NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

Mathematics strengthens logical reasoning and analytical thinking, and Chapter 2: Polynomials plays a crucial role in building strong algebraic foundations in Class 10 Maths. This chapter focuses on polynomials and their zeros, the relationship between zeros and coefficients, graphical representation of polynomials, and forming polynomials when the zeros are given.

These NCERT Solutions for Class 10 Maths Chapter 2 include important CBSE board questions asked between 2020 and 2025. All solutions are explained step by step in clear and simple language, helping students understand concepts thoroughly, avoid common mistakes, and score well in board examinations.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

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


​(i) 14, −1          (ii) 2,13          (iii) 0, 5(iv) 1, 1             (v) −14, 14      (vi) 4, 1 (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}​​​
Polynomials
Easy
Q.
The graph of y = p(x) is given, for a polynomial p(x). The number of zeroes of p(x) from the graph is
[CBSE - 2023]
Polynomials
Medium
Q.
If α\alpha and β\beta are the zeroes of the quadratic polynomial p(x)=x2+x−1p(x) = x^2 + x - 1, then determine the value of 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}.
Polynomials
Medium
Q.
Find all zeroes of the polynomial (2x4−9x3+5x2+3x−1)\rm \left(2 x^4-9 x^3+5 x^2+3 x-1\right)​ if two of its zeroes are (2+3) and (2−3)(2+\sqrt{3}) \rm\ { and\ }(2-\sqrt{3})​.
[CBSE - 2018]
Polynomials
Medium
Q.
Find the value of k such that the polynomial x2−(k+6)x+2(2k−1)\rm x^2-(k+6) x+2(2 k-1)​ has sum of its zeroes equal to half of their product.
[CBSE - 2019]
Polynomials
Medium
Q.
Divide the polynomial f(x)=3x2−x3−3x+5\rm f(x)=3 x^2-x^3-3 x+5​ by the polynomial g(x)=x−1−x2\rm g(x)=x-1-x^2​ and verify the division algorithm.
[CBSE - 2020]
Polynomials
Medium
Q.
Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x)=ax2+bx+c,a≠0,c≠0.\rm f(x)=a x^2+b x+c, a \neq 0, c \neq 0 .​
[CBSE - 2020]
Polynomials
Medium
Q.
Assertion (A): If the graph of a polynomial touches x-axis at only one point, then the polynomial cannot be a quadratic polynomial.
Reason (R): A polynomial of degree n (n > 1) can have at most n zeroes.
[CBSE - 2024]
Polynomials
Medium
Q.
If α, β\alpha,\ \beta ​ are the zeroes of a polynomial p(x)=x2 + x – 1, then 1α+1β \rm p(x)=x^2\ +\ x\ –\ 1,\ then\ \frac{1}{\alpha}+\frac{1}{ \beta }​ equals to
[CBSE - 2023]
Polynomials
Medium
Q.
The zeroes of a polynomial x2+px+q\rm x^2+px+q​ are twice the zeroes of the polynomial 4x2 – 5x – 6\rm 4x^2\ –\ 5x\ –\ 6​. The value of p is: 
[CBSE - 2024]
Polynomials
Medium
Q.
On dividingx3−3x2+x+2by a polynomial g(x),the quotient and remainder werex−2and−2x+4,respectively. Findg(x). \begin{array}{l}{On dividing\hspace{0.33em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2{\hspace{0.33em}by a polynomial }\mathrm{g}\left(\mathrm{x}\right),\\ {the quotient and remainder were\hspace{0.33em}}\mathrm{x}-2{\hspace{0.33em}and\hspace{0.33em}}-2\mathrm{x}+4,\\ {respectively. Find\hspace{0.33em}}\mathrm{g}\left(\mathrm{x}\right){.}\end{array}
Polynomials
Easy
Q.
If the sum of zeroes of the polynomial p(x)=2x2−k2x+1 is 2,\rm p(x)=2 x^2-k \sqrt{2} x+1 \text { is } \sqrt{2},​ then the value of k is :
[CBSE - 2024]
Polynomials
Medium
Q.
For 3x2+6x+k\rm 3x^2+6x+k; α\alpha​ and β\beta​ are the zeroes of polynomial such that α+β+αβ=–23\alpha+\beta +\alpha\beta=– \frac{2}{3}​, then the value of k is:
[CBSE - 2025]
Polynomials
Easy
Q.
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:
[CBSE - 2025]
Polynomials
Medium
Q.
If the polynomialx4−6x3+16x2−25x+10is dividedby another polynomialx2−2x+k,the remainder comesout to bex+a,findkanda. \begin{array}{l}{If the polynomial\hspace{0.33em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10{\hspace{0.33em}is divided}\\ {by another polynomial\hspace{0.33em}}{\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}{,\hspace{0.33em}the remainder comes}\\ {out to be\hspace{0.33em}}\mathrm{x}+\mathrm{a}{,\hspace{0.33em}find\hspace{0.33em}}\mathrm{k}{\hspace{0.33em}and\hspace{0.33em}}\mathrm{a}{.}\end{array}
Polynomials
Medium
Q.
If two zeroes of the polynomial  x 4 −6 x 3 −26 x 2 +138x−35 are 2± 3 , find other zeroes. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeqabeqadiWaceGabeqabeWaaqaafeaakq aabeqaaiaabMeacaqGMbGaaeiiaiaabshacaqG3bGaae4Baiaabcca caqG6bGaaeyzaiaabkhacaqGVbGaaeyzaiaabohacaqGGaGaae4Bai aabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabchacaqGVbGa aeiBaiaabMhacaqGUbGaae4Baiaab2gacaqGPbGaaeyyaiaabYgaca aMe8UaamiEamaaCaaaleqabaGaaGinaaaakiabgkHiTiaaiAdacaWG 4bWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaGOmaiaaiAdacaWG4b WaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaiaaiodacaaI4aGa amiEaiabgkHiTiaaiodacaaI1aaabaGaaeyyaiaabkhacaqGLbGaaG jbVlaaikdacqGHXcqSdaGcaaqaaiaaiodaaSqabaGccaqGSaGaaGjb VlaabAgacaqGPbGaaeOBaiaabsgacaqGGaGaae4BaiaabshacaqGOb GaaeyzaiaabkhacaqGGaGaaeOEaiaabwgacaqGYbGaae4Baiaabwga caqGZbGaaeOlaaaaaa@80A2@
Polynomials
Medium
Q.
If the zeroes of the polynomialx3−3x2+x+1are a−b, a,a+b, find a and b.  \begin{array}{l}{If the zeroes of the polynomial\hspace{0.33em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\\ {are a}-{b, a,\hspace{0.33em}}\mathrm{a}+\mathrm{b}{, find a and b. }\end{array}
Polynomials
Medium
Q.
Find a cubic polynomial with the sumofzeroes, sum of theproduct of its zeroes taken two at a time, and theproduct of its zeroes as 2, −7, −14 respectively. \begin{array}{l}{Find a cubic polynomial with the sum of zeroes , sum of the}\\ {product of its zeroes taken two at a time, and the}\\ {product of its zeroes as 2, }-{7, }-{14 respectively.}\end{array}
Polynomials
Medium
Q.
Give examples of polynomials p(x),g(x),q(x) and r(x)which satisfy the division algorithm and   (i) deg p(x)=deg q(x)         (ii) deg q(x)=deg r(x) (iii) deg r(x)=0 \begin{array}{l}{Give examples of polynomials  }\mathrm{p}\left(\mathrm{x}\right),\mathrm{g}\left(\mathrm{x}\right),\mathrm{q}\left(\mathrm{x}\right){ and }\mathrm{r}\left(\mathrm{x}\right)\\ {which satisfy the division algorithm and }\\ {  (i) deg  }\mathrm{p}\left(\mathrm{x}\right)={deg  }\mathrm{q}\left(\mathrm{x}\right){         (ii) deg  }\mathrm{q}\left(\mathrm{x}\right)={deg  }\mathrm{r}\left(\mathrm{x}\right)\\ { (iii) deg  }\mathrm{r}\left(\mathrm{x}\right)=0\end{array}
Polynomials
Medium
Q.
If α\alpha and β\beta are the zeroes of the polynomial 3x2+6x+k3x^2+6x+k, and it is given that α+β+αβ=–23\alpha+\beta +\alpha\beta = – \frac{2}{3}, then find the value of kk.

