NCERT Class 10 Maths Chapter 2 — Polynomials: Complete Notes, Solutions & Shortcut Tricks (Convex Classes, Jaipur)

Polynomials is the chapter where a lot of Class 10 students in Jaipur either build real confidence in Algebra — or start avoiding Maths altogether. The reason is simple: the formulas (sum of zeroes, product of zeroes, division algorithm) look easy to mug up, but the moment the question is twisted slightly — a graph-based question, a “find the other zeroes” question, or a division-algorithm proof — students freeze.

At Convex Classes, Jaipur, this is one of the first chapters we spend extra time on in our Class 10 Maths batches, because it decides how comfortable a student is with Algebra for the rest of the year (Chapter 4 Quadratic Equations directly builds on this). This guide is written the way we actually teach it — concept, geometry, shortcut, and then the complete NCERT solution — so you understand every step instead of memorising it.

Why Polynomials Deserves Extra Attention

• It carries 6–8 marks directly in most CBSE Class 10 board papers (1 MCQ/VSA + 1 short/long answer is common).
• The “relationship between zeroes and coefficients” concept comes back in Chapter 4 (Quadratic Equations) and again in Class 11 Complex Numbers & Quadratic Equations — so mastering it now saves you effort later.
• Graph-based zero-counting questions are pure logic, zero calculation — meaning they’re free marks if you understand the concept, but easy to lose if you don’t.

Concept Breakdown — Explained the Convex Classes Way

1. What Is a Polynomial, Really?

A polynomial is simply an algebraic expression where every power of the variable is a whole number (0, 1, 2, 3…) — never negative, never a fraction. So x² + 3x⁻¹ is not a polynomial (negative power), and √x + 5 is not a polynomial either (power ½).

Classification by degree — the table students always mix up:

Degree	Name	General Form	Example	Max Zeroes
0	Constant	k	7	0
1	Linear	ax + b	3x + 5	1
2	Quadratic	ax² + bx + c	x² – 4x + 3	2
3	Cubic	ax³ + bx² + cx + d	x³ – 2x² + 5	3
4	Biquadratic	ax⁴+bx³+cx²+dx+e	x⁴ – 6x³ – 26x² + 138x – 35	4

Rule to remember: A polynomial of degree n can have at most n zeroes — never more. This single rule solves almost every graph-based question in Exercise 2.1.

2. The Geometrical Meaning of Zeroes (Where Most Marks Are Lost)

A “zero” of a polynomial p(x) is the value of x where p(x) = 0. Graphically, this is exactly the point(s) where the curve touches or crosses the x-axis. That’s it — no calculation needed for graph questions.

Convex Classes’ 3-second trick for graph questions: Count the number of times the curve touches/cuts the x-axis. That count is the number of zeroes. A parabola that doesn’t touch the x-axis at all → 0 zeroes (this trips up most students, who assume every graph must have at least 1 zero).

3. Relationship Between Zeroes and Coefficients — Don’t Memorise, Understand the Pattern

For a quadratic polynomial ax² + bx + c with zeroes α and β:$$\alpha + \beta = \frac{-b}{a} \qquad \alpha \beta = \frac{c}{a}$$α+β=a−b​αβ=ac​

For a cubic polynomial ax³ + bx² + cx + d with zeroes α, β, γ:$$\alpha + \beta + \gamma = \frac{-b}{a} \qquad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \qquad \alpha\beta\gamma = \frac{-d}{a}$$α+β+γ=a−b​αβ+βγ+γα=ac​αβγ=a−d​

The pattern that makes this easy to remember: the sign alternates starting from negative, and the coefficient you use moves one step down every time (b → c → d), always divided by ‘a’. Once you see this pattern, you never need to “mug up” the formula again — you can rebuild it in 5 seconds during your exam.

