NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.2

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.2 are provided here to assist students in their Class 10 exam preparations. These solutions are designed by subject experts to help students understand key polynomial operations, including addition, subtraction, and multiplication of polynomials, and solving problems on the zeroes of polynomials.

Exercise 2.2 in Polynomials focuses on finding the zeroes of polynomials and understanding how to factorize quadratic polynomials. This exercise is essential for mastering the concept of roots of polynomials and applying them in solving various problems. The solutions are explained in step-by-step language, making it easier for students to practice and grasp the concepts.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.2

Polynomials

Medium

Q.

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

​$$(i) \dfrac{1}{4},{ }-1{          (ii) }\sqrt{2},{\hspace{0.17em}\hspace{0.17em}}\frac{1}{3}{          (iii) 0, }\sqrt{5}\\ {(iv) 1, 1             (v) }\frac{-1}{4},{ }\frac{1}{4}{      \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(vi) 4, 1}$$​​​

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Polynomials

Medium

Q.

For $$\rm 3x^2+6x+k$$; $$\alpha$$​ and $$\beta$$​ are the zeroes of polynomial such that $$\alpha+\beta +\alpha\beta=– \frac{2}{3}$$​, then the value of k is:

[CBSE - 2025]

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Polynomials

Easy

Q.

If the sum of zeroes of the polynomial $$\rm p(x)=2 x^2-k \sqrt{2} x+1 \text { is } \sqrt{2},$$​ then the value of k is :

[CBSE - 2024]

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Polynomials

Medium

Q.

The zeroes of a polynomial $$\rm x^2+px+q$$​ are twice the zeroes of the polynomial $$\rm 4x^2\ –\ 5x\ –\ 6$$​. The value of p is: 

[CBSE - 2024]

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Polynomials

Medium

Q.

If $$\alpha,\ \beta$$​ are the zeroes of a polynomial $$\rm p(x)=x^2\ +\ x\ –\ 1,\ then\ \frac{1}{\alpha}+\frac{1}{ \beta }$$​ equals to

[CBSE - 2023]

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Polynomials

Medium

Q.

Find all zeroes of the polynomial $$\rm \left(2 x^4-9 x^3+5 x^2+3 x-1\right)$$​ if two of its zeroes are $$(2+\sqrt{3}) \rm\ { and\ }(2-\sqrt{3})$$​.

[CBSE - 2018]

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The NCERT Solutions are aligned with the latest CBSE syllabus and help students solve problems efficiently, ensuring they are well-prepared for the Class 10 exams.

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Given quadratic polynomials:

1. $x^2 - 2x - 8$x2−2x−8

2. $4s^2 - 4s + 1$4s2−4s+1

3. $6x^2 - 3x - 7x$6x2−3x−7x

4. $4u^2 + 8u$4u2+8u

5. $t^2 - 15$t2−15

6. $3x^2 - x - 4$3x2−x−4

Explanation:

• The general quadratic equation is

  $ax^2 + bx + c = 0$ax2+bx+c=0.

• The sum and product of the zeroes of the quadratic polynomial are given by:

  • Sum of zeroes:

    $\alpha + \beta = -\frac{b}{a}$α+β=−ab​

  • Product of zeroes:

    $\alpha \beta = \frac{c}{a}$αβ=ac​

Solution:

(i)

$x^2 - 2x - 8$

x2−2x−8
Factoring the quadratic equation:

$$x^2 - 2x - 8 = (x - 4)(x + 2) = 0$$

x2−2x−8=(x−4)(x+2)=0

The zeroes are

$x = 4$

x=4 and

$x = -2$

x=−2.

Sum of zeroes:

$$4 + (-2) = 2 \quad \text{which is} \quad -\frac{-2}{1}$$

4+(−2)=2which is−1−2​

Product of zeroes:

$$4 \times (-2) = -8 \quad \text{which is} \quad \frac{-8}{1}$$

4×(−2)=−8which is1−8​

(ii)

$4s^2 - 4s + 1$

4s2−4s+1
Factoring the quadratic equation:

$$4s^2 - 4s + 1 = (2s - 1)^2 = 0$$

4s2−4s+1=(2s−1)2=0

The zero is

$s = \frac{1}{2}$

s=21​.

