Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 2: Polynomials
Chapter 2 covers the relationship between zeros and coefficients of polynomials, graphical representation of polynomials, and finding zeros of quadratic polynomials. Students learn to find zeros, verify the sum and product relationships, and form polynomials from given zeros. This chapter is part of the Algebra unit and carries 3-4 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 2 Polynomials has only 2 exercises - Exercise 2.1 (Geometrical meaning of zeros using graphs) and Exercise 2.2 (Relationship between zeros and coefficients of quadratic polynomials). The older exercises on Division Algorithm for Polynomials (previously Exercise 2.3 and 2.4) have been removed from the CBSE syllabus and will NOT be asked in the 2027 board exam. However, solutions for the older exercises are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still follow the 4-exercise format.
Table of Contents
Exercise 2.1 - Geometrical Meaning of the Zeros of a Polynomial
1 questions
Exercise 2.2 - Relationship Between Zeros and Coefficients
2 questions
Old Exercise 2.3 - Division Algorithm for Polynomials (For State Boards)
5 questions • This exercise has been removed from the CBSE 2026-
Old Exercise 2.4 - Summary Exercise (For State Boards)
5 questions • This exercise has been removed from the CBSE 2026-
Previous Year Questions
16 questions (CBSE 2019-2026)
MCQ Practice
12 questions
2.1Exercise 2.1 - Geometrical Meaning of the Zeros of a Polynomial
Q1The graphs of y = p(x) are given below, for some polynomials p(x). Find the number of zeros of p(x), in each case. (i) Graph is a straight line parallel to the x-axis above it. (ii) Graph intersects the x-axis at one point. (iii) Graph intersects the x-axis at three points. (iv) Graph intersects the x-axis at two points. (v) Graph intersects the x-axis at four points. (vi) Graph intersects the x-axis at three points.
The number of zeros of a polynomial p(x) is equal to the number of points where the graph y = p(x) intersects or touches the x-axis.
(i) The graph does not intersect the x-axis at all. It is a horizontal line above the x-axis.
Number of zeros = 0
(ii) The graph intersects the x-axis at exactly one point.
Number of zeros = 1
(iii) The graph intersects the x-axis at three points.
Number of zeros = 3
(iv) The graph intersects the x-axis at two points.
Number of zeros = 2
(v) The graph intersects the x-axis at four points.
Number of zeros = 4
(vi) The graph intersects the x-axis at three points.
Number of zeros = 3
2.2Exercise 2.2 - Relationship Between Zeros and Coefficients
Q1Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients: (i) x^2 - 2x - 8 (ii) 4s^2 - 4s + 1 (iii) 6x^2 - 3 - 7x (iv) 4u^2 + 8u (v) t^2 - 15 (vi) 3x^2 - x - 4
(i) x^2 - 2x - 8:
x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)
Zeros: x = 4 and x = -2
Verification: Sum of zeros = 4 + (-2) = 2 = -(-2)/1 = -(coefficient of x)/(coefficient of x^2) ✓
Product of zeros = 4 x (-2) = -8 = -8/1 = (constant term)/(coefficient of x^2) ✓
(ii) 4s^2 - 4s + 1:
4s^2 - 4s + 1 = (2s - 1)^2
Zeros: s = 1/2 and s = 1/2 (repeated root)
Verification: Sum of zeros = 1/2 + 1/2 = 1 = -(-4)/4 = 1 ✓
Product of zeros = 1/2 x 1/2 = 1/4 = 1/4 ✓
(iii) 6x^2 - 3 - 7x = 6x^2 - 7x - 3:
6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)
Zeros: x = 3/2 and x = -1/3
Verification: Sum of zeros = 3/2 + (-1/3) = 9/6 - 2/6 = 7/6 = -(-7)/6 = 7/6 ✓
Product of zeros = 3/2 x (-1/3) = -1/2 = -3/6 = -1/2 ✓
(iv) 4u^2 + 8u:
4u^2 + 8u = 4u(u + 2)
Zeros: u = 0 and u = -2
Verification: Sum of zeros = 0 + (-2) = -2 = -(8)/4 = -2 ✓
Product of zeros = 0 x (-2) = 0 = 0/4 = 0 ✓
(v) t^2 - 15:
t^2 - 15 = (t - sqrt(15))(t + sqrt(15))
Zeros: t = sqrt(15) and t = -sqrt(15)
Verification: Sum of zeros = sqrt(15) + (-sqrt(15)) = 0 = -(0)/1 = 0 ✓
Product of zeros = sqrt(15) x (-sqrt(15)) = -15 = -15/1 = -15 ✓
(vi) 3x^2 - x - 4:
3x^2 - x - 4 = 3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)
Zeros: x = 4/3 and x = -1
Verification: Sum of zeros = 4/3 + (-1) = 4/3 - 3/3 = 1/3 = -(-1)/3 = 1/3 ✓
Product of zeros = 4/3 x (-1) = -4/3 = -4/3 ✓
Q2Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively: (i) 1/4, -1 (ii) sqrt(2), 1/3 (iii) 0, sqrt(5) (iv) 1, 1 (v) -1/4, 1/4 (vi) 4, 1
If the sum of zeros is S and the product of zeros is P, then the quadratic polynomial is x^2 - Sx + P (or any scalar multiple of it).
(i) Sum = 1/4, Product = -1:
p(x) = x^2 - (1/4)x + (-1) = x^2 - x/4 - 1
Multiplying by 4: 4x^2 - x - 4
(ii) Sum = sqrt(2), Product = 1/3:
p(x) = x^2 - sqrt(2)x + 1/3
Multiplying by 3: 3x^2 - 3sqrt(2)x + 1
(iii) Sum = 0, Product = sqrt(5):
p(x) = x^2 - 0x + sqrt(5)
x^2 + sqrt(5)
(iv) Sum = 1, Product = 1:
p(x) = x^2 - 1x + 1
x^2 - x + 1
(v) Sum = -1/4, Product = 1/4:
p(x) = x^2 - (-1/4)x + 1/4 = x^2 + x/4 + 1/4
Multiplying by 4: 4x^2 + x + 1
(vi) Sum = 4, Product = 1:
p(x) = x^2 - 4x + 1
x^2 - 4x + 1
old-2.3Old Exercise 2.3 - Division Algorithm for Polynomials (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students who still follow the older textbook.