4. Division Algorithm — The Concept Behind “Dividend = Divisor × Quotient + Remainder”

This is nothing but the polynomial version of what you already know from Class 5 division of numbers: 17 ÷ 5 gives quotient 3 and remainder 2, and 17 = 5×3 + 2. Polynomials follow exactly the same logic:$$p(x) = g(x) \times q(x) + r(x)$$p(x)=g(x)×q(x)+r(x)

where r(x) = 0, or degree of r(x) < degree of g(x). This single formula is used to (a) verify division, (b) find unknown polynomials, and (c) find remaining zeroes of higher-degree polynomials — Q3, Q4, and Q5 of Exercise 2.3 & 2.4 are all this one idea applied differently.

Fully Worked NCERT Solutions

Exercise 2.1 — Finding Zeroes from Graphs

Using the “count the touch/cut points” rule:

(i) 0 zeroes — the graph does not touch the x-axis at all (ii) 1 zero — touches the x-axis at exactly one point (iii) 3 zeroes — cuts the x-axis at three points (iv) 2 zeroes — cuts the x-axis at two points (v) 4 zeroes — cuts the x-axis at four points (vi) 3 zeroes — cuts the x-axis at three points

Exercise 2.2 — Zeroes of Quadratic Polynomials & Verification

Q1(i) x² – 2x – 8 Splitting the middle term:

x² – 4x + 2x – 8 = x(x–4) + 2(x–4) = (x–4)(x+2) Zeroes: 4, –2 Verify: Sum = 4 + (–2) = 2 = –(–2)/1 ✓ | Product = 4×(–2) = –8 = –8/1 ✓

Q1(ii) 4s² – 4s + 1 = 4s² – 2s – 2s + 1 = 2s(2s–1) – 1(2s–1) = (2s–1)(2s–1) Zeroes: 1/2, 1/2 Verify: Sum = 1 = –(–4)/4 ✓ | Product = 1/4 = 1/4 ✓

Q1(iii) 6x² – 3 – 7x → 6x² – 7x – 3 = 6x² – 9x + 2x – 3 = 3x(2x–3) + 1(2x–3) = (3x+1)(2x–3) Zeroes: –1/3, 3/2 Verify: Sum = 7/6 = –(–7)/6 ✓ | Product = –1/2 = –3/6 ✓

Q1(iv) 4u² + 8u = 4u(u+2) Zeroes: 0, –2 Verify: Sum = –2 = –8/4 ✓ | Product = 0 = 0/4 ✓

Q1(v) t² – 15 → t² = 15 → t = ±√15 Zeroes: √15, –√15 Verify: Sum = 0 = –0/1 ✓ | Product = –15 = –15/1 ✓

Q1(vi) 3x² – x – 4 = 3x² – 4x + 3x – 4 = x(3x–4) + 1(3x–4) = (3x–4)(x+1) Zeroes: 4/3, –1 Verify: Sum = 1/3 = –(–1)/3 ✓ | Product = –4/3 = –4/3 ✓

Q2. Forming a quadratic polynomial from sum and product of zeroes

Formula: x² – (sum)x + (product) = 0

(i) Sum = 1/4, Product = –1 → x² – (1/4)x – 1 = 0 → 4x² – x – 4 (ii) Sum = √2, Product = 1/3 → 3x² – 3√2x + 1 (iii) Sum = 0, Product = √5 → x² + √5 (iv) Sum = 1, Product = 1 → x² – x + 1 (v) Sum = –1/4, Product = 1/4 → 4x² + x + 1 (vi) Sum = 4, Product = 1 → x² – 4x + 1

Exercise 2.3 — Division Algorithm

Q1(i) p(x) = x³–3x²+5x–3 ÷ g(x) = x²–2 Quotient = x – 3, Remainder = 7x – 9

Q1(ii) p(x) = x⁴–3x²+4x+5 ÷ g(x) = x²+1–x Quotient = x² + x – 3, Remainder = 8

Q1(iii) p(x) = x⁴–5x+6 ÷ g(x) = 2–x² Quotient = –x² – 2, Remainder = –5x + 10

Q2. Checking if the first polynomial is a factor of the second (by dividing)

