Class 10 Mathematics Notes Chapter 2 (Chapter 2) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 2: Polynomials from the NCERT Exemplar. This is a crucial chapter, not just for your board exams but also frequently tested in various government entrance exams. We need a solid understanding of the concepts and how to apply them.

Chapter 2: Polynomials - Detailed Notes for Competitive Exams

1. What is a Polynomial?

• A polynomial in one variable 'x' is an algebraic expression of the form:
  p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0
• Conditions:
  • a_0, a_1, ..., a_n are real numbers (these are called coefficients).
  • a_n ≠ 0 (the coefficient of the highest power term, called the leading coefficient, is non-zero).
  • The powers of the variable 'x' (n, n-1, ..., 1, 0) must be non-negative integers (0, 1, 2, 3,...). This is a critical point! Expressions with negative or fractional powers like x⁻¹, √x (which is x^(1/2)), 1/x are NOT polynomials.
• Degree: The highest power of 'x' in the polynomial p(x) is called its degree (denoted as deg(p(x))). In the general form above, the degree is 'n'.
• Types based on Degree:
  • Linear Polynomial: Degree 1 (e.g., 3x - 5, y + √2). General form: ax + b, where a ≠ 0.
  • Quadratic Polynomial: Degree 2 (e.g., 2x² + 3x - 1, y² - 4). General form: ax² + bx + c, where a ≠ 0.
  • Cubic Polynomial: Degree 3 (e.g., x³ - 7x + 6, 5z³ - z²). General form: ax³ + bx² + cx + d, where a ≠ 0.
  • Constant Polynomial: Degree 0 (e.g., 7, -3/2, π). General form: p(x) = k, where k is any non-zero real number.
  • Zero Polynomial: The polynomial consisting of only one term, which is 0 (p(x) = 0). The degree of the zero polynomial is not defined.

2. Zeroes of a Polynomial

• A real number 'k' is called a zero (or sometimes a root) of the polynomial p(x) if substituting 'k' for the variable 'x' makes the polynomial equal to zero, i.e., p(k) = 0.
• Finding the zeroes of a polynomial is equivalent to solving the equation p(x) = 0.
• Geometrical Meaning: The real zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of the corresponding equation y = p(x) intersects the x-axis.
  • Linear (Graph: Straight Line): y = ax + b. Intersects the x-axis at exactly one point, (-b/a, 0). Hence, a linear polynomial has exactly one zero: -b/a.
  • Quadratic (Graph: Parabola): y = ax² + bx + c.
    • The parabola opens upwards if a > 0.
    • The parabola opens downwards if a < 0.
    • The graph can intersect the x-axis in three possible ways:
      • Two distinct points: The quadratic polynomial has two distinct real zeroes. (This happens when the discriminant, D = b² - 4ac > 0).
      • Exactly one point (touches the x-axis): The quadratic polynomial has two equal real zeroes (or one distinct real zero). (This happens when D = b² - 4ac = 0).
      • No points: The quadratic polynomial has no real zeroes. (The zeroes are complex/imaginary). (This happens when D = b² - 4ac < 0).
  • Cubic (Graph: Curve): y = ax³ + bx² + cx + d. Can intersect the x-axis at a maximum of 3 points. A cubic polynomial always has at least one real zero.
• Fundamental Theorem of Algebra (Extended Idea): A polynomial of degree 'n' has exactly 'n' zeroes, counting multiplicity and including complex zeroes. However, for real polynomials in Class 10 scope, we focus on real zeroes. A polynomial of degree 'n' can have at most 'n' real zeroes.

3. Relationship Between Zeroes and Coefficients

This relationship provides a powerful way to find information about zeroes without actually calculating them, and vice-versa.

• Linear Polynomial (ax + b):

  • Zero = k = -b/a
  • Zero = -(Constant term) / (Coefficient of x)

• Quadratic Polynomial (ax² + bx + c, a ≠ 0):

  • Let the zeroes be α (alpha) and β (beta).
  • Sum of zeroes: α + β = -b/a = -(Coefficient of x) / (Coefficient of x²)
  • Product of zeroes: αβ = c/a = (Constant term) / (Coefficient of x²)
  • Forming the polynomial: If the zeroes α, β are given, the quadratic polynomial can be written as k[x² - (α + β)x + αβ], where k is any non-zero real constant. For simplicity, we often take k=1, giving the polynomial x² - (Sum of zeroes)x + (Product of zeroes).

• Cubic Polynomial (ax³ + bx² + cx + d, a ≠ 0):

  • Let the zeroes be α, β, and γ (gamma).
  • Sum of zeroes: α + β + γ = -b/a = -(Coefficient of x²) / (Coefficient of x³)
  • Sum of the product of zeroes taken two at a time: αβ + βγ + γα = c/a = (Coefficient of x) / (Coefficient of x³)
  • Product of zeroes: αβγ = -d/a = -(Constant term) / (Coefficient of x³)
  • Forming the polynomial: If the zeroes α, β, γ are given, the cubic polynomial can be written as k[x³ - (α + β + γ)x² + (αβ + βγ + γα)x - αβγ], where k is any non-zero real constant.

