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Polynomials Notes: Class 9 Maths Chapter 2

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.
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Class 9 › Mathematics › Polynomials
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Topics in this chapter: Polynomials in One VariableZeroes of a PolynomialRemainder Theorem & Factor TheoremAlgebraic Identities