Q1Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: (i) p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2 (ii) p(x) = x^4 - 3x^2 + 4x + 5, g(x) = x^2 + 1 - x (iii) p(x) = x^4 - 5x + 6, g(x) = 2 - x^2
(i) p(x) = x^3 - 3x^2 + 5x - 3, g(x) = x^2 - 2:
Dividing x^3 - 3x^2 + 5x - 3 by x^2 - 2:
x^3 - 3x^2 + 5x - 3 = (x^2 - 2)(x - 3) + (7x - 9)
Quotient = x - 3, Remainder = 7x - 9
(ii) p(x) = x^4 - 3x^2 + 4x + 5, g(x) = x^2 - x + 1:
x^4 - 3x^2 + 4x + 5 = (x^2 - x + 1)(x^2 + x - 3) + (8)
Quotient = x^2 + x - 3, Remainder = 8
(iii) p(x) = x^4 - 5x + 6, g(x) = -x^2 + 2:
Rearranging g(x) = -x^2 + 2. We divide by -x^2 + 2:
x^4 - 5x + 6 = (-x^2 + 2)(-x^2 - 2) + (-5x + 10)
Quotient = -x^2 - 2, Remainder = -5x + 10
Q2Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: (i) t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12 (ii) x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2 (iii) x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1
(i) Dividing 2t^4 + 3t^3 - 2t^2 - 9t - 12 by t^2 - 3:
2t^4 + 3t^3 - 2t^2 - 9t - 12 = (t^2 - 3)(2t^2 + 3t + 4) + 0
Remainder = 0. Yes, t^2 - 3 is a factor.
(ii) Dividing 3x^4 + 5x^3 - 7x^2 + 2x + 2 by x^2 + 3x + 1:
3x^4 + 5x^3 - 7x^2 + 2x + 2 = (x^2 + 3x + 1)(3x^2 - 4x + 2) + 0
Remainder = 0. Yes, x^2 + 3x + 1 is a factor.
(iii) Dividing x^5 - 4x^3 + x^2 + 3x + 1 by x^3 - 3x + 1:
x^5 - 4x^3 + x^2 + 3x + 1 = (x^3 - 3x + 1)(x^2 - 1) + 2
Remainder = 2 (not zero). No, x^3 - 3x + 1 is not a factor.
Q3Obtain all other zeros of 3x^4 + 6x^3 - 2x^2 - 10x - 5, if two of its zeros are sqrt(5/3) and -sqrt(5/3).
Since sqrt(5/3) and -sqrt(5/3) are zeros, (x - sqrt(5/3))(x + sqrt(5/3)) = x^2 - 5/3 is a factor.
So (3x^2 - 5) is a factor of 3x^4 + 6x^3 - 2x^2 - 10x - 5.
Dividing: 3x^4 + 6x^3 - 2x^2 - 10x - 5 = (3x^2 - 5)(x^2 + 2x + 1) = (3x^2 - 5)(x + 1)^2
From (x + 1)^2 = 0: x = -1 (repeated)
The other two zeros are -1 and -1.
Q4On dividing x^3 - 3x^2 + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4 respectively. Find g(x).
Using the division algorithm: p(x) = g(x) x q(x) + r(x)
x^3 - 3x^2 + x + 2 = g(x)(x - 2) + (-2x + 4)
g(x)(x - 2) = x^3 - 3x^2 + x + 2 - (-2x + 4)
g(x)(x - 2) = x^3 - 3x^2 + x + 2 + 2x - 4
g(x)(x - 2) = x^3 - 3x^2 + 3x - 2
Dividing x^3 - 3x^2 + 3x - 2 by (x - 2):
x^3 - 3x^2 + 3x - 2 = (x - 2)(x^2 - x + 1)
g(x) = x^2 - x + 1
Q5Give 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
(i) deg p(x) = deg q(x):
This happens when g(x) is a constant.
p(x) = 2x^2 + 2x + 8, g(x) = 2, q(x) = x^2 + x + 4, r(x) = 0
Check: 2(x^2 + x + 4) + 0 = 2x^2 + 2x + 8 ✓
(ii) deg q(x) = deg r(x):
p(x) = x^3 + x^2 + x + 1, g(x) = x^2 - 1
q(x) = x + 1, r(x) = 2x + 2 (both degree 1)
Check: (x^2 - 1)(x + 1) + (2x + 2) = x^3 + x^2 - x - 1 + 2x + 2 = x^3 + x^2 + x + 1 ✓
(iii) deg r(x) = 0:
p(x) = x^3 + 2x^2 - x + 2, g(x) = x^2 - 1
q(x) = x + 2, r(x) = 4 (degree 0, a constant)
Check: (x^2 - 1)(x + 2) + 4 = x^3 + 2x^2 - x - 2 + 4 = x^3 + 2x^2 - x + 2 ✓
old-2.4Old Exercise 2.4 - Summary Exercise (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. Provided for state board students.
Q1Verify that the numbers given alongside of the cubic polynomials below are their zeros. Also verify the relationship between the zeros and the coefficients in each case: (i) 2x^3 + x^2 - 5x + 2; 1/2, 1, -2 (ii) x^3 - 4x^2 + 5x - 2; 2, 1, 1
(i) p(x) = 2x^3 + x^2 - 5x + 2; zeros: 1/2, 1, -2
Verification:
p(1/2) = 2(1/8) + 1/4 - 5/2 + 2 = 1/4 + 1/4 - 5/2 + 2 = 1/2 - 5/2 + 2 = -2 + 2 = 0 ✓
p(1) = 2 + 1 - 5 + 2 = 0 ✓
p(-2) = 2(-8) + 4 + 10 + 2 = -16 + 16 = 0 ✓
Relationship (a = 2, b = 1, c = -5, d = 2):
Sum of zeros = 1/2 + 1 + (-2) = -1/2 = -b/a = -1/2 ✓
Sum of product of zeros taken two at a time = (1/2)(1) + (1)(-2) + (-2)(1/2) = 1/2 - 2 - 1 = -5/2 = c/a = -5/2 ✓
Product of zeros = (1/2)(1)(-2) = -1 = -d/a = -2/2 = -1 ✓
(ii) p(x) = x^3 - 4x^2 + 5x - 2; zeros: 2, 1, 1
p(2) = 8 - 16 + 10 - 2 = 0 ✓
p(1) = 1 - 4 + 5 - 2 = 0 ✓
Relationship (a = 1, b = -4, c = 5, d = -2):
Sum = 2 + 1 + 1 = 4 = -(-4)/1 = 4 ✓
Sum of products (two at a time) = 2(1) + 1(1) + 1(2) = 2 + 1 + 2 = 5 = 5/1 ✓
Product = 2(1)(1) = 2 = -(-2)/1 = 2 ✓
Q2Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as 2, -7, -14 respectively.