Class 10 Maths Chapter 2 Questions & Answers – Polynomials

Important Board Questions (CBSE 2020–2025)

Q1. (Medium)

Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255

Solution:
(i) 225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
∴ HCF = 45

(ii) 38220 = 196 × 195 + 0
∴ HCF = 196

(iii) 867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
∴ HCF = 51

Q2. (Medium)

Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.

Solution:
Any integer can be written as 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5.
Odd integers cannot be of the form 6q, 6q + 2, or 6q + 4.
Hence, any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.

Q3. (Medium)

An army contingent of 616 members is to march behind an army band of 32 members. Find the maximum number of columns.

Solution:
Maximum number of columns = HCF of 616 and 32
616 = 32 × 19 + 8
32 = 8 × 4 + 0
∴ HCF = 8

Q4. (Medium)

Use Euclid’s division lemma to show that the square of any positive integer is of the form 3m or 3m + 1.

Solution:
Let x be any positive integer. Then x = 3q, 3q + 1, or 3q + 2.

Squaring each:
(3q)² = 9q² = 3(3q²)
(3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1
(3q + 2)² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1

Hence proved.

Q5. (Medium)

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:
Let x = 3q, 3q + 1, or 3q + 2.

Cubing each:
(3q)³ = 27q³ = 9(3q³)
(3q + 1)³ = 9m + 1
(3q + 2)³ = 9m + 8

Hence proved.

