Examora | MCQ & Notes
Back to Home

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials covers polynomials in one variable, zeroes of a polynomial, the Remainder Theorem, the Factor Theorem, and algebraic identities used for factorisation.

Overview of Topics & Subtopics (Click to Jump)

1. Polynomials in One Variable

Short Note:

• A polynomial in one variable x is an expression of the form anxn + ... + a1x + a0, where a0, a1, ... are real numbers and the exponents are non-negative integers.

• Degree: the highest power of the variable in a polynomial.

• Based on degree: Linear (degree 1), Quadratic (degree 2), Cubic (degree 3) polynomials.

• Based on number of terms: Monomial (1 term), Binomial (2 terms), Trinomial (3 terms).

2. Zeroes of a Polynomial

Short Note:

• A real number k is a zero of polynomial p(x) if p(k) = 0.

• A linear polynomial has exactly one zero, a quadratic polynomial has at most two zeroes, and in general a polynomial of degree n has at most n zeroes.

3. Remainder Theorem & Factor Theorem

Short Note:

• Remainder Theorem: If p(x) is divided by (x − a), the remainder is p(a).

• Factor Theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.

4. Algebraic Identities

Short Note:

• (x + y)² = x² + 2xy + y²

• (x − y)² = x² − 2xy + y²

• x² − y² = (x + y)(x − y)

• (x + y)³ = x³ + y³ + 3xy(x + y)

• (x − y)³ = x³ − y³ − 3xy(x − y)

• x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

NCERT Textbook Exercise Questions

Which of the following expressions are polynomials in one variable: (i) 3x² − 4x + 15 (ii) y + 2/y?
Answer: (i) 3x² − 4x + 15 is a polynomial in one variable x, since all exponents of x are whole numbers.
(ii) y + 2/y is not a polynomial because it contains y−1, a negative exponent.
Find the zero of the polynomial p(x) = 2x + 5.
Answer: Put p(x) = 0: 2x + 5 = 0 ⇒ x = −5/2. So the zero of the polynomial is −5/2.
Find the remainder when x³ + 3x² + 3x + 1 is divided by x + 1, using the Remainder Theorem.
Answer: Here the divisor is (x − (−1)), so remainder = p(−1) = (−1)³ + 3(−1)² + 3(−1) + 1 = −1 + 3 − 3 + 1 = 0.
Factorise x² + 9x + 18.
Answer: We need two numbers whose sum is 9 and product is 18: 3 and 6. So x² + 9x + 18 = x² + 3x + 6x + 18 = x(x + 3) + 6(x + 3) = (x + 3)(x + 6).
Without actually calculating the cubes, find the value of 12³ + (−5)³ + (−7)³.
Answer: Since 12 + (−5) + (−7) = 0, using the identity x³ + y³ + z³ = 3xyz when x + y + z = 0: value = 3 × 12 × (−5) × (−7) = 1260.
Continue learning with this chapter:
Practice MCQsTake Chapter Test
Notes, MCQs and test for the same chapter
Chapter Test

How to use these notes

These notes are designed for quick revision and should be used alongside the textbook. Read the chapter once for full understanding, then return here to recall definitions, formulas, key ideas and important examples.

Topics covered on this page

Quick revision tip

Finish one small section, close the page for a moment, and try to recall the main points in your own words. Then check the page again. This simple recall step can make revision more active.

More for this chapter

Class 9 › Mathematics › Polynomials
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Polynomials in One VariableZeroes of a PolynomialRemainder Theorem & Factor TheoremAlgebraic Identities