NCERT Class 10 - Polynomials | Interactive Learning

Important Notes & Formulas

Definition: A polynomial p(x) 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 powers of x are non-negative integers.

Key Fact: The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.

Degree & Zeroes Relationship

Degree	Name	General Form	Max Zeroes
1	Linear	ax + b	1
2	Quadratic	ax² + bx + c	2
3	Cubic	ax³ + bx² + cx + d	3

Relationship between Zeroes and Coefficients

For Quadratic: ax² + bx + c with zeroes α, β

Sum of zeroes: α + β = -b/a
Product of zeroes: αβ = c/a

For Cubic: ax³ + bx² + cx + d with zeroes α, β, γ

Sum: α + β + γ = -b/a
Sum of products (two at a time): αβ + βγ + γα = c/a
Product: αβγ = -d/a

To form a quadratic polynomial: If sum = S and product = P, then:
x² - Sx + P or k(x² - Sx + P) where k ≠ 0

Remember: A polynomial of degree n can have at most n real zeroes.

Exercise 2.1 - Zeroes from Graphs

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

Q1. Graph (i) - A horizontal line not touching the x-axis

A 1 zero

B 0 zeroes

C 2 zeroes

D Infinite zeroes

✅ Correct Answer: 0 zeroes

The graph is a horizontal line parallel to the x-axis and does NOT intersect it.

Since there is no point where y = 0, the polynomial has no real zeroes.

This represents a constant polynomial like p(x) = k where k ≠ 0.

Q2. Graph (ii) - A cubic curve crossing the x-axis at one point only

A 0 zeroes

B 1 zero

C 2 zeroes

D 3 zeroes

✅ Correct Answer: 1 zero

The graph intersects the x-axis at exactly one point.

The x-coordinate of this intersection point is the only zero of the polynomial.

Even though it may be a cubic polynomial, it has only 1 real zero (the other two zeroes are complex/not real).

Q3. Graph (iii) - A curve crossing the x-axis at three distinct points

A 1 zero

B 2 zeroes

C 3 zeroes

D 4 zeroes

✅ Correct Answer: 3 zeroes

The graph intersects the x-axis at three distinct points.

Each intersection point gives one zero of the polynomial.

This is characteristic of a cubic polynomial with 3 real zeroes.

Q4. Graph (iv) - A straight line passing through the origin

A 1 zero

B 0 zeroes

C 2 zeroes

D Infinite zeroes

✅ Correct Answer: 1 zero

A straight line (linear polynomial) intersects the x-axis at exactly one point.

In this case, the line passes through the origin (0, 0), so the zero is x = 0.

For a linear polynomial ax + b, the zero is always -b/a.

Q5. Graph (v) - A parabola touching the x-axis at exactly one point (vertex on x-axis)

A 1 zero

B 0 zeroes

C 2 zeroes

D 3 zeroes

✅ Correct Answer: 1 zero

The parabola touches the x-axis at exactly one point (its vertex lies on the x-axis).

This means the quadratic has equal real roots (discriminant = 0).

Example: y = x² touches x-axis only at x = 0.

We count this as 1 zero (even though the multiplicity is 2).

Q6. Graph (vi) - A curve crossing the x-axis at four distinct points

A 2 zeroes

B 3 zeroes

C 4 zeroes

D 5 zeroes

✅ Correct Answer: 4 zeroes

The graph intersects the x-axis at four distinct points.

This indicates a polynomial of degree at least 4 with 4 real zeroes.

Each crossing point represents one real root of the polynomial equation p(x) = 0.

