomial separated by '+' or '-' signs (a_n x^n, a_{n-1} x^{n-1}, etc.) are called the terms of the polynomial.
• Coefficient: The numerical part associated with each term is its coefficient. a_0 is the constant term.
• Degree of a Polynomial: The highest power of the variable in a polynomial is called its degree.
  • Example: In p(x) = 5x³ - 2x² + 7x - 1, the degree is 3.
  • The degree of a non-zero constant polynomial (e.g., p(x) = 7) is 0.
  • The degree of the zero polynomial (p(x) = 0) is not defined.

2. Types of Polynomials

• Based on Degree:
  • Linear Polynomial: Degree 1 (e.g., ax + b, where a ≠ 0). Graph is a straight line.
  • Quadratic Polynomial: Degree 2 (e.g., ax² + bx + c, where a ≠ 0). Graph is a parabola.
  • Cubic Polynomial: Degree 3 (e.g., ax³ + bx² + cx + d, where a ≠ 0).
  • Constant Polynomial: Degree 0 (e.g., p(x) = 5).
• Based on Number of Terms:
  • Monomial: One term (e.g., 5x², 7, -3y).
  • Binomial: Two terms (e.g., x + 1, 3y² - 5).
  • Trinomial: Three terms (e.g., x² + 2x + 1, 4a³ - a + 9).

3. Zeroes of a Polynomial

• Definition: A real number k is called a zero (or root) of the polynomial p(x) if p(k) = 0.
• Finding Zeroes:
  • For a linear polynomial p(x) = ax + b, the zero is x = -b/a.
  • For higher degree polynomials, finding zeroes involves methods like factorization, the Factor Theorem, or graphical interpretation (where the graph intersects the x-axis).
• Important Properties:
  • A non-zero constant polynomial has no zero.
  • Every real number is a zero of the zero polynomial.
  • A polynomial of degree n can have at most n real zeroes.

4. Remainder Theorem

• Statement: 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).
• Application: This theorem allows us to find the remainder without actually performing the long division.
  • Example: Find the remainder when p(x) = x³ + 2x² - 5x + 8 is divided by (x - 1).
    • Here, a = 1.
    • Remainder = p(1) = (1)³ + 2(1)² - 5(1) + 8 = 1 + 2 - 5 + 8 = 6.
  • Example: Find the remainder when p(x) = 2x³ - x² + 3x - 1 is divided by (x + 2).
    • Here, x + 2 = x - (-2), so a = -2.
    • Remainder = p(-2) = 2(-2)³ - (-2)² + 3(-2) - 1 = 2(-8) - 4 - 6 - 1 = -16 - 4 - 6 - 1 = -27.
  • Exam Tip: Be careful when the divisor is like (ax - b). Set ax - b = 0, so x = b/a. The remainder is p(b/a).

5. Factor Theorem

• Statement: Let p(x) be a polynomial of degree n ≥ 1 and a be any real number.
  • (i) If p(a) = 0, then (x - a) is a factor of p(x).
  • (ii) If (x - a) is a factor of p(x), then p(a) = 0.
• Application: This theorem connects the zeroes of a polynomial with its factors. It's extremely useful for factorizing polynomials, especially cubic and higher-degree ones.
  • Example: Check if (x - 2) is a factor of p(x) = x³ - 3x² + 4x - 4.
    • Find p(2) = (2)³ - 3(2)² + 4(2) - 4 = 8 - 3(4) + 8 - 4 = 8 - 12 + 8 - 4 = 16 - 16 = 0.
    • Since p(2) = 0, (x - 2) is a factor of p(x).
  • Example: Find the value of k if (x + 1) is a factor of p(x) = x² + kx + 6.
    • Since (x + 1) is a factor, p(-1) = 0.
    • p(-1) = (-1)² + k(-1) + 6 = 1 - k + 6 = 7 - k.
    • Setting p(-1) = 0: 7 - k = 0, so k = 7.