Sum of zeroes:

$$\frac{1}{2} + \frac{1}{2} = 1 \quad \text{which is} \quad -\frac{-4}{4}$$

21​+21​=1which is−4−4​

Product of zeroes:

$$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \quad \text{which is} \quad \frac{1}{4}$$

21​×21​=41​which is41​

(iii)

$6x^2 - 3x - 7x$

6x2−3x−7x
Simplifying:

$$6x^2 - 10x = 0 \quad \Rightarrow \quad x(6x - 10) = 0$$

6x2−10x=0⇒x(6x−10)=0

The zeroes are

$x = 0$

x=0 and

$x = \frac{5}{3}$

x=35​.

Sum of zeroes:

$$0 + \frac{5}{3} = \frac{5}{3} \quad \text{which is} \quad -\frac{-10}{6}$$

0+35​=35​which is−6−10​

Product of zeroes:

$$0 \times \frac{5}{3} = 0 \quad \text{which is} \quad \frac{-7}{6}$$

0×35​=0which is6−7​

(iv)

$4u^2 + 8u$

4u2+8u
Factoring the quadratic equation:

$$4u(u + 2) = 0$$

4u(u+2)=0

The zeroes are

$u = 0$

u=0 and

$u = -2$

u=−2.

Sum of zeroes:

$$0 + (-2) = -2 \quad \text{which is} \quad -\frac{8}{4}$$

0+(−2)=−2which is−48​

Product of zeroes:

$$0 \times (-2) = 0 \quad \text{which is} \quad \frac{0}{4}$$

0×(−2)=0which is40​

(v)

$t^2 - 15$

t2−15
Factoring the quadratic equation:

$$t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15}) = 0$$

t2−15=(t−15​)(t+15​)=0

The zeroes are

$t = -\sqrt{15}$

t=−15​ and

$t = \sqrt{15}$

t=15​.

Sum of zeroes:

$$-\sqrt{15} + \sqrt{15} = 0 \quad \text{which is} \quad -\frac{0}{1}$$

−15​+15​=0which is−10​

Product of zeroes:

$$-\sqrt{15} \times \sqrt{15} = -15 \quad \text{which is} \quad \frac{-15}{1}$$

−15​×15​=−15which is1−15​

(vi)

$3x^2 - x - 4$

3x2−x−4
Factoring the quadratic equation:

$$3x^2 - x - 4 = (3x + 4)(x - 1) = 0$$

3x2−x−4=(3x+4)(x−1)=0

The zeroes are

$x = -\frac{4}{3}$

x=−34​ and

$x = 1$

x=1.

Sum of zeroes:

$$-\frac{4}{3} + 1 = \frac{-4 + 3}{3} = -\frac{1}{3} \quad \text{which is} \quad -\frac{-1}{3}$$

−34​+1=3−4+3​=−31​which is−3−1​

Product of zeroes:

$$-\frac{4}{3} \times 1 = -\frac{4}{3} \quad \text{which is} \quad \frac{-4}{3}$$

−34​×1=−34​which is3−4​

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Given:

• (i) Sum = 1, Product = -1

• (ii) Sum =

  $\sqrt{2}, \frac{1}{3}$2​,31​

• (iii) Sum = 0, Product = 5

• (iv) Sum = 1, Product = 1

• (v) Sum =

  $-\frac{1}{4}, \frac{1}{4}$−41​,41​

• (vi) Sum = 4, Product = 1

FAQs: Class 10 Maths Chapter 2 – Polynomials Exercise 2.2

Q1. What are the zeroes of a polynomial?
Answer:
The zeroes of a polynomial are the values of x for which the polynomial equals zero. For example, if the polynomial is p(x) = x² - 5x + 6, then the zeroes are the values of x that satisfy p(x) = 0.

Q2. How do I find the zeroes of a polynomial?
Answer:
To find the zeroes of a polynomial, set the polynomial equal to zero and solve for x. For example, to find the zeroes of x² - 5x + 6, solve x² - 5x + 6 = 0.

Q3. How do I multiply two polynomials?
Answer:
To multiply two polynomials, distribute each term of the first polynomial to each term of the second polynomial and then combine like terms. For example, to multiply (x + 1) by (x - 2), use the distributive property:
(x + 1)(x - 2) = x² - 2x + x - 2 = x² - x - 2.

Q4. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step explanations for performing polynomial operations and finding zeroes, which are key topics in Class 10 exams. By practicing these solutions, students gain confidence in solving polynomial equations and improve their problem-solving skills.

Q5. Are there any important tips for solving problems in Exercise 2.2?
Answer:

• Factorize quadratic polynomials where possible.

• Use the quadratic formula if factorization is difficult.

• Carefully check your calculations when solving for zeroes.