For a cubic polynomial ax^3 + bx^2 + cx + d with zeros alpha, beta, gamma:
alpha + beta + gamma = -b/a = 2
alpha.beta + beta.gamma + gamma.alpha = c/a = -7
alpha.beta.gamma = -d/a = -14
Taking a = 1:
-b = 2, so b = -2
c = -7
-d = -14, so d = 14
The cubic polynomial is x^3 - 2x^2 - 7x + 14.
Q3If the zeros of the polynomial x^3 - 3x^2 + x + 1 are a - b, a, a + b, find a and b.
Given zeros: (a - b), a, (a + b)
For p(x) = x^3 - 3x^2 + x + 1 (coefficients: leading = 1, second = -3, third = 1, constant = 1):
Sum of zeros = (a - b) + a + (a + b) = 3a
By Vieta's: 3a = -(-3)/1 = 3
So a = 1
Product of zeros = (a - b)(a)(a + b) = a(a^2 - b^2)
By Vieta's: a(a^2 - b^2) = -(1)/1 = -1
1(1 - b^2) = -1
1 - b^2 = -1
b^2 = 2
b = sqrt(2) or b = -sqrt(2)
Q4If two zeros of the polynomial x^4 - 6x^3 - 26x^2 + 138x - 35 are 2 + sqrt(3) and 2 - sqrt(3), find the other zeros.
Given zeros: 2 + sqrt(3) and 2 - sqrt(3).
Their factor: [x - (2 + sqrt(3))][x - (2 - sqrt(3))] = (x - 2)^2 - 3 = x^2 - 4x + 1
Dividing x^4 - 6x^3 - 26x^2 + 138x - 35 by x^2 - 4x + 1:
x^4 - 6x^3 - 26x^2 + 138x - 35 = (x^2 - 4x + 1)(x^2 - 2x - 35)
Factoring x^2 - 2x - 35 = (x - 7)(x + 5)
The other two zeros are 7 and -5.
Q5If the polynomial x^4 - 6x^3 + 16x^2 - 25x + 10 is divided by another polynomial x^2 - 2x + k, the remainder comes out to be x + a, find k and a.
Dividing x^4 - 6x^3 + 16x^2 - 25x + 10 by x^2 - 2x + k:
Quotient = x^2 - 4x + (8 - k)
Remainder = (-9 + 2k)x + (10 - k(8 - k)) = (-9 + 2k)x + (10 - 8k + k^2)
Given remainder = x + a:
Comparing coefficient of x: -9 + 2k = 1, so 2k = 10, k = 5
Comparing constant: 10 - 8(5) + 5^2 = 10 - 40 + 25 = -5
So a = -5
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Ask AI TutorPrevious Year Board Exam Questions
If one zero of the polynomial 2x^2 - 3x + k is the reciprocal of the other, find the value of k.
Let the zeros be alpha and 1/alpha.
Product of zeros = alpha x (1/alpha) = 1
Using the formula: product of zeros = c/a = k/2
So k/2 = 1, which gives k = 2.
If alpha and beta are the zeros of the polynomial x^2 - 5x + 6, find the value of alpha + beta and alpha x beta.
Given polynomial: x^2 - 5x + 6
Here a = 1, b = -5, c = 6.
Sum of zeros = alpha + beta = -b/a = -(-5)/1 = 5
Product of zeros = alpha x beta = c/a = 6/1 = 6
We can verify: x^2 - 5x + 6 = (x - 2)(x - 3), so zeros are 2 and 3.
Sum = 2 + 3 = 5 ✓, Product = 2 x 3 = 6 ✓
Find the zeros of the quadratic polynomial 6x^2 - 7x - 3 and verify the relationship between the zeros and the coefficients.
6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)
Zeros: x = -1/3 and x = 3/2
Verification (a = 6, b = -7, c = -3):
Sum of zeros = -1/3 + 3/2 = -2/6 + 9/6 = 7/6 = -b/a = -(-7)/6 = 7/6 ✓
Product of zeros = (-1/3)(3/2) = -1/2 = c/a = -3/6 = -1/2 ✓
Zeros are -1/3 and 3/2. Relationship verified.
If alpha and beta are zeros of the polynomial p(x) = x^2 - 7x + 10, find the value of 1/alpha + 1/beta.
Given polynomial: x^2 - 7x + 10. Here a = 1, b = -7, c = 10.
alpha + beta = -b/a = 7
alpha x beta = c/a = 10
1/alpha + 1/beta = (alpha + beta)/(alpha x beta) = 7/10
1/alpha + 1/beta = 7/10
Find a quadratic polynomial, the sum and product of whose zeros are -3 and 2 respectively.
If sum of zeros = S = -3 and product of zeros = P = 2, then the quadratic polynomial is:
p(x) = x^2 - Sx + P = x^2 - (-3)x + 2 = x^2 + 3x + 2
Verification: x^2 + 3x + 2 = (x + 1)(x + 2), zeros are -1 and -2.
Sum = -1 + (-2) = -3 ✓, Product = (-1)(-2) = 2 ✓
The required polynomial is x^2 + 3x + 2.
The graph of y = p(x) is a parabola which intersects the x-axis at (-2, 0) and (4, 0). The number of zeros of p(x) is:
The graph intersects the x-axis at two points: (-2, 0) and (4, 0). The number of zeros of a polynomial equals the number of points where its graph intersects the x-axis. Since the parabola crosses the x-axis at exactly two points, the polynomial has 2 zeros.
If the sum of the zeros of the polynomial p(x) = (k^2 - 14)x^2 - 2x - 12 is 1, then find the value of k.
For the polynomial (k^2 - 14)x^2 - 2x - 12:
a = k^2 - 14, b = -2, c = -12
Sum of zeros = -b/a = -(-2)/(k^2 - 14) = 2/(k^2 - 14)
Given sum = 1:
2/(k^2 - 14) = 1
k^2 - 14 = 2
k^2 = 16
k = 4 or k = -4
If alpha and beta are the zeros of the polynomial f(x) = x^2 - px + q, find the value of alpha^2 + beta^2.
