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📐 CBSE Class 10 Maths · Chapter 2

Polynomials — Complete NCERT Solutions

Every Example + Every Exercise Question — Solved Step-by-Step. Free Formula Sheet, Exam Tips & FAQs.

All NCERT Examples Solved Exercise 2.1 & 2.2 Solved Board 2026–27 Manish Sir Explained
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👋 Introduction

What is this chapter about? Class 10 Chapter 2 — Polynomials — builds on what you studied in Class IX and takes it deeper. You will explore the geometrical meaning of zeroes of a polynomial (what they look like on a graph), the powerful relationship between zeroes and coefficients of quadratic and cubic polynomials, and how to find and verify zeroes step-by-step.

The chapter covers three key ideas: understanding zeroes from graphs (parabolas and curves), the sum/product formulas for quadratic polynomials (α + β and αβ), and the three relations for cubic polynomials (α + β + γ, αβ + βγ + γα, and αβγ).

This chapter carries 6–8 marks in the board exam. Questions on finding zeroes and verifying relationships are asked every single year. Master these solutions and you are guaranteed full marks on Chapter 2.

Zeroes of Polynomials Geometrical Meaning Quadratic Polynomials Sum & Product of Zeroes Cubic Polynomials Splitting the Middle Term
📐 Formula Sheet

🔢 1. Types of Polynomials

Linear: ax + b (degree 1)
Quadratic: ax² + bx + c (degree 2)
Cubic: ax³ + bx² + cx + d (degree 3)

Zeroes of a polynomial p(x): A real number k is a zero if p(k) = 0. Geometrically, zeroes are the x-coordinates where the graph of y = p(x) crosses the x-axis.

e.g., p(x) = x² − 3x − 4 → zeroes are −1 and 4 since p(−1) = 0, p(4) = 0

🔑 2. Quadratic — Sum & Product of Zeroes

For ax² + bx + c, zeroes α and β:

α + β = −b/a (Sum of zeroes)
α × β = c/a (Product of zeroes)

How to remember: Sum = −(coeff of x) ÷ (coeff of x²). Product = constant term ÷ (coeff of x²).

e.g., x² + 7x + 10: α+β = −7/1 = −7, αβ = 10/1 = 10

🔑 3. Forming a Quadratic from Zeroes

p(x) = x² − (α + β)x + αβ

or k[x² − (Sum)x + (Product)]
where k is any non-zero real constant

Use this when: Sum and product of zeroes are given and you need to find the polynomial. Substitute values directly.

e.g., Sum = −3, Product = 2 → p(x) = x² + 3x + 2

🔑 4. Cubic — Three Zeroes Relations

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

α + β + γ = −b/a
αβ + βγ + γα = c/a
α × β × γ = −d/a

Note the signs: Sum = −b/a, Sum of products two at a time = c/a, Product of all three = −d/a. The sign on the product flips!

e.g., 2x³ − 5x² − 14x + 8: α+β+γ = 5/2, αβγ = −8/2 = −4

📊 5. Geometrical Meaning of Zeroes

Linear: graph is a straight line → 1 zero
Quadratic: parabola → 0, 1, or 2 zeroes
Cubic: S-curve → 1, 2, or 3 zeroes

Number of zeroes = number of times graph cuts x-axis. A polynomial of degree n has at most n zeroes.

Parabola opens up if a > 0 (∪ shape), down if a < 0 (∩ shape)

📐 6. Splitting the Middle Term

For ax² + bx + c:
Find p, q such that p + q = b
and p × q = ac
Then split bx = px + qx and factorise

This is the standard method to factorise quadratics and find their zeroes. Always verify zeroes by substituting back.

