Polynomials Chapter 2 Class 9 Maths Revision Notes

What is a Polynomial?

An algebraic expression f(x) = a₀ + a₁x + a₂x² + … + aₙxⁿ, where a₀, a₁, …, aₙ are real numbers and all powers of x are non-negative integers, is called a polynomial in x.

Each part — a₀, a₁x, a₂x², … is called a term. The real numbers a₀, a₁, … are the coefficients. The degree is the highest power of x with a non-zero coefficient.

Special Cases

Zero Degree Polynomial: Any non-zero constant, e.g. f(x) = 7, is degree 0 (since 7 = 7x⁰).

Zero Polynomial: f(x) = 0. All coefficients are zero. Its degree is undefined.

Standard Form: A polynomial is in standard form when terms are written in increasing or decreasing order of powers of x.

Classification

Types of Polynomials

By Degree

Linear:

ax + b

Degree 1. Exactly one root. e.g. 3x + 5

Quadratic:

ax² + bx + c

Degree 2. At most 2 roots. e.g. x² − 5x + 6

Cubic:

ax³ + bx² + cx + d

Degree 3. At most 3 roots.

Biquadratic:

ax⁴ + bx³ + cx² + dx + e

Degree 4. Also called quartic.

By Number of Terms

Name	Terms	Example
Monomial	1	9x²
Binomial	2	2x² + 3x
Trinomial	3	3x³ − 8x + 5
Polynomial	4 or more	No special name

Note: Polynomials of degree 5 or more have no special name they are simply called "degree-n polynomials."

Algebraic Identities

An algebraic identity is true for all values of its variables. These are tools for expansion, simplification, and factorisation not equations to solve.

Standard Identities

(a + b)² = a² + 2ab + b² (Square of sum)

(a − b)² = a² − 2ab + b² (Square of difference)

a² − b² = (a + b)(a − b) (Difference of squares)

a³ + b³ = (a + b)(a² − ab + b²) (Sum of cubes)

a³ − b³ = (a − b)(a² + ab + b²) (Difference of cubes)

(a + b)³ = a³ + b³ + 3ab(a + b) (Cube of sum)

(a − b)³ = a³ − b³ − 3ab(a − b) (Cube of difference)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca (Trinomial square)

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc −ca)(Three-cube identity)

(x + a)(x + b) = x² + (a + b)x + ab (Product of linear binomials)

Special Case: If a + b + c = 0, then a³ + b³ + c³ = 3abc. A very common exam shortcut.

Derived Value Forms

a² + b² = (a+b)² − 2ab (When (a+b) and ab are known)

a² + b² = (a−b)² + 2ab (When (a−b) and ab are known)

a² + 1/a² = (a + 1/a)² − 2 (When (a + 1/a) is given)

a² + 1/a² = (a − 1/a)² + 2 (When (a − 1/a) is given)

x³ + 1/x³ = (x+1/x)³ − 3(x+1/x) (Cubic reciprocal sum)

x³ − 1/x³ = (x−1/x)³ + 3(x−1/x) (Cubic reciprocal difference)

Zeros (Roots) of a Polynomial

Definition

A real number α is a zero or root of f(x) if f(α) = 0 — substituting x = α into the polynomial gives zero.

For example, x = 3 is a root of f(x) = x³ − 6x² + 11x − 6, because f(3) = 27 − 54 + 33 − 6 = 0.

Value at a Point

The value of f(x) at x = α is found by substituting x = α. This value may or may not be zero. Only when it equals zero is α a root.

Important points: An nᵗʰ degree polynomial has at most n real roots.

Theorem

Remainder Theorem

Remainder Theorem

Statement

Let p(x) be a polynomial of degree ≥ 1. When p(x) is divided by (x − a), the remainder equals p(a).

p(x) = (x − a) · q(x) + r where r = p(a)

Quick Reference

Divisor	Remainder
(x − a)	p(a)
(x + a)	p(−a)
(ax − b)	p(b/a)
(ax + b)	p(−b/a)

Theorem

Factor Theorem

Factor Theorem

Statement

Let p(x) be a polynomial of degree ≥ 1 and 'a' be a real number. Then:

(x − a) is a factor of p(x) if and only if p(a) = 0.

p(a) = 0 ⟺ (x − a) divides p(x) exactly (zero remainder)

The Factor Theorem is an extension of the Remainder Theorem. It identifies exactly when the remainder is zero confirming a factor without full polynomial long division.

Methods

Factorisation Methods

1 — Perfect Square Trinomial

Recognise expressions of the form A² ± 2AB + B² = (A ± B)². Then apply the difference-of-squares identity if needed.

2 — Difference of Two Squares

Apply a² − b² = (a+b)(a−b). Rearrange the expression first to reveal the pattern.

