NCERT Solutions Class 10 Mathematics Polynomials

Mastering NCERT Solutions Class 10 Mathematics for Polynomials is the key to building a rock-solid foundation in algebra for your CBSE Class 10 Board Examination. Polynomials is one of the most scoring chapters in the Class 10 Mathematics syllabus, bridging foundational concepts of quadratic equations with coordinate geometry and algebraic identities. Whether you are aiming for a perfect 100/100 in board exams or seeking clarity on graphical zeroes and coefficient relationships, these NCERT Solutions step by step provide complete, error-free, and comprehensive explanations for every exercise question strictly aligned with the latest CBSE syllabus.

Chapter Overview

A polynomial in one variable x is an algebraic expression of the form p(x) = a_nxⁿ + a_n-1x^n-1 + … + a₁x + a₀, where a_n ≠ 0 and each power of x is a non-negative integer. The highest power of x is known as the degree of the polynomial.

Classification of Polynomials by Degree

Linear Polynomial: A polynomial of degree 1. Standard form: p(x) = ax + b, where a ≠ 0. It has at most 1 real zero.
Quadratic Polynomial: A polynomial of degree 2. Standard form: p(x) = ax² + bx + c, where a ≠ 0. It has at most 2 real zeroes. Its graph is a parabola opening upwards (if a > 0) or downwards (if a < 0).
Cubic Polynomial: A polynomial of degree 3. Standard form: p(x) = ax³ + bx² + cx + d, where a ≠ 0. It has at most 3 real zeroes.

Geometrical Meaning of Zeroes

A real number k is called a zero of a polynomial p(x) if p(k) = 0. Geometrically, the real zeroes of p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects or touches the x-axis. Thus, a polynomial of degree n can intersect the x-axis at a maximum of n points and therefore can have at most n real zeroes.

Real-World Applications

Polynomial functions are widely applied in real life. Quadratic polynomials model projectile trajectories (such as the path of a basketball or rocket), satellite dish curvatures, suspension bridge cables, and profit-maximization curves in economics.

Important Formulas and Theorems

Below is the complete summary of formulas required for solving NCERT Mathematics Class 10 solutions for this chapter:

1. Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

For a quadratic polynomial p(x) = ax² + bx + c (where a ≠ 0), let α and β be its two zeroes:

Sum of Zeroes: α + β = -b / a = -(Coefficient of x) / (Coefficient of x²)
Product of Zeroes: αβ = c / a = (Constant term) / (Coefficient of x²)

2. Forming a Quadratic Polynomial

If α and β are the zeroes of a quadratic polynomial, the polynomial can be written as:

p(x) = k [x² - (α + β)x + αβ] = k [x² - (Sum of zeroes)x + (Product of zeroes)], where k is any non-zero real constant.

3. Relationship Between Zeroes and Coefficients of a Cubic Polynomial

For a cubic polynomial p(x) = ax³ + bx² + cx + d (where a ≠ 0), let α, β, and γ be its zeroes:

Sum of Zeroes: α + β + γ = -b / a
Sum of the Products of Zeroes taken two at a time: αβ + βγ + γα = c / a
Product of Zeroes: αβγ = -d / a

4. Division Algorithm for Polynomials

If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) (quotient) and r(x) (remainder) such that:

p(x) = g(x) × q(x) + r(x), where either r(x) = 0 or deg(r(x)) < deg(g(x)).

Exercise Solutions

Here are the step-by-step, fully explained CBSE NCERT solutions for all textbook exercises.

Exercise 2.1 Solutions

Question 1: The graphs of y = p(x) are given in the figures below for some polynomials p(x). Find the number of zeroes of p(x) in each case.

(i) Graph is a straight line parallel to the x-axis:

Method: The number of zeroes equals the number of times the graph of y = p(x) intersects the x-axis. Here, the line does not intersect the x-axis at all.

Answer: The number of zeroes is 0.
(ii) Graph intersects the x-axis at 1 point:

Method: The curve cuts the x-axis at exactly one point.

Answer: The number of zeroes is 1.
(iii) Graph intersects the x-axis at 3 points:

Method: The curve intersects the x-axis at three distinct points.

Answer: The number of zeroes is 3.
(iv) Graph is a U-shaped curve (parabola) cutting the x-axis at 2 points:

Method: The curve intersects the x-axis at two distinct points.

Answer: The number of zeroes is 2.
(v) Graph intersects the x-axis at 4 points:

Method: The curve intersects the x-axis at four distinct points.

Answer: The number of zeroes is 4.
(vi) Graph touches/intersects the x-axis at 3 points:

Method: The curve intersects at one point and touches the x-axis at two other points (total 3 points of contact).

Answer: The number of zeroes is 3.

