Detailed NCERT Solutions for 10 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Detailed NCERT Solutions for 10 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Preparing for exams becomes easier with Exercise 2.2. Whether you are studying for board exams or mid-term exams, 10 Mathematics Chapter 2. Polynomials solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.
ncert_solutionsChapter 2. Polynomials
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
Solution:
Using Middle term splitting method:
x2 – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 x + 2 = 0
⇒ x = 4, x = – 2
Zeroes; α = 4 β = – 2
Verifying the relationship between the zeroes and the coefficients.
a = 1, b = – 2, and c = – 8
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(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Solution:​
(ii) 4s2 – 4s + 1
Solution:
4s2– 4s + 1 = 0
⇒ 4s2 – 2s – 2s + 1 = 0
⇒ 2s (2s – 1) –1(2s – 1) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1= 0, 2s – 1= 0
⇒ 2s = 1, 2s = 1

Coefficients
a = 4, b = – 4, c = 1
Relationship between zeroes and coefficients

L.H.S = R.H.S
​
L.H.S = R.H.S
Hence verified the relationship between coefficients and the zeroes in both cases.
(iii) 6x2 – 3 – 7x
Solution:
⇒ 6x2 – 3 – 7x = 0
⇒ 6x2 – 7x – 3 = 0 (After rearranging the equation)
⇒ 6x2 – 9x + 2x – 3 = 0
⇒ 3x (2x – 3) +1 (2x – 3) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ 2x – 3 = 0, 3x + 1 = 0
⇒ 2x = 3, 3x = - 1


Hence verified the relationship between coefficients and the zeroes in both cases.
(iv) 4u2 + 8u
Solution: 4u(u + 2) = 0
4u = 0, u + 2 = 0

u = 0, u = – 2
Zeroes: a = 0, b = – 2
Coefficients:
a = 4, b = 8, c = 0
Verifying relationship between zeroes and coefficients

⇒ 0 = 0
⇒ L.H.S = R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
(v) t2 – 15
Solution:
t2 – 15 = 0
t2 = 15
t = ±√15
t = √15, t = – √15
Zeroes: a = √15, b = – √15
Coefficients a = 1, b = 0, c = – 15
Verifying relationship between zeroes and Coefficients

Hence verified, the relationship between coefficients and zeroes in both cases.
(vi) 3x2 – x – 4
Solution:
3x2 – x – 4 = 0
⇒ 3x2 + 3x – 4x – 4 = 0
⇒ 3x (x + 1) – 4 (x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0, 3x – 4 = 0
⇒ x = –1 and 3x = 4

L.H.S= R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
Q2. . Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
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Hence, required quadratic polynomial is 4x2 – x – 4

Hence, required quadratic polynomial is 3x2 –3 x + 1
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Hence, required quadratic polynomial is x2 + √5

Hence, required quadratic polynomial is x2 – x + 1

Hence, required quadratic polynomial is 4X2 + x + 1
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Hence, required quadratic polynomial is x2 – 4 + 1
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In short, Exercise 2.2 for 10 Mathematics Chapter 2. Polynomials are a complete study package. With quick revising points, NCERT exercises, and additional important questions, you can prepare effectively for exams. Make these solutions your study companion and excel in your academic journey.
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