3 — Sum / Difference of Cubes

Use a³ + b³ = (a+b)(a²−ab+b²) or a³ − b³ = (a−b)(a²+ab+b²).

4 — Completion of the Square

For ax² + bx + c, add and subtract (b/2a)² to create a perfect square, then factor as a difference of squares.

5 — Splitting the Middle Term

For ax² + bx + c, find p and q such that p + q = b and pq = ac. Rewrite bx = px + qx, then factor by grouping.

6 — Factor Theorem Method

Use the Integral Root Theorem to find candidate roots from factors of the constant term. Confirm with f(a) = 0, then divide to reduce the degree and repeat.

Worked Problems

Solved Examples

Ex. 01 Evaluate using an algebraic identity

Find the value of 36x² + 49y² + 84xy when x = 3, y = 6.

1. Recognise the pattern. 36x² + 84xy + 49y² = (6x)² + 2·(6x)·(7y) + (7y)² = (6x + 7y)²

2. Substitute x = 3, y = 6: (6×3 + 7×6)² = (18 + 42)² = (60)²

3. Calculate: 60² = 3600

Answer

36x² + 49y² + 84xy = 3600

Ex. 02 Find (x + 1/x) given x² + 1/x² = 23

If x² + 1/x² = 23, find the value of (x + 1/x).

1. Use the identity: (x + 1/x)² = x² + 2 + 1/x²

2. Substitute: (x + 1/x)² = 23 + 2 = 25

3. Solve: x + 1/x = √25 = 5

Answer

x + 1/x = 5

Ex. 03 Evaluate (997)³ using binomial cube identity

Evaluate (997)³ without a calculator.

1. Rewrite: 997 = 1000 − 3, so (997)³ = (1000 − 3)³

2. Apply (a−b)³ = a³ − b³ − 3ab(a−b):
= 1,000,000,000 − 27 − 9000×(997)

3. = 1,000,000,000 − 27 − 8,973,000 = 991,026,973

Answer

(997)³ = 991,026,973

Ex. 04 Find x³ − 1/x³ given x − 1/x = 5

If x − 1/x = 5, find x³ − 1/x³.

1. Cube both sides: (x − 1/x)³ = 125

2. Expand: x³ − 1/x³ − 3(x − 1/x) = 125

3. Substitute (x − 1/x) = 5: x³ − 1/x³ − 15 = 125

4. Solve: x³ − 1/x³ = 140

Answer

x³ − 1/x³ = 140

Ex. 05 Remainder Theorem — Find the remainder

Find the remainder when f(x) = x³ − 6x² + 2x − 4 is divided by g(x) = 1 − 2x.

1. Zero of divisor: 1 − 2x = 0 → x = 1/2

2. Remainder = f(1/2): (1/2)³ − 6(1/2)² + 2(1/2) − 4

3. = 1/8 − 3/2 + 1 − 4 = (1 − 12 + 8 − 32)/8 = −35/8

Answer

Remainder = −35/8

Ex. 06 Factor Theorem — Factorise a quartic polynomial

Factorise p(x) = 2x⁴ − 7x³ − 13x² + 63x − 45 using the Factor Theorem.

1. Test x = 1: p(1) = 2 − 7 − 13 + 63 − 45 = 0 ✓ → (x − 1) is a factor

2. Test x = 3: p(3) = 162 − 189 − 117 + 189 − 45 = 0 ✓ → (x − 3) is a factor

3. Divide out: p(x) = (x − 1)(x − 3)(2x² + x − 15)

4. Factor quadratic: 2x² + x − 15 = (x + 3)(2x − 5)

Answer

p(x) = (x − 1)(x − 3)(x + 3)(2x − 5)

Ex. 07 Prove an identity using the a+b+c = 0 shortcut

Prove: (x−y)³ + (y−z)³ + (z−x)³ = 3(x−y)(y−z)(z−x)

1. Let a = x−y, b = y−z, c = z−x

2. Check sum: a + b + c = (x−y) + (y−z) + (z−x) = 0

3. Apply special case: When a+b+c = 0, then a³+b³+c³ = 3abc

4. Substitute back: (x−y)³+(y−z)³+(z−x)³ = 3(x−y)(y−z)(z−x) ∎

Result

Proved. Key insight: three differences always sum to zero.

Ex. 08 Find k if x = 4/3 is a root

If x = 4/3 is a root of f(x) = 6x³ − 11x² + kx − 20, find k.

1. Since x = 4/3 is a root, f(4/3) = 0

2. Substitute and expand: 6·(64/27) − 11·(16/9) + k·(4/3) − 20 = 0

3. Multiply through by 27/2: 128 − 176 + 12k − 180 = 0 → 12k = 228

4. Solve: k = 19

Answer

k = 19