Exercise 2.2 - Find Zeroes & Verify Relationships

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

Q1(i). Find zeroes of x² - 2x - 8

Factorize by splitting the middle term: Find two numbers whose sum = -2 and product = -8

A Zeroes are 1 and -8

B Zeroes are -2 and 4

C Zeroes are 4 and -2

D Zeroes are -4 and -2

✅ Correct Answer: Zeroes are 4 and -2

Step 1: Split middle term: x² - 2x - 8 = x² - 4x + 2x - 8

Step 2: Factor by grouping: = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)

Step 3: Zeroes: x - 4 = 0 → x = 4 and x + 2 = 0 → x = -2

Step 4 - Verify: Sum = 4 + (-2) = 2 = -(-2)/1 = -b/a ✅

Step 5 - Verify: Product = 4 × (-2) = -8 = -8/1 = c/a ✅

Q1(ii). Find zeroes of 4s² - 4s + 1

This is a perfect square trinomial. Look for (a - b)² pattern.

A Zeroes are 1 and 1/4

B Zeroes are 1/2 and 1/2

C Zeroes are -1/2 and -1/2

D Zeroes are 2 and -2

✅ Correct Answer: Zeroes are 1/2 and 1/2 (Equal roots)

Step 1: Recognize perfect square: 4s² - 4s + 1 = (2s)² - 2(2s)(1) + 1²

Step 2: Apply (a - b)² = a² - 2ab + b²: = (2s - 1)²

Step 3: Zero: 2s - 1 = 0 → s = 1/2 (repeated root)

Step 4 - Verify: Sum = 1/2 + 1/2 = 1 = -(-4)/4 = -b/a ✅

Step 5 - Verify: Product = 1/2 × 1/2 = 1/4 = 1/4 = c/a ✅

Q1(iii). Find zeroes of 6x² - 3 - 7x (or 6x² - 7x - 3)

First write in standard form: 6x² - 7x - 3. Split -7x such that product = 6×(-3) = -18

A Zeroes are 1/2 and -3

B Zeroes are 3/2 and -1/3

C Zeroes are 2/3 and -3/2

D Zeroes are 1/3 and -3/2

✅ Correct Answer: Zeroes are 3/2 and -1/3

Step 1: Standard form: 6x² - 7x - 3

Step 2: Split -7x: Need numbers with sum = -7, product = -18. Numbers: -9 and +2

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

Step 4: Zeroes: 3x + 1 = 0 → x = -1/3 and 2x - 3 = 0 → x = 3/2

Step 5 - Verify: Sum = 3/2 + (-1/3) = 7/6 = -(-7)/6 = -b/a ✅

Step 6 - Verify: Product = (3/2)(-1/3) = -1/2 = -3/6 = c/a ✅

Q1(iv). Find zeroes of 4u² + 8u

Factor out the common term first!

A Zeroes are 0 and -2

B Zeroes are 0 and 2

C Zeroes are -2 and -2

D Zeroes are 2 and -4

✅ Correct Answer: Zeroes are 0 and -2

Step 1: Factor out 4u: 4u² + 8u = 4u(u + 2)

Step 2: Zeroes: 4u = 0 → u = 0 and u + 2 = 0 → u = -2

Step 3 - Verify: Sum = 0 + (-2) = -2 = -8/4 = -b/a ✅

Step 4 - Verify: Product = 0 × (-2) = 0 = 0/4 = c/a ✅

Note: Here c = 0, so one zero is always 0.

Q1(v). Find zeroes of t² - 15

This is a difference of squares: a² - b² = (a+b)(a-b)

A Zeroes are 5 and -3

B Zeroes are √15 and -√15

C Zeroes are 15 and -15

D Zeroes are 3 and -5

✅ Correct Answer: Zeroes are √15 and -√15

Step 1: Difference of squares: t² - 15 = t² - (√15)²

Step 2: = (t - √15)(t + √15)

Step 3: Zeroes: t = √15 and t = -√15

Step 4 - Verify: Sum = √15 + (-√15) = 0 = -0/1 = -b/a ✅

Step 5 - Verify: Product = (√15)(-√15) = -15 = -15/1 = c/a ✅

Q1(vi). Find zeroes of 3x² - x - 4

Split -x such that product = 3×(-4) = -12. Numbers with sum = -1 and product = -12 are -4 and +3.