4. Division Algorithm for Polynomials

This is analogous to Euclid's division lemma for numbers.

• Statement: For any two polynomials p(x) (dividend) and g(x) (divisor), where g(x) ≠ 0, there exist unique polynomials q(x) (quotient) and r(x) (remainder) such that:
  p(x) = g(x) × q(x) + r(x)
  where either r(x) = 0 (meaning g(x) divides p(x) exactly) or degree(r(x)) < degree(g(x)).
• Algorithm: This relationship is verified using the process of polynomial long division.
• Applications:
  • Checking for Factors (Factor Theorem): A polynomial g(x) is a factor of the polynomial p(x) if and only if the remainder r(x) is 0 when p(x) is divided by g(x). A special case is: (x - k) is a factor of p(x) if and only if p(k) = 0 (Remainder Theorem extension).
  • Finding Remaining Zeroes: If some zeroes of a polynomial p(x) are known, we can construct the corresponding factors. For instance, if k₁ and k₂ are zeroes, then (x - k₁) and (x - k₂) are factors, and so is their product (x - k₁)(x - k₂). We can divide p(x) by this product. The quotient q(x) will be a polynomial of a lower degree, and its zeroes will be the remaining zeroes of p(x).
    • Common Example: If √a and -√a are two zeroes of a polynomial p(x), then (x - √a)(x + √a) = x² - a must be a factor of p(x). We can divide p(x) by (x² - a) to find the other factors/zeroes.

Key Problem Types & Tips for Government Exams:

• Identification: Quickly identify if an expression is a polynomial (check powers!). Determine the degree.
• Graphical Interpretation: Be comfortable reading graphs to find the number of zeroes (x-intercepts). Recognize the shape associated with linear and quadratic polynomials.
• Sum & Product Formulas: These are heavily tested. Practice applying α + β = -b/a, αβ = c/a and the cubic equivalents directly.
• Constructing Polynomials: Use the k[x² - (Sum)x + Product] structure. Be mindful that multiple polynomials can have the same zeroes (differing by the constant k).
• Division Algorithm: Practice polynomial long division accurately. Use it to verify factors or find remaining zeroes.
• Special Conditions on Zeroes (Very Important):
  • If zeroes of ax² + bx + c are reciprocals of each other (α and 1/α), their product α * (1/α) = 1. So, c/a = 1 => a = c.
  • If zeroes of ax² + bx + c are equal in magnitude but opposite in sign (α and -α), their sum α + (-α) = 0. So, -b/a = 0 => b = 0.
  • Similar conditions can be applied to cubic polynomials using their sum/product relations.

Now, let's test our understanding with some multiple-choice questions typical of what you might encounter.

Multiple Choice Questions (MCQs)

1. Which of the following expressions is a polynomial in one variable?
   (A) 3/x + x² - 1
   (B) √x + 5x - 7
   (C) y² + √3y - π
   (D) z³ - z^(1/3) + 2

2. The graph of a polynomial p(x) cuts the x-axis at 2 points and touches the x-axis at 1 point. What is the number of zeroes of p(x)?
   (A) 1
   (B) 2
   (C) 3
   (D) 4

3. If α and β are the zeroes of the polynomial 2x² - 5x + 7, then the value of α + β is:
   (A) -5/2
   (B) 5/2
   (C) 7/2
   (D) -7/2

4. Find a quadratic polynomial whose sum and product of zeroes are √2 and -3/2, respectively.
   (A) x² - √2x - 3/2
   (B) x² + √2x + 3/2
   (C) 2x² - 2√2x - 3
   (D) 2x² + 2√2x - 3

5. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is -1, then the product of the other two zeroes is:
   (A) b - a + 1
   (B) b - a - 1
   (C) a - b + 1
   (D) a - b - 1

6. The zeroes of the quadratic polynomial x² + kx + k, where k ≠ 0:
   (A) cannot both be positive
   (B) cannot both be negative
   (C) are always unequal
   (D) are always equal

7. If the zeroes of the quadratic polynomial ax² + bx + c, where c ≠ 0, are equal, then:
   (A) c and a have opposite signs
   (B) c and b have opposite signs
   (C) c and a have the same sign
   (D) c and b have the same sign

8. If α, β, γ are the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, then 1/α + 1/β + 1/γ is equal to:
   (A) -b/d
   (B) c/d
   (C) -c/d
   (D) -c/a

9. What must be subtracted from p(x) = 8x⁴ + 14x³ - 2x² + 7x - 8 so that the resulting polynomial is exactly divisible by g(x) = 4x² + 3x - 2?
   (A) 14x - 10
   (B) 14x + 10
   (C) -14x + 10
   (D) -14x - 10

10. A polynomial of degree n has:
    (A) only 1 zero
    (B) exactly n zeroes
    (C) at most n zeroes
    (D) more than n zeroes

Answer Key for MCQs:

1. (C) Only in (C) are the powers of the variable (y) non-negative integers (2 and 1). π and √3 are just real coefficients. (A) has 3x⁻¹, (B) has x^(1/2), (D) has z^(1/3).