Q6. (Medium)

Express each number as a product of its prime factors:
(i) 140   (ii) 156   (iii) 3825   (iv) 5005   (v) 7429

Solution:
(i) 140 = 2² × 5 × 7
(ii) 156 = 2² × 3 × 13
(iii) 3825 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

Q7. (Medium)

Find the LCM and HCF and verify that LCM × HCF = Product of numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Solution:
(i) HCF = 13, LCM = 182
13 × 182 = 26 × 91 ✔

(ii) HCF = 2, LCM = 23460
2 × 23460 = 510 × 92 ✔

(iii) HCF = 6, LCM = 3024
6 × 3024 = 336 × 54 ✔

Q8. (Easy) [CBSE – 2025]

If HCF(98, 28) = m and LCM(98, 28) = n, find n – 7m.

Solution:
HCF = 14, LCM = 196
n – 7m = 196 – 7(14) = 98

Q9. (Easy) [CBSE – 2025]

Find a rational number between √3 and √5.

Solution:
√3 ≈ 1.732 and √5 ≈ 2.236.
The rational number 2 lies between them.

Q10. (Easy) [CBSE – 2024]

If p = 18a²b⁴ and q = 20a³b², find LCM(p, q).

Solution:
LCM = 2² × 3² × 5 × a³ × b⁴

Q11. (Medium) [CBSE – 2023]

Assertion (A): The perimeter of △ABC is a rational number.
Reason (R): The sum of squares of two rational numbers is rational.

Solution:
Assertion is false.
Reason is true.

Q12. (Medium) [CBSE – 2024]

In a teachers' workshop, the number of teachers teaching French, Hindi and English are 48, 80 and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject.

Solution:
HCF of 48, 80 and 144 = 16
Minimum rooms = (48 + 80 + 144) ÷ 16 = 17

Q13. (Medium) [CBSE – 2023]

Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

Solution:
Required number = HCF(85 − 1, 72 − 2) = HCF(84, 70) = 14

Q14. (Easy) [CBSE – 2023]

Find the HCF of 1260 and 7344 using Euclid’s algorithm.

Solution:
HCF = 36

Q15. (Medium) [CBSE – 2023]

Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer.

Solution:
Let n be any positive odd integer.
On dividing n by 4, the remainder can be 1 or 3.
Hence, n = 4q + 1 or 4q + 3.

Q16. (Medium) [CBSE – 2025]

Prove that 1/√5 is an irrational number.

Solution:
Assume that 1/√5 is rational. Then its reciprocal √5 would also be rational.
But √5 is irrational. This is a contradiction.
Hence, 1/√5 is irrational.

Q17. (Medium) [CBSE – 2024]

Prove that 5 − 2√3 is an irrational number. It is given that √3 is irrational.

Solution:
Assume 5 − 2√3 is rational.
Then 2√3 = 5 − (rational) becomes rational, so √3 becomes rational.
This contradicts the given fact.
Therefore, 5 − 2√3 is irrational.

Q18. (Medium) [CBSE – 2024]

Show that the number 5 × 11 × 17 + 3 × 11 is a composite number.

Solution:
5 × 11 × 17 + 3 × 11 = 11(5 × 17 + 3)
= 11(88) = 11 × 88
Since it is a product of two integers greater than 1, the number is composite.

Q19. (Difficult) [CBSE – 2020]

Show that the square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q.

Solution:
Any integer n can be written as 5q, 5q + 1, 5q + 2, 5q + 3, or 5q + 4.
Squaring each case, the remainder when divided by 5 can only be 0, 1, or 4.
Hence, n² cannot be of the form 5q + 2 or 5q + 3.

Q20. (Medium) [CBSE – 2020]

Prove that one of every three consecutive positive integers is divisible by 3.

Solution:
Let the three consecutive integers be n, n+1 and n+2.
One of these must be divisible by 3.
Hence proved.

NCERT Solutions for Class 10 Maths Chapter 2 – FAQs

Q1. Why is Chapter 2 Polynomials important for the Class 10 board exam?

Chapter 2 Polynomials is a high-scoring chapter in CBSE Class 10 Maths. Questions are frequently asked on zeros of polynomials and the relationship between zeros and coefficients, making it important for board preparation.

Q2. Which topics from Polynomials are most important for CBSE exams?

  • Zeros of a polynomial
  • Relationship between zeros and coefficients
  • Finding a quadratic polynomial when zeros are given
  • Graph-based questions (number of zeros)
  • Case-study and assertion–reason questions

Q3. How many marks does Chapter 2 Polynomials usually carry in the board exam?

Chapter 2 typically carries 3 to 5 marks in the CBSE Class 10 Maths board exam. Questions may appear as short answers, long answers, or case-study based questions.

Q4. Are graphs important in Polynomials for the board exam?

Yes. Students must understand how the number of zeros of a polynomial is represented graphically. Graph-based conceptual questions are frequently asked.

Q5. How do NCERT Solutions help in preparing Polynomials?

NCERT Solutions provide step-by-step explanations and correct methods. Since CBSE follows NCERT closely, practicing these solutions helps ensure accuracy and better marks.

Q6. How should students prepare Chapter 2 Polynomials for exams?

Students should practice all NCERT examples and exercises, revise the relationship formulas regularly, and solve previous year CBSE questions to improve speed and accuracy.