Exercise 2.2 Solutions

Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² - 2x - 8

Step 1: Finding Zeroes by Splitting the Middle Term:

We need two numbers whose sum is -2 and product is -8. These numbers are -4 and 2.

x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)

To find zeroes, set p(x) = 0 ⇒ (x - 4)(x + 2) = 0 ⇒ x = 4 or x = -2.

Let α = 4 and β = -2.

Step 2: Verification with Coefficients:

Comparing x² - 2x - 8 with ax² + bx + c, we have a = 1, b = -2, c = -8.
- Sum of zeroes = α + β = 4 + (-2) = 2. Also, -b/a = -(-2)/1 = 2. ⇒ α + β = -b/a (Verified).
- Product of zeroes = αβ = 4 × (-2) = -8. Also, c/a = -8/1 = -8. ⇒ αβ = c/a (Verified).
Answer: Zeroes are 4 and -2.
(ii) 4s² - 4s + 1

Step 1: Finding Zeroes:

4s² - 4s + 1 = 4s² - 2s - 2s + 1 = 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1) = (2s - 1)²

Setting p(s) = 0 ⇒ (2s - 1) = 0 ⇒ s = 1/2, 1/2.

Let α = 1/2 and β = 1/2.

Step 2: Verification with Coefficients:

Comparing with as² + bs + c: a = 4, b = -4, c = 1.
- Sum of zeroes = 1/2 + 1/2 = 1. Also, -b/a = -(-4)/4 = 4/4 = 1. ⇒ α + β = -b/a (Verified).
- Product of zeroes = (1/2) × (1/2) = 1/4. Also, c/a = 1/4. ⇒ αβ = c/a (Verified).
Answer: Zeroes are 1/2 and 1/2.
(iii) 6x² - 3 - 7x

Step 1: Rearranging in Standard Form:

p(x) = 6x² - 7x - 3

Splitting middle term (-7x = -9x + 2x, since (-9)(2) = -18 = (6)(-3)):

6x² - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)

Setting p(x) = 0 ⇒ 2x - 3 = 0 ⇒ x = 3/2 or 3x + 1 = 0 ⇒ x = -1/3.

Let α = 3/2 and β = -1/3.

Step 2: Verification:

Here a = 6, b = -7, c = -3.
- Sum of zeroes = 3/2 + (-1/3) = (9 - 2)/6 = 7/6. Also, -b/a = -(-7)/6 = 7/6. ⇒ α + β = -b/a (Verified).
- Product of zeroes = (3/2) × (-1/3) = -3/6 = -1/2. Also, c/a = -3/6 = -1/2. ⇒ αβ = c/a (Verified).
Answer: Zeroes are 3/2 and -1/3.
(iv) 4u² + 8u

Step 1: Finding Zeroes:

4u² + 8u = 4u(u + 2)

Setting p(u) = 0 ⇒ 4u = 0 ⇒ u = 0 or u + 2 = 0 ⇒ u = -2.

Let α = 0 and β = -2.

Step 2: Verification:

Here a = 4, b = 8, c = 0.
- Sum of zeroes = 0 + (-2) = -2. Also, -b/a = -8/4 = -2. ⇒ α + β = -b/a (Verified).
- Product of zeroes = 0 × (-2) = 0. Also, c/a = 0/4 = 0. ⇒ αβ = c/a (Verified).
Answer: Zeroes are 0 and -2.
(v) t² - 15

Step 1: Finding Zeroes:

Using identity a² - b² = (a - b)(a + b):

t² - 15 = t² - (√15)² = (t - √15)(t + √15)

Setting p(t) = 0 ⇒ t = √15 or t = -√15.

Let α = √15 and β = -√15.

Step 2: Verification:

Here a = 1, b = 0, c = -15.
- Sum of zeroes = √15 + (-√15) = 0. Also, -b/a = -0/1 = 0. ⇒ α + β = -b/a (Verified).
- Product of zeroes = (√15)(-√15) = -15. Also, c/a = -15/1 = -15. ⇒ αβ = c/a (Verified).
Answer: Zeroes are √15 and -√15.
(vi) 3x² - x - 4

Step 1: Finding Zeroes:

Splitting middle term (-x = -4x + 3x, since (-4)(3) = -12 = (3)(-4)):

3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)

Setting p(x) = 0 ⇒ 3x - 4 = 0 ⇒ x = 4/3 or x + 1 = 0 ⇒ x = -1.

Let α = 4/3 and β = -1.

Step 2: Verification:

Here a = 3, b = -1, c = -4.
- Sum of zeroes = 4/3 + (-1) = 1/3. Also, -b/a = -(-1)/3 = 1/3. ⇒ α + β = -b/a (Verified).
- Product of zeroes = (4/3) × (-1) = -4/3. Also, c/a = -4/3. ⇒ αβ = c/a (Verified).
Answer: Zeroes are 4/3 and -1.

Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) Sum = 1/4, Product = -1

Formula: p(x) = k [x² - (Sum)x + (Product)]

p(x) = k [x² - (1/4)x + (-1)] = k [x² - (1/4)x - 1]

Taking k = 4 to eliminate fractions:

Answer: Required quadratic polynomial is 4x² - x - 4.
(ii) Sum = √2, Product = 1/3

p(x) = k [x² - (√2)x + 1/3]

Taking k = 3:

Answer: Required quadratic polynomial is 3x² - 3√2x + 1.
(iii) Sum = 0, Product = √5

p(x) = k [x² - (0)x + √5] = k [x² + √5]

Taking k = 1:

Answer: Required quadratic polynomial is x² + √5.
(iv) Sum = 1, Product = 1

p(x) = k [x² - (1)x + 1] = k [x² - x + 1]

Taking k = 1:

Answer: Required quadratic polynomial is x² - x + 1.
(v) Sum = -1/4, Product = 1/4

p(x) = k [x² - (-1/4)x + 1/4] = k [x² + (1/4)x + 1/4]

Taking k = 4:

Answer: Required quadratic polynomial is 4x² + x + 1.
(vi) Sum = 4, Product = 1

p(x) = k [x² - (4)x + 1] = k [x² - 4x + 1]

Taking k = 1:

Answer: Required quadratic polynomial is x² - 4x + 1.

High-Yield Board Questions: Higher Degree Polynomials

Example Question: Obtain all other zeroes of 3x⁴ + 6x³ - 2x² - 10x - 5, if two of its zeroes are √(5/3) and -√(5/3).

Solution:

Step 1: Form the quadratic factor: Since x = √(5/3) and x = -√(5/3) are zeroes, (x - √(5/3)) and (x + √(5/3)) are factors.
Factor = (x - √(5/3))(x + √(5/3)) = x² - 5/3 = (1/3)(3x² - 5).
Hence, g(x) = 3x² - 5 is a factor of p(x).
Step 2: Polynomial Long Division:
Dividing 3x⁴ + 6x³ - 2x² - 10x - 5 by 3x² - 5:
- (3x⁴) / (3x²) = x² ⇒ Subtracting (3x⁴ - 5x²) gives remainder 6x³ + 3x² - 10x - 5.
- (6x³) / (3x²) = +2x ⇒ Subtracting (6x³ - 10x) gives remainder 3x² - 5.
- (3x²) / (3x²) = +1 ⇒ Subtracting (3x² - 5) gives remainder 0.
Quotient q(x) = x² + 2x + 1.
Step 3: Factorize the quotient:
x² + 2x + 1 = (x + 1)² = 0 ⇒ x = -1, -1.

Answer: The other zeroes of the given polynomial are -1 and -1.

Common Mistakes and Tips

During the evaluation of board exam answer scripts, examiners frequently notice the following errors. Keep these tips in mind to avoid losing marks:

Sign Error in Sum of Zeroes: Students often write α + β = b/a instead of -b/a. Remember that the negative sign is part of the formula. For p(x) = x² - 5x + 6, b = -5, so α + β = -(-5)/1 = +5.
Counting Y-intercepts as Zeroes: In graphical questions (Exercise 2.1), only count the points where the graph intersects or touches the x-axis. Do not count intersections with the y-axis.
Standard Form Rearrangement: When given an expression like 6x² - 3 - 7x, students frequently identify b = -3 and c = -7. Always rearrange polynomials into standard descending power order ax² + bx + c first (i.e., 6x² - 7x - 3).
Missing the Constant Multiplier 'k': When forming a quadratic polynomial from given sum and product, writing x² - Sx + P instead of k[x² - Sx + P] can lead to fractional coefficients if not cleared properly. Always specify k as a non-zero real constant.
Quadratic Polynomial vs Quadratic Equation: Do not equate a polynomial to 0 unless you are explicitly finding its zeroes. A polynomial is an expression, whereas an equation contains an equality sign.

Board Exam Relevance

In the CBSE Class 10 Board Examinations, the chapter on Polynomials carries a weightage of approximately 4 to 6 marks in the Algebra unit. The standard question distribution follows this pattern:

1 Mark (MCQ / Assertion-Reason): Graphical zeroes identification from y = p(x) or finding the value of k given that one zero is the reciprocal or negative of the other.
2 Marks (Short Answer Type I): Finding zeroes of a standard quadratic polynomial and verifying the relationship with coefficients.
3 Marks (Short Answer Type II): Forming a new quadratic polynomial whose zeroes are related to the zeroes of a given polynomial (e.g., zeroes are 1/α, 1/β or α², β²).
4 Marks (Case-Study Based Question): Practical scenarios involving parabolic arches, projectile heights, or roller coaster tracks requiring students to interpret the vertex, axis of symmetry, and zeroes.

NCERT Solutions Class 10 Mathematics Chapter 1 Polynomials