6. Factorization of Polynomials

• Splitting the Middle Term (for Quadratic Polynomials ax² + bx + c):
  • Find two numbers, say p and q, such that p + q = b (coefficient of x) and p * q = ac (product of coefficient of x² and constant term).
  • Rewrite the middle term bx as px + qx.
  • Factor by grouping.
  • Example: Factorize 6x² + 17x + 5.
    • a = 6, b = 17, c = 5. ac = 30.
    • Find two numbers whose product is 30 and sum is 17. These are 15 and 2.
    • 6x² + 15x + 2x + 5
    • 3x(2x + 5) + 1(2x + 5)
    • (2x + 5)(3x + 1)
• Using the Factor Theorem (for Cubic Polynomials):
  • Find one zero a by testing factors of the constant term (using integer root theorem hint).
  • Since p(a) = 0, (x - a) is a factor.
  • Divide the polynomial p(x) by (x - a) to get a quadratic quotient.
  • Factorize the quadratic quotient using splitting the middle term.
  • Example: Factorize x³ - 2x² - x + 2.
    • Possible rational roots are factors of 2: ±1, ±2.
    • Test x = 1: p(1) = 1³ - 2(1)² - 1 + 2 = 1 - 2 - 1 + 2 = 0. So, (x - 1) is a factor.
    • Divide (x³ - 2x² - x + 2) by (x - 1) (using long division or synthetic division). The quotient is x² - x - 2.
    • Factorize the quotient: x² - x - 2 = x² - 2x + x - 2 = x(x - 2) + 1(x - 2) = (x - 2)(x + 1).
    • Therefore, x³ - 2x² - x + 2 = (x - 1)(x - 2)(x + 1).

7. Algebraic Identities

These are essential for factorization and simplification. Memorize them thoroughly.

• (x + y)² = x² + 2xy + y²

• (x - y)² = x² - 2xy + y²

• x² - y² = (x + y)(x - y)

• (x + a)(x + b) = x² + (a + b)x + ab

• (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

• (x + y)³ = x³ + y³ + 3xy(x + y) = x³ + y³ + 3x²y + 3xy²

• (x - y)³ = x³ - y³ - 3xy(x - y) = x³ - y³ - 3x²y + 3xy²

• x³ + y³ = (x + y)(x² - xy + y²)

• x³ - y³ = (x - y)(x² + xy + y²)

• x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

  • Special Case: If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

• Exam Tip: Be able to recognize when to apply these identities, both for expansion and factorization. Problems often involve substituting expressions like (2a) for x or (3b) for y.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts we've discussed. Try to solve them yourself first.

1. Which of the following expressions is a polynomial in one variable?
   (a) x² + y²
   (b) √t + 5t - 1
   (c) 3x² - 2/x + 5
   (d) y³ - 7y + √2

2. The degree of the polynomial p(x) = 5x⁴ - 0x⁵ + 3x + 7 is:
   (a) 5
   (b) 4
   (c) 1
   (d) 0

3. The zero of the linear polynomial p(x) = 3x + 5 is:
   (a) 5/3
   (b) -5/3
   (c) 3/5
   (d) -3/5

4. If p(x) = x² - 2√2 x + 1, then p(2√2) is equal to:
   (a) 0
   (b) 1
   (c) 4√2
   (d) 8√2 + 1

5. When p(x) = x³ - ax² + 6x - a is divided by (x - a), the remainder is:
   (a) 0
   (b) 5a
   (c) -5a
   (d) 6a

6. If (x + 1) is a factor of the polynomial 2x² + kx, then the value of k is:
   (a) -3
   (b) 4
   (c) 2
   (d) -2

7. The factorization of 4x² - 9y² is:
   (a) (4x - 9y)(4x + 9y)
   (b) (2x - 3y)(2x + 3y)
   (c) (2x - 3y)(2x - 3y)
   (d) (4x - 9y)(x + y)

8. The value of 102 × 98 using a suitable identity is:
   (a) 9996
   (b) 10004
   (c) 9986
   (d) 10016

9. If x + y + z = 0, then x³ + y³ + z³ is equal to:
   (a) x² + y² + z²
   (b) 3xyz
   (c) xyz
   (d) 0

10. The coefficient of x in the expansion of (x - 3)³ is:
    (a) 9
    (b) -9
    (c) 27
    (d) -27

Answers to MCQs:

1. (d)
2. (b) [Note: The term 0x⁵ is zero, so the highest power is 4]
3. (b)
4. (b) [p(2√2) = (2√2)² - 2√2(2√2) + 1 = 8 - 8 + 1 = 1]
5. (b) [Remainder = p(a) = a³ - a(a²) + 6a - a = a³ - a³ + 5a = 5a]
6. (c) [p(-1) = 0 => 2(-1)² + k(-1) = 0 => 2 - k = 0 => k = 2]
7. (b) [Using a² - b² = (a - b)(a + b) where a = 2x, b = 3y]
8. (a) [102 × 98 = (100 + 2)(100 - 2) = 100² - 2² = 10000 - 4 = 9996]
9. (b) [Direct application of the identity]
10. (c) [(x - 3)³ = x³ - 3³ - 3(x)(3)(x - 3) = x³ - 27 - 9x(x - 3) = x³ - 27 - 9x² + 27x. Coefficient of x is 27]

Revise these notes thoroughly. Practice problems from the Exemplar book, focusing on the application of theorems and identities. Understanding these basics well will greatly help in tackling complex problems in your exams. Good luck!