NCERT Solutions for Exercise 2.2 Class 10 Maths Chapter 2 - Polynomials
To study polynomials, one must first master algebraic expression behaviour. The description continues through quadratic polynomials that contain expressions with variables elevated to the degree of two. An analysis of polynomial structure combined with the study of their real-world use enables us to discover expression component relationships. The path to mathematical discovery leads to useful solutions through factorisation methods.
This Story also Contains
- NCERT Solutions Class 10 Maths Chapter 2: Exercise 2.2
- Access Solution of Polynomials Class 10 Chapter 2 Exercise: 2.2
- Topics covered in Chapter 2 Polynomials: Exercise 2.2
- NCERT Solutions of Class 10 Subject Wise
- NCERT Exemplar Solutions of Class 10 Subject Wise
The NCERT Solutions present an exercise to determine the zeroes found in quadratic polynomials as well as the connection between their coefficients and zeroes. Students will understand quadratic expressions better when they solve standard-form-based polynomial problems because that process reveals operational relations between quadratics. These NCERT Books for Class 10 Maths provide questions that build algebraic reasoning while developing problem-solving abilities through logical factorisation and identity implementation.
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Access Solution of Polynomials Class 10 Chapter 2 Exercise: 2.2
Answer:
First, factorise the polynomial to know about the zeroes.
We get, x2 - 2x - 8 = 0
x2 - 4x + 2x - 8 = 0
x(x-4) + 2(x-4) = 0
(x+2)(x-4) = 0
Therefore, the zeroes of the given quadratic polynomial are -2 and 4
$\\\alpha =-2\\ , \beta =4$
VERIFICATION:
Sum of roots:
$\\\alpha+\beta =-2+4=2$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-2}{1}\\ =2\\=\alpha +\beta$
Hence Verified
Product of roots:
$\\\alpha \beta =-2\times 4=-8$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-8}{1}\\ =-8\\=\alpha \beta$
Hence Verified
Answer:
First, factorise the polynomial to know about the zeroes.
We get, $\\4s^2 - 4s + 1 = 0 $
$\\4s^2 - 2s - 2s + 1 = 0 $
$2s(2s-1) - 1(2s-1) = 0 $
$(2s-1)(2s-1) = 0$
Therefore, the zeroes of the given quadratic polynomial are 1/2 and 1/2
$\\\alpha =\frac{1}{2}\\, \beta =\frac{1}{2}$
VERIFICATION:
Sum of roots:
$\\\alpha +\beta =\frac{1}{2}+\frac{1}{2}=1$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-4}{4}\\ =1\\=\alpha +\beta$
Hence Verified
Product of roots:
$\alpha \beta =\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{1}{4}\\ =\alpha \beta$
Hence Verified
Answer:
First, factorise the polynomial to know about the zeroes.
We get, 6x2 - 3 - 7x = 0
6x2 - 7x - 3 = 0
6x2 - 9x + 2x - 3 = 0
3x(2x - 3) + 1(2x - 3) = 0
(3x + 1)(2x - 3) = 0
Therefore, the zeroes of the given quadratic polynomial are -1/3 and 3/2
$\\\alpha =-\frac{1}{3}\\ , \beta =\frac{3}{2}$
VERIFICATION:
Sum of roots:
$\\\alpha+\beta =-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-7}{6}\\ =\frac{7}{6}\\=\alpha +\beta$
Hence Verified
Product of roots:
$\\\alpha \beta =-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-3}{6}\\ =-\frac{1}{2}\\=\alpha \beta$
Hence Verified
Answer:
First, factorise the polynomial to know about the zeroes.
We get, 4u 2 + 8u = 0
4u(u + 2) = 0
Therefore, the zeroes of the given quadratic polynomial are 0 and -2
$\\\alpha =0\\, \beta =-2$
VERIFICATION:
Sum of roots:
$\\\alpha +\beta =0+(-2)=-2$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{8}{4}\\ =-2\\=\alpha +\beta$
Hence Verified
Product of roots:
$\alpha \beta =0\times-2=0$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{0}{4}\\=0\\ =\alpha \beta$
Hence Verified
Answer:
First, factorise the polynomial to know about the zeroes.
We get, t2 - 15 = 0
$(t-\sqrt{15})(t+\sqrt{15})=0$
Therefore, the zeroes of the given quadratic polynomial are $-\sqrt{15}$ and $\sqrt{15}$
$\\\alpha =-\sqrt{15}\\, \beta =\sqrt{15}$
VERIFICATION:
Sum of roots:
$\\\alpha +\beta =-\sqrt{15}+\sqrt{15}=0$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{0}{1}\\ =0\\=\alpha +\beta$
Hence Verified
Product of roots:
$\alpha \beta =-\sqrt{15}\times \sqrt{15}=-15$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-15}{1}\\=-15\\ =\alpha \beta$
Hence Verified
Answer:
First, factorise the polynomial to know about the zeroes.