e.g., 2x² − 8x + 6: p+q = −8, pq = 12 → p = −6, q = −2
📊 Visual Diagrams — Geometrical Meaning of Zeroes
Why diagrams matter: The geometrical meaning of zeroes is one of the most visually tested concepts in Class 10 boards. You must be able to look at any graph of y = p(x) and immediately tell the number of zeroes. The diagrams below cover every possible case for linear, quadratic, and cubic polynomials — exactly as in NCERT.
📈 Linear Polynomial — ax + b (Degree 1)
X Y O −b/a y = ax + b
1 Zero (Always)
Straight line cuts x-axis at exactly one point. Zero = −b/a
X Y O y = c (constant) parallel to x-axis → 0 zeroes Not a polynomial of degree 1
Key Rule: Degree n → At Most n Zeroes
A non-zero constant has no zero. A linear polynomial always has exactly 1 zero.
📉 Quadratic Polynomial — ax² + bx + c (Degree 2) — 3 Cases
X Y O α β a > 0
Case 1: 2 Distinct Zeroes
Parabola (opens up, a > 0) cuts x-axis at two distinct points α and β
X Y O α β a < 0
Case 1: 2 Zeroes (a < 0)
Parabola opens downward (a < 0), still cuts x-axis at two points α and β
X Y O α = β a > 0
Case 2: 1 Zero (Repeated)
Parabola touches x-axis at exactly one point. Two equal zeroes: α = β
X Y O No intersection → 0 zeroes
Case 3: No Zeroes
Parabola is completely above x-axis (a > 0) — does not cut it. 0 real zeroes.
X Y O No intersection a < 0 → 0 zeroes
Case 3: No Zeroes (a < 0)
Parabola opens downward, entirely below x-axis — 0 real zeroes.
📐 Cubic Polynomial — ax³ + bx² + cx + d (Degree 3)
X Y α β γ
3 Distinct Zeroes
Cubic curve cuts x-axis at 3 different points α, β, γ (maximum possible)
X Y α=β γ
2 Zeroes (1 Repeated)
Curve touches x-axis at α (repeated zero) and crosses at γ — total 2 distinct values
X Y α only y = x³
1 Zero Only
Cubic cuts x-axis at only one point (e.g., y = x³ has zero only at x = 0)
📋 Summary: Number of Zeroes by Polynomial Type
Polynomial Type Degree Graph Shape Possible Zeroes
Linear (ax + b) 1 Straight line Exactly 1
Quadratic (ax² + bx + c) 2 Parabola (∪ or ∩) 0, 1, or 2
Cubic (ax³ + bx² + cx + d) 3 S-shaped curve 1, 2, or 3
General (degree n) n Any curve At most n
📖 NCERT Solutions — All Examples
📘 Section 2.2 — Geometrical Meaning Examples

The number of zeroes of p(x) equals the number of times the graph of y = p(x) intersects (cuts) the x-axis.

(i) The graph intersects the x-axis at one point only.
∴ Number of zeroes = 1
(ii) The graph intersects the x-axis at two points.
∴ Number of zeroes = 2
(iii) The graph intersects the x-axis at three points.
∴ Number of zeroes = 3
(iv) The graph intersects the x-axis at one point only (touches but doesn't cross).
∴ Number of zeroes = 1
(v) The graph intersects the x-axis at one point.
∴ Number of zeroes = 1
(vi) The graph intersects the x-axis at four points.
∴ Number of zeroes = 4
💡 Key rule: A polynomial of degree n has at most n zeroes. The number of zeroes ≤ degree of polynomial.
📘 Section 2.3 — Zeroes & Coefficients Examples

Step 1 — Factorise by splitting the middle term:

x² + 7x + 10
We need p + q = 7 and p × q = 10
→ p = 5, q = 2 works: 5 + 2 = 7, 5 × 2 = 10

= x² + 5x + 2x + 10
= x(x + 5) + 2(x + 5)
= (x + 2)(x + 5)
Step 2 — Find zeroes:
p(x) = 0 when x + 2 = 0 or x + 5 = 0
→ x = −2 or x = −5
∴ Zeroes are α = −2 and β = −5
Step 3 — Verify (a = 1, b = 7, c = 10):

Sum of zeroes: α + β = −2 + (−5) = −7 = −(7)/1 = −b/a ✓

Product of zeroes: α × β = (−2)(−5) = 10 = 10/1 = c/a ✓
Zeroes: −2 and −5. Both relationships verified.

Using the identity a² − b² = (a − b)(a + b):

x² − 3 = x² − (√3)²
= (x − √3)(x + √3)
Zeroes: p(x) = 0 when x − √3 = 0 or x + √3 = 0
→ α = √3 and β = −√3
Verify (a = 1, b = 0, c = −3):

Sum of zeroes: α + β = √3 + (−√3) = 0 = −(0)/1 = −b/a ✓

Product of zeroes: α × β = (√3)(−√3) = −3 = (−3)/1 = c/a ✓
Zeroes: √3 and −√3. Both relationships verified.

Let the quadratic polynomial be ax² + bx + c with zeroes α and β.

Given:
α + β = −3 = −b/a → with a = 1, b = 3
α × β = 2 = c/a → with a = 1, c = 2
Using formula: p(x) = x² − (Sum)x + (Product)
= x² − (−3)x + 2
= x² + 3x + 2
Required quadratic polynomial: x² + 3x + 2

Note: Any polynomial of the form k(x² + 3x + 2), where k is a non-zero real number, is also a valid answer.