Given: f(x) = x^2 - px + q
alpha + beta = p (sum of zeros)
alpha x beta = q (product of zeros)
alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha)(beta)
= p^2 - 2q
alpha^2 + beta^2 = p^2 - 2q
Find the zeros of the quadratic polynomial 4x^2 - 4x + 1 and verify the relationship between the zeros and the coefficients.
4x^2 - 4x + 1 = (2x - 1)^2
Zeros: x = 1/2 and x = 1/2 (both equal)
Here a = 4, b = -4, c = 1.
Sum of zeros = 1/2 + 1/2 = 1 = -(-4)/4 = 1 ✓
Product of zeros = (1/2)(1/2) = 1/4 = 1/4 ✓
Zeros are 1/2 and 1/2. Both the sum and product relationships are verified.
If alpha and beta are zeros of x^2 + 7x + 12, then find the value of 1/alpha + 1/beta - 2(alpha)(beta).
For x^2 + 7x + 12: a = 1, b = 7, c = 12
alpha + beta = -7, alpha x beta = 12
1/alpha + 1/beta - 2(alpha)(beta)
= (alpha + beta)/(alpha x beta) - 2(alpha x beta)
= (-7)/12 - 2(12)
= -7/12 - 24
= -7/12 - 288/12
= -295/12
The value is -295/12.
If one zero of the polynomial (a^2 + 9)x^2 + 13x + 6a is the reciprocal of the other, find the value of a.
Let zeros be alpha and 1/alpha.
Product of zeros = alpha x (1/alpha) = 1
Using the formula: product of zeros = c/a_coeff = 6a/(a^2 + 9)
6a/(a^2 + 9) = 1
6a = a^2 + 9
a^2 - 6a + 9 = 0
(a - 3)^2 = 0
a = 3
A quadratic polynomial, whose zeros are -3 and 4, is:
Sum of zeros = -3 + 4 = 1
Product of zeros = (-3)(4) = -12
Quadratic polynomial = x^2 - (sum)x + (product) = x^2 - x - 12
The polynomial is x^2 - x - 12.
Find the zeros of the quadratic polynomial x^2 + 7x + 10 and verify the relationship between the zeros and the coefficients.
x^2 + 7x + 10 = x^2 + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 2)(x + 5)
Zeros: x = -2 and x = -5
Here a = 1, b = 7, c = 10.
Sum of zeros = -2 + (-5) = -7 = -b/a = -7/1 = -7 ✓
Product of zeros = (-2)(-5) = 10 = c/a = 10/1 = 10 ✓
Zeros are -2 and -5. The sum and product relationships are verified.
If one zero of the polynomial 2x^2 + 3x + k is 1/2, find the value of k and the other zero.
Given: one zero is 1/2. Substituting in 2x^2 + 3x + k:
2(1/4) + 3(1/2) + k = 0
1/2 + 3/2 + k = 0
2 + k = 0
k = -2
Now the polynomial is 2x^2 + 3x - 2.
Sum of zeros = -b/a = -3/2
If one zero is 1/2, other zero = -3/2 - 1/2 = -2
Find the zeros of the polynomial x^2 - 3 and verify the relationship between zeros and coefficients.
x^2 - 3 = (x - sqrt(3))(x + sqrt(3))
Zeros: x = sqrt(3) and x = -sqrt(3)
Here a = 1, b = 0, c = -3.
Sum = sqrt(3) + (-sqrt(3)) = 0 = -b/a = 0/1 = 0 ✓
Product = sqrt(3)(-sqrt(3)) = -3 = c/a = -3/1 = -3 ✓
Zeros are sqrt(3) and -sqrt(3). Relationship verified.
If alpha and beta are the zeros of the polynomial 2x^2 - 5x + 7, find the value of alpha^(-1) + beta^(-1).
For 2x^2 - 5x + 7: a = 2, b = -5, c = 7.
alpha + beta = -b/a = 5/2
alpha x beta = c/a = 7/2
alpha^(-1) + beta^(-1) = 1/alpha + 1/beta = (alpha + beta)/(alpha x beta)
= (5/2)/(7/2) = 5/7
1/alpha + 1/beta = 5/7
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The polynomial x^2 - 2x + 3 has no real zeros.
Reason (R): If the discriminant (b^2 - 4ac) of a quadratic polynomial is negative, it has no real zeros.
Answer: (A)
Both are true. For x^2 - 2x + 3: discriminant = (-2)^2 - 4(1)(3) = 4 - 12 = -8, which is negative. Hence no real zeros. The reason correctly explains the assertion.
AR2
Assertion (A): The graph of a quadratic polynomial touches the x-axis at exactly one point when both zeros are equal.
Reason (R): When both zeros of a quadratic polynomial are equal, the polynomial is a perfect square.
Answer: (A)
Both are true. When zeros are equal (say alpha = beta), the graph touches the x-axis at one point. For example, (x - 2)^2 = x^2 - 4x + 4 touches x-axis at x = 2. The reason correctly explains the assertion.
AR3
Assertion (A): A polynomial of degree 3 can have at most 4 zeros.
Reason (R): A polynomial of degree n has at most n zeros.
Answer: (D)
Assertion is FALSE: a degree 3 polynomial has at most 3 zeros, not 4. Reason is TRUE: a polynomial of degree n has at most n zeros.
AR4
Assertion (A): If one zero of 2x^2 - 5x + k is the reciprocal of the other, then k = 2.
Reason (R): If one zero is alpha and the other is 1/alpha, then their product is 1.
Answer: (A)
Both are true. Product of zeros = alpha x (1/alpha) = 1. Also product = k/2. So k/2 = 1, k = 2. The reason correctly explains the assertion.
AR5
Assertion (A): The sum of the zeros of 3x^2 - 5x + 2 is 5/3.
Reason (R): For a quadratic polynomial ax^2 + bx + c, the sum of zeros is c/a.
Answer: (C)
Assertion is TRUE: sum of zeros = -b/a = -(-5)/3 = 5/3. Reason is FALSE: the sum of zeros is -b/a, not c/a. The formula c/a gives the product of zeros.
AR6
Assertion (A): The number of zeros of the polynomial p(x) as shown in the graph is 3, where the graph intersects the x-axis at three distinct points.