(i) t²–3 divides 2t⁴+3t³–2t²–9t–12 exactly → remainder = 0, so t²–3 IS a factor (ii) x²+3x+1 divides 3x⁴+5x³–7x²+2x+2 exactly → remainder = 0, so it IS a factor (iii) x³–3x+1 divides x⁵–4x³+x²+3x+1 → remainder ≠ 0, so it is NOT a factor

Q3. Finding all zeroes of 3x⁴+6x³–2x²–10x–5, given two zeroes are √(5/3) and –√(5/3)

Since these are zeroes: (x–√(5/3))(x+√(5/3)) = x² – 5/3, i.e., (3x²–5) is a factor.

Dividing 3x⁴+6x³–2x²–10x–5 by (3x²–5): 3x⁴+6x³–2x²–10x–5 = (3x²–5)(x²+2x+1)

Factorising x²+2x+1 = (x+1)(x+1) → zeroes –1, –1

All four zeroes: √(5/3), –√(5/3), –1, –1

Q4. Finding g(x) when dividend, quotient and remainder are known

Dividend = x³–3x²+x+2, Quotient = x–2, Remainder = –2x+4

Using Dividend = Divisor × Quotient + Remainder: g(x) × (x–2) = x³–3x²+x+2 – (–2x+4) = x³–3x²+3x–2

Dividing this by (x–2): g(x) = x² – x + 1

Q5. Examples satisfying the division algorithm for three degree cases

(i) deg p(x) = deg q(x) happens only when the divisor is a constant. Example: p(x) = 3x²+3x+3, g(x) = 3 → q(x) = x²+x+1 (degree 2 = degree 2)

(ii) deg q(x) = deg r(x) Example: p(x) = x²+3, g(x) = x–1 → x²+3 = (x–1)(x) + (x+3), so q(x)=x, r(x)=x+3 (both degree 1)

(iii) deg r(x) = 0 (remainder is a constant) Example: p(x) = x²+1, g(x) = x → x²+1 = (x)(x) + 1, so q(x) = x, r(x) = 1

Exercise 2.4 — Cubic Polynomials (Higher Order Thinking)

Q1(i). Verify zeroes –1/2, 1, –2 for 2x³+x²–5x+2

Substituting each value into p(x) gives 0 for all three — confirming they’re zeroes. Comparing with ax³+bx²+cx+d: a=2, b=1, c=–5, d=2

Sum = 1/2+1–2 = –1/2 = –b/a ✓ Sum of products two at a time = –5/2 = c/a ✓ Product = –1 = –d/a ✓

Q1(ii). Verify zeroes 2, 1, 1 for x³–4x²+5x–2

Substitution confirms all are zeroes. Here a=1, b=–4, c=5, d=–2

Sum = 4 = –b/a ✓ | Sum of products = 5 = c/a ✓ | Product = 2 = –d/a ✓

Q2. Find a cubic polynomial given sum=2, sum of products two at a time=–7, product=–14

Using ax³+bx²+cx+d with these three relations directly gives: x³ – 2x² – 7x + 14

Q3. If zeroes of x³–3x²+x+1 are a–b, a, a+b, find a and b

Sum of zeroes = 3a = –(–3)/1 = 3 → a = 1

Product of zeroes = a(a–b)(a+b) = a(a²–b²) = –(1)/1 1(1–b²) = –1 → b² = 2 → b = ±√2

Zeroes are 1–√2, 1, 1+√2

Q4. If two zeroes of x⁴–6x³–26x²+138x–35 are 2±√3, find the other zeroes

(x–(2+√3))(x–(2–√3)) = x²–4x+1 → this is a factor.