We get, 3x 2 - x - 4 = 0
3x 2 + 3x - 4x - 4 = 0
3x(x + 1) - 4(x + 1) = 0
(3x - 4)(x + 1) = 0
Therefore, the zeroes of the given quadratic polynomial are 4/3 and -1
$\\\alpha =\frac{4}{3}\\, \beta =-1$
VERIFICATION:
Sum of roots:
$\\\alpha +\beta =\frac{4}{3}+(-1)=\frac{1}{3}$
By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-1}{3}\\ =\frac{1}{3}\\=\alpha +\beta$
Hence Verified
Product of roots:
$\alpha \beta =\frac{4}{3}\times -1=-\frac{4}{3}$
By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-4}{3}\\ =\alpha \beta$
Hence Verified
Answer:
Given: $\\\alpha +\beta =\frac{1}{4}$ and $\alpha \beta$ = -1
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-\frac{1}{4}x-1= 4x^{2}-x-4=0$
Thus, the required quadratic polynomial is $4x^{2}-x-4$
Answer:
Given: $\\\alpha +\beta =\sqrt{2}$ and $\alpha \beta =\frac{1}{3}$
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-\sqrt{2}x+\frac{1}{3}= 3x^{2}-3\sqrt{2}x+1=0$
Thus, the required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$
Answer:
Given: $\\\alpha +\beta =0\$ and $\alpha \beta =\sqrt{5}$
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-0x+\sqrt{5}= x^{2}+\sqrt{5}=0$
Thus, the required quadratic polynomial is $x^{2}+\sqrt{5}$
Answer:
Given: $\\\alpha +\beta =1\\$ and $\alpha \beta =1$
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-1x+1= x^{2}-x+1=0$
Thus, the required quadratic polynomial is $x^{2}-x+1$
Answer:
Given: $\\\alpha +\beta =-\frac{1}{4}\$ and $\alpha \beta =\frac{1}{4}$
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-\left (-\frac{1}{4} \right )x+\frac{1}{4}= 4x^{2}+x+1=0$
Thus, the required quadratic polynomial is $4x^{2}+x+1$
Answer:
Given: $\\\alpha +\beta =4\$ and $\alpha \beta =1$
The quadratic polynomial equation is $\\x^{2}-(\alpha +\beta )+\alpha \beta $
Therefore, the polynomial = $x^{2}-4x+1=0\\$
Thus, the required quadratic polynomial is $x^{2}-4x+1$
Also Read-
Topics covered in Chapter 2 Polynomials: Exercise 2.2
1. Zeroes of Quadratic Polynomials: Zeroes of Quadratic Polynomials show how to discover variable solutions that make a polynomial equal to zero through factorisation or identities.
2. Verification of Relationships: The resulting zeroes from the polynomial need verification through substitution, followed by coefficient analysis, each step after finding all polynomial zeroes.
3. Constructing Quadratic Polynomials: Building Quadratic Polynomials Demands the development of a polynomial using the given values for its zeroes, together with their product through the general form structure, i.e. x2 − (sum)x + product.
4. Application of Algebraic Identities: When simplifying and factoring quadratic polynomials, we should apply identities including the square of a binomial, together with the difference of squares.
5. Concept Reinforcement Through Examples: Strong conceptual understanding develops through a wide range of problems which ask students to perform direct factorisation as well as conceptual reasoning.
Check Out-
NCERT Solutions of Class 10 Subject Wise
Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.
NCERT Exemplar Solutions of Class 10 Subject Wise
Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
The formula for sum of roots = α + β = -(b/a). where b and a are the coefficients of the polynomial. This formula is comprehensively covered in class 10 ex 2.2.
Ans: If sum of roots = α + β and Product of roots = α * β then
x² - (α + β)x + α * β = 0
To get deeper understanding keep reading class 10 maths ex 2.2 using the pdf provided above in this article.
The graph of the given polynomial will be Parabola but if it's upward or downward it is decided by the condition given – if a>0 the parabola is upward but if a<0 then the parabola is downward. To understanding how this is happening keep reading 10th class maths exercise 2.2 answers from the pdf provided above in this article.
There are 4 exercises in chapter 2 i.e. Ex 2.1, 2.2, 2.3, 2.4. practice problems from each exercise including ex 2.2 class 10 to command the concepts.
considering the highest degree of f(x) is 4 taking highest degree of g(x) = 2 as mentioned, if you use simple exponent rule then degree of remainder = 4-2 = 2
The graph of the given polynomial will be parabola and its orientation depends on the value of a, if a<0 then it's a downward parabola, and if a>0 then it's an upward parabola.
It comes under the ideal case when an Equation becomes an identity, which means When the discriminant becomes zero.
Questions related to CBSE Class 10th
On Question asked by student community
Hello student!
You can refer to the CBSE Class 10 Mathematics Question Paper 2026 to practise questions and prepare for the Mathematics examination.
https://school.careers360.com/boards/cbse/cbse-class-10-maths-question-paper-2026
Hope it helps!
Hi Smriti Lama
The CBSE Class 10 Board Exam 2026-27 question paper pattern includes the latest exam structure, subject-wise marking scheme and question types. You can check the detailed pattern on Careers360:
Hi Somshiv Rath
If you mean a CBSE Class 8 Science question paper, Careers360 has a sample paper section where you can check the available Class 8 papers.
https://learn.careers360.com/school/cbse-8-class-preparation-material/
Hey there,
If you mean CBSE Class 10 rechecking/re-evaluation results 2026, CBSE has already completed the main post-result process for 2026. The Class X second-board results were declared on 18 July 2026, and CBSE has also issued the post-result facilities notice.
For verification/re-evaluation, CBSE says the result/status is uploaded online
Please mention important details like your school board, exam, and class to get the exact papers.
You can explore the useful resources here:
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