Reason (R): The number of zeros of a polynomial is equal to the number of points where the graph of the polynomial intersects the y-axis.
Answer: (C)
Assertion is TRUE: if the graph intersects the x-axis at 3 points, the polynomial has 3 zeros. Reason is FALSE: zeros correspond to x-axis intersections, not y-axis intersections.
AR7
Assertion (A): If the product of zeros of the quadratic polynomial x^2 + 3x + k is 2, then k = 2.
Reason (R): For a quadratic polynomial ax^2 + bx + c, the product of zeros is c/a.
Answer: (A)
Both are true. Product of zeros = c/a = k/1 = k. Given product = 2, so k = 2. The reason correctly explains the assertion.
Case Study Questions
During a mathematics exhibition, a student designed a model of a roller coaster track. The path of the roller coaster is represented by the polynomial p(x) = x^3 - 5x^2 + 2x + 8. The graph of y = p(x) shows where the track crosses the ground level (x-axis). The points where the track meets the ground level represent the zeros of the polynomial.
(i) How many times does the roller coaster track cross the ground level? (1 mark)
The polynomial p(x) = x^3 - 5x^2 + 2x + 8 is of degree 3. Factoring: p(x) = (x + 1)(x - 2)(x - 4). Since there are 3 distinct linear factors, the graph crosses the x-axis at 3 points.
The track crosses the ground level 3 times.
(ii) Find all the zeros of the polynomial p(x) = x^3 - 5x^2 + 2x + 8. (2 marks)
p(x) = x^3 - 5x^2 + 2x + 8
Trying x = -1: p(-1) = -1 - 5 - 2 + 8 = 0 ✓
So (x + 1) is a factor.
Dividing: x^3 - 5x^2 + 2x + 8 = (x + 1)(x^2 - 6x + 8) = (x + 1)(x - 2)(x - 4)
The zeros are -1, 2, and 4.
(iii)(a) What is the sum of the zeros of p(x)? (1 mark)
Sum of zeros = (-1) + 2 + 4 = 5
Verification: -(-5)/1 = 5 ✓
Sum of zeros = 5.
(iii)(b) What is the product of the zeros of p(x)? (1 mark)
Product of zeros = (-1)(2)(4) = -8
Verification: -(8)/1 = -8 ✓
Product of zeros = -8.
A school is constructing a basketball court. The area of the rectangular court can be expressed as the polynomial A(x) = 6x^2 + x - 2, where x is a measurement in metres. The length and breadth of the court are the factors of A(x). Also, the path of a basketball thrown by a player is modelled by the quadratic polynomial h(t) = -4t^2 + 16t, where h is the height in metres and t is the time in seconds.
(i) Find the length and breadth of the court by factoring 6x^2 + x - 2. (1 mark)
6x^2 + x - 2 = 6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2)
Length = (3x + 2) metres, Breadth = (2x - 1) metres (taking the larger factor as length).
(ii) Find the zeros of the polynomial h(t) = -4t^2 + 16t and interpret their meaning. (2 marks)
h(t) = -4t^2 + 16t = -4t(t - 4)
Setting h(t) = 0: -4t(t - 4) = 0
t = 0 or t = 4
The zeros are t = 0 and t = 4. This means the basketball is at ground level at t = 0 seconds (when thrown) and t = 4 seconds (when it lands).
Zeros are 0 and 4. The ball is in the air for 4 seconds.
(iii)(a) Verify the relationship between zeros and coefficients for h(t) = -4t^2 + 16t. (1 mark)
Zeros: 0 and 4. Here a = -4, b = 16, c = 0.
Sum of zeros = 0 + 4 = 4 = -b/a = -(16)/(-4) = 4 ✓
Product of zeros = 0 x 4 = 0 = c/a = 0/(-4) = 0 ✓
Both relationships are verified.
(iii)(b) At what time does the basketball reach its maximum height? (1 mark)
For a quadratic h(t) = -4t^2 + 16t, the maximum occurs at t = -b/(2a) = -16/(2 x -4) = -16/(-8) = 2 seconds.
Maximum height = h(2) = -4(4) + 16(2) = -16 + 32 = 16 metres.
The basketball reaches maximum height of 16 metres at t = 2 seconds.
MCQ Practice
MCQ1. The graph of y = p(x) does not intersect the x-axis at any point. The number of zeros of p(x) is:
(A) 0
(B) 1
(C) 2
(D) 3
Answer: (A) - The number of zeros equals the number of x-axis intersections. If the graph does not touch or cross the x-axis, there are 0 zeros.
MCQ2. If one zero of the quadratic polynomial x^2 + 3x + k is 2, then the value of k is:
(A) 10
(B) -10
(C) 5
(D) -5
Answer: (B) - If 2 is a zero, then p(2) = 0. So (2)^2 + 3(2) + k = 0, giving 4 + 6 + k = 0, k = -10.
MCQ3. The zeros of the quadratic polynomial x^2 + 99x + 127 are:
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal
Answer: (B) - Sum of zeros = -99 (negative) and product of zeros = 127 (positive). Since the product is positive, both zeros have the same sign. Since the sum is negative, both zeros must be negative.
MCQ4. If alpha and beta are the zeros of the polynomial 2x^2 - 3x + 7, then alpha + beta is:
(A) 3/2
(B) -3/2
(C) 7/2
(D) -7/2
Answer: (A) - Sum of zeros = -b/a = -(-3)/2 = 3/2.
MCQ5. A quadratic polynomial whose sum of zeros is -5 and product of zeros is 6 is:
(A) x^2 + 5x + 6
(B) x^2 - 5x + 6
(C) x^2 - 5x - 6
(D) x^2 + 5x - 6
Answer: (A) - Polynomial = x^2 - (sum)x + (product) = x^2 - (-5)x + 6 = x^2 + 5x + 6.
MCQ6. The number of polynomials having zeros as -2 and 5 is:
(A) 1
(B) 2
(C) 3
(D) more than 3
Answer: (D) - If zeros are -2 and 5, the polynomial is k(x + 2)(x - 5) = k(x^2 - 3x - 10), where k is any non-zero constant. Since k can be any non-zero real number, there are infinitely many such polynomials.
MCQ7. If alpha and beta are zeros of x^2 - 4x + 3, then (alpha)(beta) is:
(A) -4
(B) 4
(C) -3
(D) 3
Answer: (D) - Product of zeros = c/a = 3/1 = 3.