Here a = 3, b = −5, c = −11, d = −3. We verify each value by substituting into p(x).

Verify x = 3:
p(3) = 3(3)³ − 5(3)² − 11(3) − 3
= 3(27) − 5(9) − 33 − 3
= 81 − 45 − 33 − 3 = 0 ✓
Verify x = −1:
p(−1) = 3(−1)³ − 5(−1)² − 11(−1) − 3
= −3 − 5 + 11 − 3 = 0 ✓
Verify x = −1/3:
p(−1/3) = 3(−1/3)³ − 5(−1/3)² − 11(−1/3) − 3
= 3(−1/27) − 5(1/9) + 11/3 − 3
= −1/9 − 5/9 + 11/3 − 3
= −6/9 + 11/3 − 3
= −2/3 + 11/3 − 9/3 = 0/3 = 0 ✓

∴ 3, −1 and −1/3 are zeroes. Let α = 3, β = −1, γ = −1/3.

Verify Sum (α + β + γ = −b/a):
3 + (−1) + (−1/3) = 2 − 1/3 = 5/3
−b/a = −(−5)/3 = 5/3 ✓
Verify Sum of products two at a time (αβ + βγ + γα = c/a):
αβ = 3 × (−1) = −3
βγ = (−1) × (−1/3) = 1/3
γα = (−1/3) × 3 = −1
αβ + βγ + γα = −3 + 1/3 − 1 = −4 + 1/3 = −11/3
c/a = −11/3 ✓
Verify Product (αβγ = −d/a):
3 × (−1) × (−1/3) = 3 × 1/3 = 1
−d/a = −(−3)/3 = 1 ✓
All three zeroes verified. All three coefficient relationships verified.
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📗 Exercise 2.1 — Zeroes from Graphs

The number of zeroes = number of times the graph cuts the x-axis.

(i) The graph does not intersect the x-axis at all.
∴ Number of zeroes = 0
(ii) The graph intersects the x-axis at one point.
∴ Number of zeroes = 1
(iii) The graph intersects the x-axis at three points.
∴ Number of zeroes = 3
(iv) The graph intersects the x-axis at two points.
∴ Number of zeroes = 2
(v) The graph intersects the x-axis at four points.
∴ Number of zeroes = 4
(vi) The graph intersects the x-axis at three points.
∴ Number of zeroes = 3
📗 Exercise 2.2 — Zeroes & Coefficients
x² − 2x − 8
Split: p + q = −2, p × q = −8 → p = −4, q = 2
= x² − 4x + 2x − 8
= x(x − 4) + 2(x − 4)
= (x + 2)(x − 4)
Zeroes: x + 2 = 0 → x = −2, and x − 4 = 0 → x = 4
∴ α = −2, β = 4
Verify (a = 1, b = −2, c = −8):

Sum: α + β = −2 + 4 = 2 = −(−2)/1 = −b/a ✓
Product: αβ = (−2)(4) = −8 = −8/1 = c/a ✓
Zeroes: −2 and 4. Relationships verified.
4s² − 4s + 1
Split: p + q = −4, p × q = 4 × 1 = 4 → p = −2, q = −2
= 4s² − 2s − 2s + 1
= 2s(2s − 1) − 1(2s − 1)
= (2s − 1)(2s − 1) = (2s − 1)²
Zeroes: 2s − 1 = 0 → s = 1/2 (repeated zero)
∴ α = 1/2, β = 1/2
Verify (a = 4, b = −4, c = 1):

Sum: α + β = 1/2 + 1/2 = 1 = −(−4)/4 = 4/4 = 1 = −b/a ✓
Product: αβ = (1/2)(1/2) = 1/4 = 1/4 = c/a ✓
Zeroes: 1/2 and 1/2 (equal/repeated zeroes). Relationships verified.

