CBSE • Class 9 • Mathematics • Chapter 2 (Polynomials) • Exercise 2.2

Exercise 2.2: Polynomials — NCERT Solutions

Zeros of linear and quadratic polynomials — finding roots algebraically.

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What this exercise covers

Finding zerosLinear polynomialsQuadratic polynomials

Step-by-step solutions — Exercise 2.2

3 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 2.2 Q1 • 2 marks

Find the zero of the linear polynomial p(x)=3x+12p(x) = 3x + 12, and verify your answer.
Hint (Socratic — try this first)▾
For what value of xx does the polynomial become 00?
Step-by-step solution▾

A zero of p(x)p(x) is the value of xx for which p(x)=0p(x) = 0.

Set 3x+12=03x + 12 = 0: 3x=−12  ⟹  x=−4.3x = -12 \implies x = -4.

Verification: p(−4)=3(−4)+12=−12+12=0.p(-4) = 3(-4) + 12 = -12 + 12 = 0. ✓

The zero of p(x)p(x) is x=−4x = -4.

Common mistake:
Sign errors while transposing — writing x=4x = 4 instead of x=−4x = -4.
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Exercise 2.2 Q2 • 3 marks

Find the value of the polynomial p(x)=x2−5x+6p(x) = x^2 - 5x + 6 at x=0x = 0, x=2x = 2 and x=3x = 3. Hence state which of these are zeros of p(x)p(x).
Hint (Socratic — try this first)▾
Substitute each value and check whether the result is zero.
Step-by-step solution▾

Substitute each value into p(x)=x2−5x+6p(x) = x^2 - 5x + 6.

p(0)=0−0+6=6p(0) = 0 - 0 + 6 = 6.

p(2)=4−10+6=0p(2) = 4 - 10 + 6 = 0.

p(3)=9−15+6=0p(3) = 9 - 15 + 6 = 0.

Since p(2)=0p(2) = 0 and p(3)=0p(3) = 0, the numbers 22 and 33 are zeros of p(x)p(x), while 00 is not.

Common mistake:
Arithmetic slips in x2x^2 (e.g. computing (3)2(3)^2 as 66), leading to wrong conclusions about the zeros.
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Exercise 2.2 Q3 • 3 marks

Factorise the quadratic polynomial x2−7x+10x^2 - 7x + 10 by splitting the middle term, and find its zeros.
Hint (Socratic — try this first)▾
Which two numbers multiply to 1010 and add up to −7-7?
Step-by-step solution▾

We split the middle term −7x-7x into two terms whose product of coefficients equals 1×10=101 \times 10 = 10 and whose sum is −7-7.

The numbers are −5-5 and −2-2 (since (−5)(−2)=10(-5)(-2)=10, −5+(−2)=−7-5+(-2)=-7).

x2−7x+10=x2−5x−2x+10=x(x−5)−2(x−5)=(x−5)(x−2).x^2 - 7x + 10 = x^2 - 5x - 2x + 10 = x(x - 5) - 2(x - 5) = (x - 5)(x - 2).

Setting each factor to zero: x−5=0⇒x=5x - 5 = 0 \Rightarrow x = 5 and x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2.

The zeros are x=5x = 5 and x=2x = 2.

Common mistake:
Choosing numbers with the correct product but wrong signs (e.g. +5+5 and +2+2), so the middle term comes out as +7x+7x instead of −7x-7x.
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How to approach Exercise 2.2

  1. Re-read the chapter summary first. Open Polynomials and refresh the key concepts: Polynomial, Degree, Remainder theorem, Factor theorem.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Polynomials

  1. Exercise 2.1 — Polynomials in one variable — degree, coefficients and zeros.
  2. Exercise 2.2 — Zeros of linear and quadratic polynomials — finding roots algebraically.
  3. Exercise 2.3 — The Remainder Theorem — evaluating remainders without long division.
  4. Exercise 2.4 — The Factor Theorem — detecting factors (x − a) from functional values.
  5. Exercise 2.5 — Algebraic identities — expanding and factorising cubic-type expressions.

← Exercise 2.1

Polynomials in one variable — degree, coefficients and zeros.

Exercise 2.3 →

The Remainder Theorem — evaluating remainders without long division.

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