MCQ8. The degree of the polynomial (x + 1)(x^2 - x - x^4 + 1) is:
(A) 2
(B) 3
(C) 4
(D) 5
Answer: (D) - The highest degree term in (x^2 - x - x^4 + 1) is -x^4 (degree 4). Multiplying by (x + 1) gives maximum degree 4 + 1 = 5.
MCQ9. If the graph of a polynomial intersects the x-axis at exactly two points, it is necessarily a:
(A) linear polynomial
(B) quadratic polynomial
(C) polynomial of degree 2 or more
(D) cubic polynomial
Answer: (C) - A polynomial with exactly 2 x-axis intersections has at least 2 zeros. It could be degree 2 (parabola crossing twice), degree 3 (cubic touching at one, crossing at another), or higher. It must be degree 2 or more.
MCQ10. If alpha and beta are zeros of 5x^2 - 7x + 1, then the value of (alpha)(beta) + (beta)(alpha) is:
(A) 7/5
(B) -7/5
(C) 39/5
(D) -39/5
Answer: (C) - alpha/beta + beta/alpha = (alpha^2 + beta^2)/(alpha x beta). alpha + beta = 7/5, alpha x beta = 1/5. alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha)(beta) = 49/25 - 2/5 = 49/25 - 10/25 = 39/25. So (39/25)/(1/5) = (39/25)(5/1) = 39/5.
MCQ11. If both zeros of the quadratic polynomial ax^2 + bx + c are equal and opposite in sign, then:
(A) b = 0
(B) c = 0
(C) a = 0
(D) a = c
Answer: (A) - If zeros are alpha and -alpha, sum of zeros = 0. So -b/a = 0, which means b = 0.
MCQ12. If one zero of the polynomial 3x^2 - 8x + 2k + 1 is seven times the other, then the value of k is:
(A) 2/3
(B) 1
(C) 2
(D) 3
Answer: (A) - Let zeros be alpha and 7alpha. Sum = 8alpha = 8/3, so alpha = 1/3. Product = 7alpha^2 = 7(1/9) = 7/9 = (2k + 1)/3. So 2k + 1 = 7/3, 2k = 4/3, k = 2/3.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If one zero of 2x^2 - 3x + k is the reciprocal of the other, find k.
Product of zeros = 1 (reciprocal zeros). So k/2 = 1, giving k = 2.
Write a quadratic polynomial whose zeros are 3 and -5.
Sum = -2, Product = -15. Polynomial = x^2 - (-2)x + (-15) = x^2 + 2x - 15.
What is the maximum number of zeros a cubic polynomial can have?
A polynomial of degree n has at most n zeros. A cubic polynomial (degree 3) has at most 3 zeros.
If alpha and beta are zeros of x^2 - 6x + k, and alpha - beta = 2, find k.
alpha + beta = 6, alpha - beta = 2. So alpha = 4, beta = 2. k = product = 4 x 2 = 8.
Find the sum of zeros of the polynomial 3x^2 - 5x + 2.
Sum of zeros = -b/a = -(-5)/3 = 5/3.
2-3 Mark Questions
Find the zeros of x^2 - 2x - 8 and verify the relationship between zeros and coefficients.
x^2 - 2x - 8 = (x - 4)(x + 2). Zeros: 4 and -2.
Sum = 4 + (-2) = 2 = -(-2)/1 = 2 ✓
Product = 4(-2) = -8 = -8/1 ✓
If alpha, beta are zeros of x^2 - 5x + 6, find alpha^2 + beta^2.
alpha + beta = 5, alpha x beta = 6.
alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha x beta) = 25 - 12 = 13.
Find a quadratic polynomial whose zeros are 2 + sqrt(3) and 2 - sqrt(3).
Sum = (2 + sqrt(3)) + (2 - sqrt(3)) = 4.
Product = (2 + sqrt(3))(2 - sqrt(3)) = 4 - 3 = 1.
Polynomial = x^2 - 4x + 1.
If alpha and beta are zeros of x^2 - px + q, find the value of alpha/beta + beta/alpha.
alpha/beta + beta/alpha = (alpha^2 + beta^2)/(alpha x beta) = [(alpha + beta)^2 - 2(alpha x beta)]/(alpha x beta) = (p^2 - 2q)/q.
3-4 Mark Questions
Find the zeros of 6x^2 - 3 - 7x and verify the relationship between zeros and coefficients.
6x^2 - 7x - 3 = (2x - 3)(3x + 1). Zeros: 3/2 and -1/3.
Sum = 3/2 - 1/3 = 7/6 = -(-7)/6 ✓
Product = (3/2)(-1/3) = -1/2 = -3/6 ✓
If alpha and beta are zeros of 2x^2 + 5x + k such that alpha^2 + beta^2 + alpha.beta = 21/4, find the value of k.
alpha + beta = -5/2, alpha x beta = k/2.
alpha^2 + beta^2 + alpha.beta = (alpha + beta)^2 - 2(alpha.beta) + alpha.beta = (alpha + beta)^2 - alpha.beta
= 25/4 - k/2 = 21/4.
k/2 = 25/4 - 21/4 = 4/4 = 1. So k = 2.
Find the zeros of the polynomial x^2 + 7x + 10 and form a polynomial whose zeros are the reciprocals of the zeros.
x^2 + 7x + 10 = (x + 2)(x + 5). Zeros: -2 and -5.
Reciprocals: -1/2 and -1/5.
Sum = -1/2 + (-1/5) = -7/10.
Product = (-1/2)(-1/5) = 1/10.
New polynomial = x^2 + 7x/10 + 1/10. Multiplying by 10: 10x^2 + 7x + 1.
Key Concepts
Important Formulas
Study Plan - Polynomials
Basics of Polynomials and Graphical Meaning of Zeros
Read NCERT theory on polynomials, degrees, types (linear, quadratic, cubic). Understand graphical meaning of zeros. Solve Exercise 2.1 Q1 (all 6 graphs).
Zeros and Coefficients Relationship
Learn the sum and product of zeros formulas for quadratic polynomials. Solve Exercise 2.2 Q1 (all 6 parts - find zeros and verify relationship).
Forming Polynomials from Zeros
Understand how to form a quadratic polynomial given sum and product of zeros. Solve Exercise 2.2 Q2 (all 6 parts). Practice forming polynomials with irrational zeros.
