10th Class Maths Fully solved Exercise 2.1-2.4 with explanations

Class 10 Maths – Chapter 2: Polynomials (NCERT Detailed Solutions)

Prepared by: Zayyan Kaseer

Exercise 2.1 – Geometrical Meaning of Zeroes

Q1: Graphs of y = f(x) are provided. Count how many times the graph intersects the x-axis.

Explanation: The zeroes of a polynomial are the x-values where the curve touches or crosses the x-axis. If the graph touches/cuts the x-axis:

• At 1 point → 1 zero
• At 2 points → 2 zeroes
• At 3 points → 3 zeroes
• Does not intersect → 0 zeroes

Students should count x-axis intersections visually for each graph provided in the textbook.

Exercise 2.2 – Relationship Between Zeroes and Coefficients

Concept Recap:

If a quadratic polynomial is written as ax² + bx + c, and its zeroes are α and β, then:

• Sum of the zeroes = α + β = -b/a
• Product of the zeroes = α * β = c/a

Q1: Find the zeroes and verify the relationships for the following:

(i) x² + 5x + 6

Factorise using splitting the middle term: Find two numbers that add to 5 and multiply to 6. They are 2 and 3.

So, x² + 5x + 6 = (x + 2)(x + 3)

Zeroes: -2 and -3

Sum = -2 + (-3) = -5 | -b/a = -5/1 = -5

Product = (-2) * (-3) = 6 | c/a = 6/1 = 6

✅ Relationship Verified

(ii) x² - 3x - 10

Factorise: Numbers -5 and 2 multiply to -10 and add to -3

So, x² - 3x - 10 = (x - 5)(x + 2)

Zeroes: 5 and -2

Sum = 5 + (-2) = 3 | -b/a = 3

Product = 5 * (-2) = -10 | c/a = -10

✅ Verified

(iii) x² - 4x + 4

Perfect square trinomial: (x - 2)²

Zeroes: 2 and 2 (repeated root)

Sum = 2 + 2 = 4 | -b/a = 4

Product = 2 * 2 = 4 | c/a = 4

✅ Verified

(iv) 2x² + 7x + 3

Factorising using middle term: 2x² + 6x + x + 3 = (2x + 1)(x + 3)

Zeroes: -1/2 and -3

Sum = (-1/2) + (-3) = -7/2 | -b/a = -7/2

Product = (-1/2) * (-3) = 3/2 | c/a = 3/2

✅ Verified

(v) 4s² + 4√2s + 1

Perfect square trinomial: (2s + √2)²

Zeroes: -√2 / 2 and -√2 / 2

Sum = -√2 | -b/a = -√2

Product = (√2 / 2)² = 1 | c/a = 1

✅ Verified

(vi) t² - 15

This is a difference of squares: t² - 15 = (t - √15)(t + √15)

Zeroes: √15 and -√15

Sum = 0 | -b/a = 0

Product = -15 | c/a = -15

✅ Verified

Q2: Construct a quadratic polynomial with given sum and product of its zeroes.

Formula: x² - (sum)x + product

• (i) Sum = -1, Product = 1 → x² + x + 1
• (ii) Sum = 1, Product = -6 → x² - x - 6
• (iii) Sum = 0, Product = √5 → x² - √5
• (iv) Sum = 4, Product = 1 → x² - 4x + 1

Exercise 2.3 – Find Polynomial from Given Zeroes

Q1: Use the standard form x² - (sum)x + product

(i) Zeroes = 2 and -2 → Sum = 0, Product = -4 → Polynomial: x² - 4

(ii) Zeroes = -3 and -4 → Sum = -7, Product = 12 → Polynomial: x² + 7x + 12

(iii) Zeroes = 4 and 1 → Sum = 5, Product = 4 → Polynomial: x² - 5x + 4

Q2: Find polynomial whose zeroes are reciprocals of those of ax² + bx + c

Let original zeroes be α and β. Then new zeroes are 1/α and 1/β.

Sum of new zeroes = (1/α + 1/β) = (α + β) / (αβ) = (-b/a) / (c/a) = -b/c

Product of new zeroes = 1/(αβ) = a/c

Required Polynomial: x² + (b/c)x + (a/c)

Exercise 2.4 – Division Algorithm

Q1: Divide p(x) = x³ - 3x² + 5x - 3 by g(x) = x² - 2

Step 1: Divide first term: x³ ÷ x² = x

Multiply: x(x² - 2) = x³ - 2x

Subtract: (x³ - 3x² + 5x - 3) - (x³ - 2x) = -3x² + 7x - 3

Step 2: Divide -3x² ÷ x² = -3

Multiply: -3(x² - 2) = -3x² + 6

Subtract: (-3x² + 7x - 3) - (-3x² + 6) = 7x - 9

Quotient: x - 3, Remainder: 7x - 9

Verification: p(x) = g(x) × q(x) + r(x) → ✅ Verified

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