A Zeroes are 1 and -4/3

B Zeroes are 4/3 and -1

C Zeroes are -4/3 and -1

D Zeroes are 1 and 4/3

✅ Correct Answer: Zeroes are 4/3 and -1

Step 1: Split -x using -4 and +3: 3x² - 4x + 3x - 4

Step 2: Group: x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4)

Step 3: Zeroes: x + 1 = 0 → x = -1 and 3x - 4 = 0 → x = 4/3

Step 4 - Verify: Sum = 4/3 + (-1) = 1/3 = -(-1)/3 = -b/a ✅

Step 5 - Verify: Product = (4/3)(-1) = -4/3 = -4/3 = c/a ✅

Exercise 2.2 - Q2: Form Quadratic Polynomials

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

Formula: If sum = S and product = P, then polynomial = x² - Sx + P

Q2(i). Sum = 1/4, Product = -1

A 4x² - x - 4

B 4x² - x - 4 (or x² - (1/4)x - 1)

C x² + (1/4)x - 1

D 4x² + x - 4

✅ Correct Answer: 4x² - x - 4 (or equivalent)

Step 1: Using formula: x² - (1/4)x + (-1) = x² - (1/4)x - 1

Step 2: To eliminate fractions, multiply by 4: 4x² - x - 4

Step 3 - Verify: Sum = -(-1)/4 = 1/4 ✅

Step 4 - Verify: Product = -4/4 = -1 ✅

Q2(ii). Sum = √2, Product = 1/3

A x² + √2x + 1/3

B 3x² - 3√2x + 1

C x² - √2x - 1/3

D 3x² + 3√2x + 1

✅ Correct Answer: 3x² - 3√2x + 1 (or equivalent)

Step 1: Using formula: x² - √2x + 1/3

Step 2: Multiply by 3 to clear fraction: 3x² - 3√2x + 1

Step 3 - Verify: Sum = -(-3√2)/3 = √2 ✅

Step 4 - Verify: Product = 1/3 ✅

Q2(iii). Sum = 0, Product = √5

A x² + √5 (or x² - 0x + √5)

B x² - √5

C x² + √5x

D x² - √5x

✅ Correct Answer: x² + √5

Step 1: Using formula: x² - (0)x + √5 = x² + √5

Step 2: Since sum = 0, there is no x term!

Step 3 - Verify: Sum = -(0)/1 = 0 ✅

Step 4 - Verify: Product = √5/1 = √5 ✅

This polynomial has no real zeroes (since x² = -√5 has no real solution).

Q2(iv). Sum = 1, Product = 1

A x² + x + 1

B x² - x + 1

C x² - x - 1

D x² + x - 1

✅ Correct Answer: x² - x + 1

Step 1: Using formula: x² - (1)x + (1) = x² - x + 1

Step 2 - Verify: Sum = -(-1)/1 = 1 ✅

Step 3 - Verify: Product = 1/1 = 1 ✅

Note: Discriminant = (-1)² - 4(1)(1) = 1 - 4 = -3 < 0, so no real zeroes.

Q2(v). Sum = -1/4, Product = 1/4

A 4x² - x + 1

B 4x² + x + 1 (or x² + (1/4)x + 1/4)

C 4x² - x - 1

D x² - (1/4)x + 1/4

✅ Correct Answer: 4x² + x + 1 (or equivalent)

Step 1: Using formula: x² - (-1/4)x + 1/4 = x² + (1/4)x + 1/4

Step 2: Multiply by 4: 4x² + x + 1

Step 3 - Verify: Sum = -(1)/4 = -1/4 ✅

Step 4 - Verify: Product = 1/4 ✅

Q2(vi). Sum = 4, Product = 1

A x² + 4x + 1

B x² - 4x + 1

C x² - 4x - 1

D x² + 4x - 1

✅ Correct Answer: x² - 4x + 1

Step 1: Using formula: x² - (4)x + (1) = x² - 4x + 1

Step 2 - Verify: Sum = -(-4)/1 = 4 ✅

Step 3 - Verify: Product = 1/1 = 1 ✅

Zeroes would be: x = [4 ± √(16-4)]/2 = [4 ± √12]/2 = 2 ± √3

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