Previous Year Questions and MCQs
Solve all 15 PYQs from this page. Practice all 12 MCQs and 7 assertion-reason questions. Focus on alpha/beta relationship-based calculations.
Case Study + Revision Test
Attempt both case study questions under timed conditions. Solve all important questions (mark-wise) as a mini-test. Review common mistakes.
Common Mistakes to Avoid
- ✗Confusing the sum of zeros formula: it is -b/a (not b/a). Many students forget the negative sign and lose marks
- ✗Confusing sum of zeros (-b/a) with product of zeros (c/a) - remember 'S for Sum uses second coefficient b, P for Product uses constant c'
- ✗When forming a polynomial from given zeros, writing x^2 + (sum)x + product instead of x^2 - (sum)x + product. The sum has a MINUS sign in front
- ✗Not rearranging the polynomial in standard form (descending powers of x) before applying the formulas. For example, 6x^2 - 3 - 7x must be written as 6x^2 - 7x - 3 first
- ✗In graph-based questions, counting the number of turning points instead of x-axis intersections. Zeros correspond only to points where the graph crosses or touches the x-axis
- ✗Forgetting to verify the relationship after finding zeros - CBSE awards 1 mark for the verification step, so skipping it means losing marks
Tips for Board Exam
- ✓Memorise the two key formulas: sum of zeros = -b/a and product of zeros = c/a. These are tested in every exam without fail, either directly or through MCQs
- ✓For 'find zeros and verify' questions (3 marks), always show three parts clearly: (1) factorisation, (2) finding zeros, (3) verification using -b/a and c/a. Each part earns marks
- ✓When asked 'find a quadratic polynomial whose zeros are...', always use x^2 - (sum)x + (product). Multiply through if there are fractions to get integer coefficients
- ✓For graph-based MCQs (Exercise 2.1 type), remember: number of zeros = number of x-axis intersections. A graph that touches the x-axis at one point still counts as a zero (repeated zero)
- ✓Practice 'reciprocal zeros' and 'one zero is k times the other' type questions - these are high-frequency PYQ patterns that appear almost every year
Practice Polynomials with AI Tutor
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Start Polynomials - Choose Your ModeAbout Polynomials - Class 10 Mathematics
Chapter 2 Polynomials is the second chapter of CBSE Class 10 Mathematics and belongs to the Algebra unit, which carries 20 marks in total across Chapters 2 through 5. Polynomials typically carries 3 to 4 marks in the CBSE Class 10 board exam 2027. This chapter builds on the polynomial concepts introduced in Class 9 and focuses on the relationship between zeros and coefficients of quadratic polynomials, graphical interpretation of zeros, and forming polynomials from given zeros.
What Are Polynomials?
A polynomial is an algebraic expression made up of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. For example, 2x^3 - 5x^2 + 3x - 7 is a polynomial in x. The highest power of the variable is called the degree of the polynomial. A polynomial of degree 1 is called linear (e.g., 3x + 2), degree 2 is called quadratic (e.g., x^2 - 5x + 6), and degree 3 is called cubic (e.g., x^3 - 4x).
The general form of a quadratic polynomial is ax^2 + bx + c, where a is not equal to 0. In Class 10, most of the work focuses on quadratic polynomials because the relationship between zeros and coefficients is especially elegant and frequently tested in board exams.
Zeros of a Polynomial
A zero of a polynomial p(x) is a value of x for which p(x) = 0. For example, if p(x) = x^2 - 5x + 6, then p(2) = 4 - 10 + 6 = 0 and p(3) = 9 - 15 + 6 = 0. So 2 and 3 are the zeros of this polynomial. Finding zeros means solving the equation p(x) = 0.
For quadratic polynomials, zeros are found by factorisation. The standard method is to split the middle term. For example, to factorise x^2 - 2x - 8, we look for two numbers that multiply to give -8 and add to give -2. These numbers are -4 and 2. So x^2 - 2x - 8 = (x - 4)(x + 2), and the zeros are 4 and -2.
An important property is that a polynomial of degree n has at most n zeros. A linear polynomial has exactly 1 zero, a quadratic has at most 2 zeros, a cubic has at most 3 zeros, and so on. Some quadratic polynomials like x^2 + 1 have no real zeros because x^2 + 1 is always positive for real values of x.
Relationship Between Zeros and Coefficients
This is the most important concept in Chapter 2 and the one tested most frequently. For a quadratic polynomial ax^2 + bx + c with zeros alpha and beta:
Sum of zeros: alpha + beta = -b/a Product of zeros: alpha x beta = c/a
These formulas come from the factorisation ax^2 + bx + c = a(x - alpha)(x - beta). Expanding the right side and comparing coefficients gives the relationships.
For example, for the polynomial 3x^2 - 5x + 2: a = 3, b = -5, c = 2. The sum of zeros = -(-5)/3 = 5/3 and the product = 2/3. Students can verify by factoring: 3x^2 - 5x + 2 = (3x - 2)(x - 1), zeros are 2/3 and 1. Sum = 2/3 + 1 = 5/3 and product = 2/3 x 1 = 2/3.
Conversely, if the sum of zeros is S and the product is P, then the quadratic polynomial is x^2 - Sx + P or any non-zero scalar multiple k(x^2 - Sx + P). This allows us to form a polynomial from its zeros.
For cubic polynomials ax^3 + bx^2 + cx + d with zeros alpha, beta, gamma: alpha + beta + gamma = -b/a alpha.beta + beta.gamma + gamma.alpha = c/a alpha.beta.gamma = -d/a
These cubic relationships are tested in state board exams but not in CBSE 2026-27 since the related exercises have been removed.
Graphical Meaning of Zeros
The zeros of a polynomial p(x) are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. This is a simple but powerful concept that appears as MCQs in every board exam.
For a linear polynomial ax + b, the graph is a straight line that intersects the x-axis at one point (-b/a, 0). So it has exactly one zero.
For a quadratic polynomial, the graph is a parabola. If it intersects the x-axis at two distinct points, the polynomial has 2 distinct zeros. If it touches the x-axis at one point, both zeros are equal (the parabola just touches). If it does not intersect the x-axis at all, the polynomial has no real zeros.
The shape of the parabola depends on the sign of the leading coefficient. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward.
