NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4

In this section, we provide clear and step-by-step NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4. These solutions are designed to help students understand the concepts of polynomials, algebraic identities, and their applications in a simple and easy-to-follow manner. Each question is solved in a detailed way to improve problem-solving skills and build a strong mathematical foundation.

These solutions are prepared according to the latest NCERT guidelines and are helpful for exam preparation, revision, and self-study.

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

Solution:
(i) Polynomial: 2x³ + x² – 5x + 2; Zeroes: 1, –2

Step 1: Verify zeroes

For x = 1:
= 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2
= 0 ✔️

For x = –2:
= 2(–2)³ + (–2)² – 5(–2) + 2
= 2(–8) + 4 + 10 + 2
= –16 + 4 + 10 + 2
= 0 ✔️

So, 1 and –2 are zeroes.

Since it is a cubic polynomial, there must be 3 zeroes.
Let the third zero be α.

Step 2: Use relationships between zeroes and coefficients

For polynomial ax³ + bx² + cx + d:

Sum of zeroes = –b/a
Product of zeroes = –d/a

Here, a = 2, b = 1, c = –5, d = 2

Sum of zeroes = –1/2

So,
1 + (–2) + α = –1/2
–1 + α = –1/2
α = 1/2

Step 3: Verify product

Product of zeroes = 1 × (–2) × (1/2) = –1

According to formula:
Product = –d/a = –2/2 = –1 ✔️

Verified

(ii) Polynomial: x³ – 4x² + 5x – 2; Zeroes: 2, 1, 1

Step 1: Verify zeroes

For x = 2:
= 8 – 16 + 10 – 2 = 0 ✔️

For x = 1:
= 1 – 4 + 5 – 2 = 0 ✔️

Since 1 is repeated, it is a double root.

Step 2: Check relationships

Here, a = 1, b = –4, c = 5, d = –2

Sum of zeroes = 2 + 1 + 1 = 4
Formula: –b/a = –(–4)/1 = 4 ✔️

Step 3: Sum of products of zeroes taken two at a time

= (2×1) + (1×1) + (2×1)
= 2 + 1 + 2 = 5

Formula: c/a = 5/1 = 5 ✔️

Step 4: Product of zeroes

= 2 × 1 × 1 = 2

Formula: –d/a = –(–2)/1 = 2 ✔️

Final Conclusion

In both cases, the given numbers are zeroes of the polynomials, and the relationships between zeroes and coefficients are verified successfully.

Question 2.
Find a cubic polynomial with the sum, some of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

Solution:
Given:
Sum of zeroes = 2
Sum of product of zeroes taken two at a time = –7
Product of zeroes = –14

For a cubic polynomial with zeroes α, β, γ, the standard form is:
x³ – (sum of zeroes)x² + (sum of product of zeroes taken two at a time)x – (product of zeroes)

Substituting the values:
x³ – 2x² – 7x + 14

Required cubic polynomial = x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.

Solution:
Polynomials Class 10 Maths NCERT Solutions Exercise 2.4 Q3

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x+ 138x – 35 are 2 ± √3, find other zeroes.
Solution:
Polynomials Exercise 2.4 Class 10 Maths NCERT Solutions Q4
Ex 2.4 Class 10 Maths Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:
Polynomials Ex 2.4 Class 10 Maths NCERT Solutions Q4

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NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4

The solutions provided for NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4 help students clearly understand the relationship between zeroes and coefficients of cubic polynomials. By practicing these questions, learners can strengthen their conceptual knowledge and improve their problem-solving skills effectively.

Even though NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4 has been recently excluded from the syllabus, it remains highly useful for building a strong mathematical foundation. Students preparing for higher classes should not skip NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2.4, as it enhances logical thinking and algebraic understanding.