Dividing the original polynomial by (x²–4x+1): x⁴–6x³–26x²+138x–35 = (x²–4x+1)(x²–2x–35)

Factorising x²–2x–35 = (x–7)(x+5) → zeroes 7, –5

All four zeroes: 2+√3, 2–√3, 7, –5

Q5. Find k and a if dividing x⁴–6x³+16x²–25x+10 by x²–2x+k leaves remainder x+a

After carrying out the division and comparing the remainder terms: 2k – 9 = 1 → k = 5 10 – 8k + k² = a → 10 – 40 + 25 = a = –5

k = 5, a = –5

Common Mistakes Convex Classes Students Make (and Fixes)

1. Assuming every graph must have at least one zero — a curve that doesn’t touch the x-axis has zero zeroes; don’t force an answer.
2. Sign errors in sum/product formulas — always write out the pattern (alternating signs, coefficient shifting down) before substituting numbers, don’t rely on pure memory under exam pressure.
3. Forgetting the division algorithm condition — remainder’s degree must always be less than the divisor’s degree; if it isn’t, your division is incomplete.
4. Not verifying factorisation — after splitting the middle term, always multiply back out mentally to check you haven’t made a sign error; this catches almost all silly mistakes.
5. Confusing “zero of a polynomial” with “root of an equation” — they’re the same value, just different vocabulary; don’t get confused when a question switches terms.

Extra Practice Questions (Beyond NCERT) — For Board-Level Practice

1. If α and β are the zeroes of x² – 5x + 6, find the value of α² + β² without finding α and β individually.
2. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of 2x² – 3x + 1.
3. If one zero of the polynomial 3x² – 8x + 2k+1 is seven times the other, find the value of k.
4. Obtain all zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if two of its zeroes are √2 and –√2.
5. If the zeroes of the polynomial x³ – 3x² – 9x – 5 are of the form a, a+b, a+2b (i.e., in arithmetic progression), find all three zeroes.

(Convex Classes students: these HOTS-level questions are solved on the board with full working in our regular batches — ask your teacher or WhatsApp us for the complete solved set.)

Convex Classes’ Quick Revision Checklist Before Your Test

• Can you state the maximum number of zeroes for any degree without hesitation?
• Can you rebuild the sum/product-of-zeroes formulas from the alternating-sign pattern, not just recall them?
• Can you apply the division algorithm to find an unknown polynomial (not just verify a division)?
• Can you solve a “find remaining zeroes” cubic/biquadratic question using the factor-and-divide method?
• Have you practiced at least one graph-based question from every pattern (0, 1, 2, 3, 4 zeroes)?

Need Personal Guidance on This Chapter?

Formulas are easy to read but hard to apply under exam pressure without practice. At Convex Classes, Jaipur, our Class 10 Maths batches focus on exactly this — building the reasoning behind each formula so you can solve any variation the board throws at you, not just the ones you’ve memorised.

For a free demo class, chapter-wise test series, or personal doubt support in Class 10 Maths, message us directly on WhatsApp: 8290601516.

FAQs

Q1. What is the maximum number of zeroes a polynomial can have?

A polynomial of degree n can have at most n zeroes — never more, though it can have fewer (including zero).

Q2. What’s the easiest way to remember the sum and product of zeroes formulas?

Notice the pattern: signs alternate starting negative, and the coefficient used shifts one step down each time (b, then c, then d), always divided by ‘a’. Once you see this, you don’t need to memorise separate formulas for quadratic and cubic polynomials.

Q3. How do I find the remaining zeroes of a cubic or biquadratic polynomial if some zeroes are already given?

Form a factor from the given zeroes, divide the original polynomial by that factor using the division algorithm, and factorise the resulting quotient to get the remaining zeroes.

Q4. Where can Jaipur students get in-person doubt support for Polynomials?

Convex Classes, Jaipur runs dedicated Class 10 Maths batches with chapter tests and doubt-clearing sessions — WhatsApp 8290601516 for details.