First, write in standard form: 6x² − 7x − 3

6x² − 7x − 3
Split: p + q = −7, p × q = 6 × (−3) = −18
→ p = −9, q = 2 (since −9 + 2 = −7, −9 × 2 = −18)
= 6x² − 9x + 2x − 3
= 3x(2x − 3) + 1(2x − 3)
= (3x + 1)(2x − 3)
Zeroes: 3x + 1 = 0 → x = −1/3, and 2x − 3 = 0 → x = 3/2
∴ α = −1/3, β = 3/2
Verify (a = 6, b = −7, c = −3):

Sum: α + β = −1/3 + 3/2 = −2/6 + 9/6 = 7/6 = −(−7)/6 = −b/a ✓
Product: αβ = (−1/3)(3/2) = −3/6 = −1/2 = (−3)/6 = c/a ✓
Zeroes: −1/3 and 3/2. Relationships verified.
4u² + 8u
= 4u(u + 2)
[Take 4u common — no splitting needed here!]
Zeroes: 4u = 0 → u = 0, and u + 2 = 0 → u = −2
∴ α = 0, β = −2
Verify (a = 4, b = 8, c = 0):

Sum: α + β = 0 + (−2) = −2 = −8/4 = −b/a ✓
Product: αβ = (0)(−2) = 0 = 0/4 = c/a ✓
Zeroes: 0 and −2. Relationships verified.
t² − 15 = t² − (√15)²
= (t − √15)(t + √15)
Zeroes: t = √15 and t = −√15
∴ α = √15, β = −√15
Verify (a = 1, b = 0, c = −15):

Sum: α + β = √15 + (−√15) = 0 = −0/1 = −b/a ✓
Product: αβ = (√15)(−√15) = −15 = −15/1 = c/a ✓
Zeroes: √15 and −√15. Relationships verified.
3x² − x − 4
Split: p + q = −1, p × q = 3 × (−4) = −12
→ p = −4, q = 3 (since −4 + 3 = −1, −4 × 3 = −12)
= 3x² − 4x + 3x − 4
= x(3x − 4) + 1(3x − 4)
= (x + 1)(3x − 4)
Zeroes: x + 1 = 0 → x = −1, and 3x − 4 = 0 → x = 4/3
∴ α = −1, β = 4/3
Verify (a = 3, b = −1, c = −4):

Sum: α + β = −1 + 4/3 = −3/3 + 4/3 = 1/3 = −(−1)/3 = −b/a ✓
Product: αβ = (−1)(4/3) = −4/3 = (−4)/3 = c/a ✓
Zeroes: −1 and 4/3. Relationships verified.
Sum of zeroes (α + β) = 1/4
Product of zeroes (αβ) = −1
p(x) = k[x² − (Sum)x + Product]
= k[x² − (1/4)x + (−1)]
= k[x² − x/4 − 1]

Multiplying by 4 (taking k = 4):
= 4x² − x − 4
Required polynomial: 4x² − x − 4
p(x) = k[x² − (√2)x + 1/3]

Multiplying by 3 (taking k = 3):
= 3x² − 3√2 x + 1
Required polynomial: 3x² − 3√2 x + 1
p(x) = k[x² − (0)x + √5]
= k[x² + √5]

Taking k = 1:
= x² + √5
Required polynomial: x² + √5
p(x) = k[x² − (1)x + 1]
= k[x² − x + 1]

Taking k = 1:
= x² − x + 1
Required polynomial: x² − x + 1
p(x) = k[x² − (−1/4)x + 1/4]
= k[x² + x/4 + 1/4]

Multiplying by 4 (taking k = 4):
= 4x² + x + 1
Required polynomial: 4x² + x + 1
p(x) = k[x² − (4)x + 1]
= k[x² − 4x + 1]

