Class 9 Maths Chapter 2: Polynomials

Polynomials is Chapter 2 of NCERT Class 9 Maths. It introduces algebraic expressions in one variable, their degree, values and zeroes, and shows how the Factor Theorem and standard algebraic identities help you factorise quadratic and cubic polynomials quickly. Use these notes, formulas and solved examples to prepare for your CBSE exams.

What you will learn

• What a polynomial in one variable is and how to find its degree
• How to classify polynomials by degree and by number of terms
• How to find the value and the zeroes of a polynomial
• The Factor Theorem and how to use it to factorise cubic polynomials
• Factorising quadratics by splitting the middle term
• The eight standard algebraic identities and their uses

Key Concepts and Formulas

Polynomials in One Variable

A polynomial in one variable x is an expression made of terms of the form a × xⁿ, where a is a real number (the coefficient) and n is a whole number. The highest power of the variable is called the degree.

4x³ − 2x² + 7   → degree 3
9 (constant)    → degree 0
0 (zero poly.)  → degree not defined

• Expressions like 1/x, √x or x^(−2) are not polynomials, because the powers are not whole numbers.
• A constant polynomial is a single number; the zero polynomial is the constant 0.

Types of Polynomials

Polynomials are named either by their degree or by the number of terms they have. A polynomial of degree n can be written in the general form shown below, where aₙ ≠ 0.

General form:
aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀,  aₙ ≠ 0

• Linear polynomial: degree 1, for example 5x − 3
• Quadratic polynomial: degree 2, for example x² + 6x + 1
• Cubic polynomial: degree 3, for example 2x³ − x
• Monomial: one term (7x²); binomial: two terms (x + 4); trinomial: three terms (x² − x + 8)

Value and Zeroes of a Polynomial

To find the value of p(x) at x = k, replace x by k everywhere and simplify. If the result is zero, k is called a zero of the polynomial.

Value of p(x) at x = k: p(k)
k is a zero of p(x) when p(k) = 0
Zero of ax + b (a ≠ 0): x = −b/a

• A zero is also called a root of the equation p(x) = 0.
• A linear polynomial has exactly one zero.
• A non-zero constant polynomial has no zero; every real number is a zero of the zero polynomial.
• A polynomial can have more than one zero.

The Factor Theorem

The Factor Theorem links zeroes and factors. It saves you from long division when you only want to check whether something is a factor. It applies to any polynomial p(x) of degree n ≥ 1 and any real number a.

If p(a) = 0, then (x − a) is a factor of p(x)
If (x − a) is a factor of p(x), then p(a) = 0
(ax − b) is a factor of p(x) when p(b/a) = 0

• To factorise a cubic, find one zero by trial (try factors of the constant term), divide, and factorise the quadratic that remains.

Factorising Quadratics by Splitting the Middle Term

For a quadratic ax² + bx + c, find two numbers p and q that satisfy the conditions below. Rewrite bx as px + qx and group the terms in pairs.

p + q = b
p × q = a × c
ax² + bx + c = ax² + px + qx + c

Algebraic Identities

These identities are true for all values of the variables and are used for expanding, factorising and quick calculation.

(x + y)² = x² + 2xy + y²
(x − y)² = x² − 2xy + y²
x² − y² = (x + y)(x − y)
(x + a)(x + b) = x² + (a + b)x + ab
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
(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)
If x + y + z = 0, then x³ + y³ + z³ = 3xyz

Solved Examples

Example 1: Find the zero of p(x) = 3x − 12.

3x − 12 = 0
3x = 12
∴ x = 4

Example 2: Find p(−1) for p(x) = x³ − 2x² + x + 5.

p(−1) = (−1)³ − 2(−1)² + (−1) + 5
= −1 − 2 − 1 + 5
∴ p(−1) = 1

Example 3: Check whether (x − 2) is a factor of x³ − 3x² + 4.

p(2) = 2³ − 3(2)² + 4
= 8 − 12 + 4
= 0
∴ (x − 2) is a factor (Factor Theorem)

Example 4: Find k if (x − 1) is a factor of 2x² + kx − 5.

p(1) = 0
2(1)² + k(1) − 5 = 0
2 + k − 5 = 0
∴ k = 3

Example 5: Factorise 6x² + 7x − 3.

Sum = 7, product = 6 × (−3) = −18
Numbers: 9 and −2
6x² + 9x − 2x − 3
= 3x(2x + 3) − 1(2x + 3)
∴ 6x² + 7x − 3 = (3x − 1)(2x + 3)

Example 6: Factorise x³ − 6x² + 11x − 6.

p(1) = 1 − 6 + 11 − 6 = 0
So (x − 1) is a factor
x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6)
x² − 5x + 6 = (x − 2)(x − 3)
∴ Answer = (x − 1)(x − 2)(x − 3)

Example 7: Evaluate (−8)³ + 5³ + 3³ without cubing.

x + y + z = −8 + 5 + 3 = 0
So x³ + y³ + z³ = 3xyz
= 3 × (−8) × 5 × 3
∴ (−8)³ + 5³ + 3³ = −360

Common Mistakes to Avoid

• Do not confuse a zero of a polynomial with the constant term; a zero is the value of x that makes p(x) = 0.
• For a factor like (x + 3), substitute x = −3, not x = 3, when applying the Factor Theorem.
• While splitting the middle term, check that the product equals a × c and not just c.
• In (x − y)³, the middle part is − 3xy(x − y); students often write + 3xy and lose the mark.

Quick Revision

k is a zero of p(x) when p(k) = 0
(x − a) is a factor of p(x) when p(a) = 0
Split middle term: p + q = b, p × q = ac
If x + y + z = 0, then x³ + y³ + z³ = 3xyz

• Degree is the highest power of the variable; the zero polynomial has no degree.
• Memorise all eight identities; they save time in factorisation and calculation questions.

Contact us to enrol in our Class 9 Maths batch at Janakpuri or to book a free demo class with our teachers.