NCERT Solutions Class 10 Maths Chapter 2 Polynomials (MP Board)

Welcome to the comprehensive study guide for Class 10 Mathematics Chapter 2: Polynomials. This page provides clear, step-by-step solutions for all questions from Exercise 2.1 and Exercise 2.2 according to the latest NCERT / MP Board syllabus.

EXERCISE 2.1

Question 1: The graphs of y = p(x) are given below for some polynomials p(x). Find the number of zeroes of p(x), in each case.

1. (i) The graph does not intersect the x-axis at all.
   Answer: The number of zeroes is 0.
2. (ii) The graph intersects the x-axis at 1 point.
   Answer: The number of zeroes is 1.
3. (iii) The graph intersects the x-axis at 3 points.
   Answer: The number of zeroes is 3.
4. (iv) The graph intersects the x-axis at 2 points.
   Answer: The number of zeroes is 2.
5. (v) The graph intersects the x-axis at 4 points.
   Answer: The number of zeroes is 4.
6. (vi) The graph touches/intersects the x-axis at 3 points.
   Answer: The number of zeroes is 3.

EXERCISE 2.2

Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² - 2x - 8

Solution:
x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)
To find zeroes, set p(x) = 0:
x - 4 = 0 or x + 2 = 0 ⇒ x = 4 or x = -2.
Zeroes (α, β) = 4, -2.
Verification:
• Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x²) = -b/a
• Product of zeroes = 4 × (-2) = -8 = -8/1 = (Constant term)/(Coefficient of x²) = c/a

(ii) 4s² - 4s + 1

Solution:
4s² - 4s + 1 = 4s² - 2s - 2s + 1 = 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1)
To find zeroes, set p(s) = 0:
2s - 1 = 0 ⇒ s = 1/2, 1/2.
Zeroes (α, β) = 1/2, 1/2.
Verification:
• Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -b/a
• Product of zeroes = (1/2) × (1/2) = 1/4 = c/a

(iii) 6x² - 3 - 7x

Solution:
Rearranging: 6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)
Setting p(x) = 0 ⇒ x = 3/2 or x = -1/3.
Zeroes (α, β) = 3/2, -1/3.
Verification:
• Sum of zeroes = 3/2 + (-1/3) = 7/6 = -(-7)/6 = -b/a
• Product of zeroes = (3/2) × (-1/3) = -3/6 = -1/2 = -3/6 = c/a

(iv) 4u² + 8u

Solution:
4u² + 8u = 4u(u + 2)
Setting p(u) = 0 ⇒ 4u = 0 or u + 2 = 0 ⇒ u = 0 or u = -2.
Zeroes (α, β) = 0, -2.
Verification:
• Sum of zeroes = 0 + (-2) = -2 = -8/4 = -b/a
• Product of zeroes = 0 × (-2) = 0 = 0/4 = c/a

(v) t² - 15

Solution:
t² - 15 = (t - √15)(t + √15)
Setting p(t) = 0 ⇒ t = √15 or t = -√15.
Zeroes (α, β) = √15, -√15.
Verification:
• Sum of zeroes = √15 + (-√15) = 0 = -0/1 = -b/a
• Product of zeroes = (√15) × (-√15) = -15 = -15/1 = c/a

(vi) 3x² - x - 4

Solution:
3x² - x - 4 = 3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)
Setting p(x) = 0 ⇒ x = 4/3 or x = -1.
Zeroes (α, β) = 4/3, -1.
Verification:
• Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -b/a
• Product of zeroes = (4/3) × (-1) = -4/3 = c/a

Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Formula: Required polynomial = k[x² - (Sum)x + Product]

(i) Sum = 1/4, Product = -1

Solution:
Polynomial = k[x² - (1/4)x + (-1)] = k[x² - x/4 - 1]
Taking k = 4, we get: 4x² - x - 4

(ii) Sum = √2, Product = 1/3

Solution:
Polynomial = k[x² - √2 x + 1/3]
Taking k = 3, we get: 3x² - 3√2 x + 1

(iii) Sum = 0, Product = √5

Solution:
Polynomial = k[x² - 0x + √5]
Taking k = 1, we get: x² + √5

(iv) Sum = 1, Product = 1

Solution:
Polynomial = k[x² - 1x + 1]
Taking k = 1, we get: x² - x + 1

(v) Sum = -1/4, Product = 1/4

Solution:
Polynomial = k[x² - (-1/4)x + 1/4] = k[x² + x/4 + 1/4]
Taking k = 4, we get: 4x² + x + 1

(vi) Sum = 4, Product = 1

Solution:
Polynomial = k[x² - 4x + 1]
Taking k = 1, we get: x² - 4x + 1