2. (C) The graph intersects the x-axis at points corresponding to distinct zeroes. Touching the x-axis corresponds to a zero with even multiplicity (here, likely 2 equal zeroes). So, 2 distinct points + 1 touching point means 2 + 1 = 3 distinct x-coordinates where y=0. Alternatively, a touch point counts as one zero location, but often implies multiplicity 2. Total zeroes = 2 (from cuts) + 1 (from touch) = 3. Self-correction: A touch point means one distinct zero value, but it counts as two equal roots. The question asks for the number of zeroes, which usually means distinct values. Let's re-read typical interpretations. NCERT usually means the number of distinct x-values. Cut at 2 points = 2 zeroes. Touch at 1 point = 1 zero. Total = 3.

3. (B) For ax² + bx + c, sum of zeroes α + β = -b/a. Here a=2, b=-5, c=7. So, α + β = -(-5)/2 = 5/2.

4. (C) Polynomial is k[x² - (Sum)x + Product] = k[x² - (√2)x + (-3/2)] = k[x² - √2x - 3/2]. To remove the fraction, let k=2. The polynomial becomes 2(x² - √2x - 3/2) = 2x² - 2√2x - 3. This matches option (C).

5. (A) Let the zeroes be α, β, γ. We are given γ = -1. We know α + β + γ = -a/1 = -a, αβ + βγ + γα = b/1 = b, αβγ = -c/1 = -c. We want the product of the other two zeroes, which is αβ. From the third relation, αβ(-1) = -c, so αβ = c. Let's re-check the relations. αβ + βγ + γα = b. Substitute γ = -1: αβ + β(-1) + α(-1) = b => αβ - β - α = b => αβ - (α + β) = b. From the first relation, α + β + (-1) = -a, so α + β = 1 - a. Substitute this into the previous equation: αβ - (1 - a) = b => αβ - 1 + a = b => αβ = b - a + 1.

6. (A) Let the zeroes be α, β. Sum α + β = -k/1 = -k. Product αβ = k/1 = k. Since k ≠ 0, αβ ≠ 0. If both zeroes were positive (α > 0, β > 0), then Sum α + β > 0 and Product αβ > 0. This means -k > 0 (so k < 0) and k > 0. This is a contradiction. Therefore, they cannot both be positive. (They could both be negative if k > 0, as Sum -k would be negative and Product k would be positive).

7. (C) If the zeroes of ax² + bx + c are equal, the discriminant D = b² - 4ac = 0. So, b² = 4ac. Since b² is always non-negative (≥ 0), 4ac must also be non-negative. As 4 > 0, ac ≥ 0. This means a and c must have the same sign (both positive or both negative). If c=0, the zeroes are 0 and -b/a, which are equal only if b=0, but the question states c ≠ 0. So ac > 0, meaning a and c have the same sign.

8. (C) 1/α + 1/β + 1/γ = (βγ + αγ + αβ) / (αβγ). We know αβ + βγ + γα = c/a and αβγ = -d/a. So, (1/α + 1/β + 1/γ) = (c/a) / (-d/a) = c/a * (a/-d) = -c/d.

9. (B) According to the Division Algorithm, p(x) = g(x)q(x) + r(x). For p(x) to be exactly divisible by g(x), the remainder r(x) must be 0. If we subtract the actual remainder r(x) from p(x), the new polynomial p(x) - r(x) will be exactly divisible by g(x). So, we need to find the remainder when p(x) is divided by g(x).
   Performing long division of 8x⁴ + 14x³ - 2x² + 7x - 8 by 4x² + 3x - 2:
   The quotient q(x) is 2x² + 2x + 1.
   The remainder r(x) is 14x - 10.
   Therefore, we must subtract the remainder (14x - 10) from p(x). Correction: The question asks what must be subtracted. If p = gq + r, then p - r = gq. So we must subtract r. The remainder is 14x - 10. So we subtract 14x - 10. Option (A). Wait, let me recheck the division.

         2x²   + 2x    + 1
       ____________________
   4x²+3x-2 | 8x⁴ + 14x³ - 2x² + 7x - 8
            -(8x⁴ + 6x³ - 4x²)
            ____________________
                  8x³ + 2x² + 7x
                -(8x³ + 6x² - 4x)
                ____________________
                      -4x² + 11x - 8
                    -(-4x² - 3x + 2)
                    ____________________
                            14x - 10
   

   Yes, the remainder is 14x - 10. We need to subtract the remainder. So, subtract 14x - 10. This matches option (A). Let me re-read the options and my result. Remainder is 14x - 10. Subtract 14x - 10. Option (A) is 14x - 10. Okay.

10. (C) A polynomial of degree n can have at most n real zeroes. It can have fewer real zeroes if some zeroes are complex or if some real zeroes have multiplicity greater than 1. Option (B) is true only if we count complex zeroes and multiplicity (Fundamental Theorem of Algebra), but in the context of real zeroes for Class 10, "at most n" is the correct statement.

Make sure you thoroughly revise these concepts and practice problems from the Exemplar book itself. Pay close attention to the relationship between zeroes and coefficients, and the division algorithm, as these are frequently tested areas. Good luck!