Division Algorithm (For State Boards)
The Division Algorithm for polynomials states that for any two polynomials p(x) and g(x) where g(x) is not zero, there exist polynomials q(x) (quotient) and r(x) (remainder) such that p(x) = g(x) x q(x) + r(x), where the degree of r(x) is less than the degree of g(x), or r(x) is zero.
This is analogous to the division algorithm for integers (Euclid's Division Lemma from Chapter 1). The process of polynomial long division is used to find the quotient and remainder.
This topic has been removed from the CBSE 2026-27 syllabus but remains relevant for state board students in UP, Bihar, MP, Rajasthan, and Haryana. Padhaao provides solutions for the older exercises (2.3 and 2.4) for these students.
How This Chapter Is Tested in Board Exams
Polynomials questions in the CBSE board exam typically follow these patterns. One MCQ worth 1 mark tests the graphical meaning of zeros (how many zeros from a given graph) or the sum/product relationship (find the sum of zeros given a polynomial). One short answer question worth 2 to 3 marks asks students to find the zeros of a given quadratic polynomial and verify the relationship between zeros and coefficients. Alternatively, students may be asked to form a quadratic polynomial given the sum and product of zeros.
In recent years, the 'reciprocal zeros' pattern has been very popular: If one zero of kx^2 + bx + c is the reciprocal of the other, find k. Since the product of reciprocal zeros is 1, c/a = 1 gives the answer directly. Questions asking for expressions like 1/alpha + 1/beta or alpha^2 + beta^2 in terms of coefficients are also common.
Assertion-Reason questions on polynomials test conceptual clarity about degree, zeros, and coefficient relationships. Case study questions typically involve a real-world scenario where a quadratic polynomial models a physical situation like the path of a ball or the area of a field.
Tips for Scoring Full Marks
To score all 3 to 4 marks from this chapter, students should focus on four things. First, memorise the sum and product formulas perfectly, including the negative sign in the sum formula (-b/a, not b/a). Second, practice factorisation of all 6 polynomials in Exercise 2.2 Q1 until it becomes automatic. Third, always write the verification step when finding zeros, since examiners award separate marks for it. Fourth, solve at least 10 previous year questions to recognise the common patterns like reciprocal zeros, alpha^2 + beta^2, and 1/alpha + 1/beta.
Common mistakes to avoid include forgetting the negative sign in the sum formula, confusing sum and product formulas, not rearranging polynomials into standard form before applying formulas, and counting turning points instead of x-axis intersections in graph questions.
Connection to Other Chapters
Polynomials connects directly to Chapter 4 (Quadratic Equations). The zeros of a quadratic polynomial ax^2 + bx + c are exactly the roots of the quadratic equation ax^2 + bx + c = 0. The sum and product of roots formulas in Chapter 4 are the same as the sum and product of zeros in Chapter 2. Students who master polynomials will find quadratic equations much easier.
The concept of factorisation used to find zeros also appears in Chapter 3 (Pair of Linear Equations) when solving systems by substitution or elimination. Understanding graphs of polynomials helps with the graphical method of solving linear equations.
For students preparing for competitive exams like NTSE and Olympiads, the polynomial relationships extend to higher-degree polynomials and connect to advanced topics like symmetric functions of roots and polynomial identities.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 2 Polynomials for CBSE 2026-27?
For CBSE 2026-27, Chapter 2 Polynomials has only 2 exercises: Exercise 2.1 (1 question with 6 graph-based parts on finding number of zeros) and Exercise 2.2 (2 questions on finding zeros and verifying the relationship, and forming polynomials from given zeros). The older exercises on Division Algorithm for Polynomials (Exercise 2.3 and 2.4) have been removed from the CBSE syllabus.
What is the relationship between zeros and coefficients of a quadratic polynomial?
For a quadratic polynomial ax^2 + bx + c with zeros alpha and beta: Sum of zeros (alpha + beta) = -b/a and Product of zeros (alpha x beta) = c/a. For example, for x^2 - 5x + 6 with zeros 2 and 3: sum = 2 + 3 = 5 = -(-5)/1 and product = 2 x 3 = 6 = 6/1.
How many marks does Polynomials carry in the CBSE Class 10 board exam?
Polynomials is part of the Algebra unit which carries 20 marks total (shared across Chapters 2, 3, 4, and 5). Polynomials typically carries 3 to 4 marks: usually 1 MCQ (1 mark) and 1 short answer question on finding zeros or forming a polynomial (2-3 marks).
How to find zeros of a quadratic polynomial for the CBSE board exam?
To find zeros of a quadratic polynomial ax^2 + bx + c: (1) Factorise by splitting the middle term - find two numbers that multiply to give a x c and add to give b. (2) Write the polynomial as a product of two linear factors. (3) Set each factor to zero to find the zeros. (4) Verify using sum = -b/a and product = c/a.
Is the Division Algorithm for Polynomials deleted from CBSE 2026-27 syllabus?
Yes, the Division Algorithm for Polynomials (old Exercise 2.3 and 2.4) has been removed from the CBSE 2026-27 syllabus. These topics will not appear in the CBSE Class 10 board exam 2027. However, state board students (UP Board, Bihar Board, MP Board) may still need to study these topics. Padhaao provides solutions for these older exercises as well.
What is the formula to form a quadratic polynomial from its zeros?
If the sum of zeros is S and the product of zeros is P, then the quadratic polynomial is x^2 - Sx + P. You can multiply by any non-zero constant k to get k(x^2 - Sx + P). For example, if zeros are 3 and -2: sum = 1, product = -6, polynomial = x^2 - x - 6.
How to find number of zeros from the graph of a polynomial?
The number of zeros of a polynomial p(x) equals the number of points where the graph of y = p(x) intersects or touches the x-axis. If the graph does not cross the x-axis at all, the number of zeros is 0. If it crosses at 2 points, the polynomial has 2 zeros. A graph touching the x-axis at one point counts as one zero (repeated).
What type of questions come from Polynomials in the CBSE board exam?
Three main types: (1) MCQs on number of zeros from graph or sum/product of zeros - 1 mark. (2) Short answer: find zeros of a quadratic polynomial and verify the relationship with coefficients - 2 to 3 marks. (3) Short answer: form a quadratic polynomial given sum and product of zeros, or find the value of k when one zero satisfies a condition (like reciprocal zeros) - 2 marks. Case study questions on polynomial models were introduced in CBSE 2023-24.