Taking k = 1:
= x² − 4x + 1
Required polynomial: x² − 4x + 1
📊 Quick Reference Table
Concept Formula / Rule Example
Linear polynomial ax + b, degree 1, exactly 1 zero 2x + 3 → zero = −3/2
Quadratic polynomial ax² + bx + c, degree 2, at most 2 zeroes x²−3x−4 → zeroes: −1, 4
Cubic polynomial ax³+bx²+cx+d, degree 3, at most 3 zeroes x³−4x → zeroes: −2, 0, 2
Sum of zeroes (quadratic) α + β = −b/a x²+7x+10: α+β = −7
Product of zeroes (quadratic) αβ = c/a x²+7x+10: αβ = 10
Form polynomial from zeroes x² − (Sum)x + Product Sum=−3, Prod=2 → x²+3x+2
Sum of zeroes (cubic) α + β + γ = −b/a 2x³−5x²−14x+8: sum = 5/2
Sum of products two at a time αβ + βγ + γα = c/a 2x³−5x²−14x+8: = −7
Product of zeroes (cubic) αβγ = −d/a 2x³−5x²−14x+8: = −4
Parabola opens up a > 0 (∪ shape) x²−3x−4 → opens upward
Parabola opens down a < 0 (∩ shape) −x²+2x+3 → opens downward
Number of zeroes rule Degree n → at most n zeroes degree 2 → max 2 zeroes
⚠️ Common Mistakes
❌ Wrong
Forgetting the negative sign: writing Sum = b/a instead of −b/a
✅ Correct
Sum of zeroes = −b/a (negative sign is essential; product = c/a has no negative)
❌ Wrong
Writing the polynomial with wrong signs: p(x) = x² + (Sum)x + Product
✅ Correct
Formula is p(x) = x² − (Sum)x + Product. The minus sign before Sum is critical.
❌ Wrong
Not writing 6x²−3−7x in standard form before factorising (missing step)
✅ Correct
Always rewrite in standard form ax²+bx+c first → 6x²−7x−3, then split.
❌ Wrong
Product of zeroes for cubic: writing αβγ = d/a (missing negative sign)
✅ Correct
Product of zeroes for cubic = −d/a. Only the product of all three has a negative sign.
❌ Wrong
Thinking 4s²−4s+1 has two different zeroes because it's a quadratic
✅ Correct
4s²−4s+1 = (2s−1)² has a repeated zero s = 1/2 twice. This counts as one unique zero.
❌ Wrong
Assuming a quadratic always has 2 distinct zeroes from a graph
✅ Correct
A quadratic can have 0, 1, or 2 zeroes. If parabola doesn't cut x-axis → 0 zeroes.
🎯 Exam Tips from Manish Sir
⚡
Zeroes from graphs = guaranteed 1 mark
CBSE asks graph-based questions every year in MCQ or 1-mark section. Remember: number of zeroes = number of times graph cuts x-axis. A parabola touching (not crossing) the x-axis = 1 zero.
📝
Always verify at the end
After finding zeroes, always verify the sum and product relationships. It takes 2 lines and can save you 1 mark. Examiners check for this verification step explicitly.
🔢
Remember the sign rule
For quadratic: Sum = −b/a (negative), Product = c/a (positive). For cubic: Sum = −b/a, Sum of pairs = c/a, Product = −d/a. The signs on Sum and cubic Product are negative — very commonly forgotten in exams.
🏆
Chapter 2 = easy 6–8 marks
Polynomials questions are highly predictable. Practice all 6 parts of Exercise 2.2 Q1 and all 6 parts of Q2. These are repeated every year with minor variations. Master these = guaranteed full marks.
⚠️
Standard form first, always
Before splitting the middle term, rewrite the polynomial in standard form ax²+bx+c. Questions like 6x²−3−7x are given out of order deliberately. Rearranging first prevents silly errors.
💡
Forming a polynomial is the easy 2-marker
Questions like "find quadratic polynomial with sum 4 and product 1" are pure formula application. Just plug into x² − (sum)x + product. Never miss these 2 marks.
❓ Frequently Asked Questions
The zeroes of a polynomial p(x) are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. For a linear polynomial, the graph (straight line) cuts the x-axis at exactly one point. For a quadratic polynomial, the parabola can cut the x-axis at 0, 1, or 2 points — giving 0, 1, or 2 zeroes respectively.
A quadratic polynomial (degree 2) can have at most 2 zeroes — it can have 0, 1, or 2 zeroes. A cubic polynomial (degree 3) can have at most 3 zeroes. In general, a polynomial of degree n has at most n zeroes.
For a quadratic polynomial ax² + bx + c with zeroes α and β:

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

These are the most important formulas in this chapter.
Use the formula: p(x) = k[x² − (Sum of zeroes)x + (Product of zeroes)], where k is any non-zero real number (usually take k = 1). For example, if sum = 4 and product = 3, then p(x) = x² − 4x + 3. Any scalar multiple k(x² − 4x + 3) is also valid.
For ax³ + bx² + cx + d with zeroes α, β, γ:

1. α + β + γ = −b/a (sum of all three zeroes)
2. αβ + βγ + γα = c/a (sum of products of zeroes taken two at a time)
3. αβγ = −d/a (product of all three zeroes)

Note: Relationships 1 and 3 have a negative sign; relationship 2 does not.
Chapter 2 typically carries 6 to 8 marks in the CBSE Class 10 board exam — usually 1 MCQ (1 mark) + 1 short answer (2 marks) + 1 long answer (3 marks). The most frequently asked questions are: finding zeroes and verifying, and forming a quadratic polynomial from given sum and product.
In the context of polynomials, zero and root mean the same thing — a value of x for which p(x) = 0. The term "zero of a polynomial" is preferred in NCERT. The term "root" is more commonly used when referring to equations (like "roots of the equation x²−5x+6 = 0"). Both are acceptable in board exams.
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