Polynomials in one variable, remainder and factor theorems, algebraic identities (NCERT Ch. 2).

Polynomials

What you'll learn

Define polynomials and identify their degree, coefficients, and types
Apply the Remainder Theorem to find remainders without long division
Use the Factor Theorem to test if a binomial is a factor
State and use five standard algebraic identities
Factorise cubic and quadratic polynomials using identities

Key concepts

Definitions — Degree, Coefficients, Types

A polynomial in variable x is an expression of the form: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀

where n is a non-negative integer and aₙ ≠ 0 is the leading coefficient.

Key terms:

Term	Definition	Example in 3x³ − 5x + 2
Degree	Highest power of variable	3
Leading coefficient	Coefficient of highest-degree term	3
Constant term	Term with no variable (a₀)	2
Coefficient of x	Number multiplied by x	−5

Types by degree:

Name	Degree	Example
Constant polynomial	0	7
Linear polynomial	1	2x + 3
Quadratic polynomial	2	x² − 4x + 1
Cubic polynomial	3	2x³ + x − 5
Biquadratic	4	x⁴ − 3x² + 2

Types by number of terms:

Name	Terms	Example
Monomial	1	5x²
Binomial	2	x³ − 4
Trinomial	3	x² + 3x − 2

Zero of a polynomial p(x): A value 'a' such that p(a) = 0.

Worked Example: Find the zero of p(x) = 3x + 6. 3x + 6 = 0 → x = −2. So zero = −2.

Note: A polynomial of degree n has at most n zeros.

Remainder Theorem

Statement: If a polynomial p(x) is divided by (x − a), the remainder is p(a).

Worked Example 1: Find the remainder when p(x) = x³ − 3x² + 5x − 2 is divided by (x − 2). p(2) = 8 − 12 + 10 − 2 = 4 Remainder = 4 (no long division needed!)

Worked Example 2: Find the remainder when 2x³ + x − 3 is divided by (x + 1). Divisor = x − (−1), so a = −1 p(−1) = 2(−1)³ + (−1) − 3 = −2 − 1 − 3 = −6

Worked Example 3: Find the value of k if (x + 2) leaves remainder 0 when dividing x³ + kx² + x − 6. a = −2: p(−2) = −8 + 4k − 2 − 6 = 0 → 4k − 16 = 0 → k = 4

Factor Theorem

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

Factor theorem = special case of remainder theorem where remainder = 0.

Worked Example 1: Is (x − 3) a factor of p(x) = x³ − 4x² + x + 6? p(3) = 27 − 36 + 3 + 6 = 0 ✓ → Yes, (x − 3) is a factor

Worked Example 2: Is (x + 2) a factor of x³ + 3x² − x − 3? p(−2) = −8 + 12 + 2 − 3 = 3 ≠ 0 → No, (x + 2) is NOT a factor

Using factor theorem to fully factorise a cubic:

Worked Example: Factorise x³ − 2x² − 5x + 6. Try x = 1: 1 − 2 − 5 + 6 = 0 ✓ → (x − 1) is a factor. Divide: x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) Factorise x² − x − 6 = (x − 3)(x + 2) Final: (x − 1)(x − 3)(x + 2)

Algebraic Identities

Five standard identities at Class 9 level:

#	Identity
1	(x + y)² = x² + 2xy + y²
2	(x − y)² = x² − 2xy + y²
3	x² − y² = (x + y)(x − y)
4	(x + a)(x + b) = x² + (a+b)x + ab
5	(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
6	(x + y)³ = x³ + 3x²y + 3xy² + y³ = x³ + y³ + 3xy(x + y)
7	(x − y)³ = x³ − 3x²y + 3xy² − y³ = x³ − y³ − 3xy(x − y)
8	x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

Key derived results:

x³ + y³ = (x + y)(x² − xy + y²)
x³ − y³ = (x − y)(x² + xy + y²)
If x + y + z = 0, then x³ + y³ + z³ = 3xyz

Using Identity 4:

(x + 3)(x − 5) = x² + (3 + (−5))x + (3 × (−5)) = x² − 2x − 15

Using Identity 5:

(a + b + c)² where a=2, b=3, c=−1: = 4 + 9 + 1 + 2(2)(3) + 2(3)(−1) + 2(2)(−1) = 14 + 12 − 6 − 4 = 16 ✓ Direct: (2+3−1)² = 4² = 16 ✓

Factorisation Using Identities

Quadratic factorisation (Identity 3 and 4):

x² − 25 = x² − 5² = (x+5)(x−5)

x² + 7x + 12 = x² + (3+4)x + (3)(4) = (x+3)(x+4)

x² − x − 6 = x² + (2+(−3))x + (2)(−3) = (x+2)(x−3)

Sum and difference of cubes:

8x³ + 27 = (2x)³ + 3³ = (2x + 3)((2x)² − (2x)(3) + 9) = (2x+3)(4x² − 6x + 9)

125a³ − 8b³ = (5a)³ − (2b)³ = (5a − 2b)(25a² + 10ab + 4b²)

Identity 8 application:

a³ + b³ + c³ − 3abc when a+b+c ≠ 0: = (a+b+c)(a²+b²+c²−ab−bc−ca)

If a + b + c = 0 → a³ + b³ + c³ = 3abc

Worked Example: Find 1³ + (−1)³ + 0³. a + b + c = 1 + (−1) + 0 = 0 So 1³ + (−1)³ + 0³ = 3(1)(−1)(0) = 0 ✓

Complete factorisation strategy:

Pattern detected	Identity/Method	Result
a² − b²	Identity 3	(a+b)(a−b)
a² + 2ab + b²	Identity 1	(a+b)²
a² − 2ab + b²	Identity 2	(a−b)²
x² + (a+b)x + ab	Identity 4	(x+a)(x+b)
a³ + b³	Sum of cubes	(a+b)(a²−ab+b²)
a³ − b³	Difference of cubes	(a−b)(a²+ab+b²)
Cubic with known zero	Factor theorem	Long divide, then factorise quotient

Quick check

Find the degree and number of zeros (maximum) of p(x) = 2x⁴ − x³ + 3x − 7.
Find the remainder when 3x³ − 4x² + 7 is divided by (x + 2) using the Remainder Theorem.
Show that (x − 2) is a factor of x³ − 6x² + 11x − 6, then fully factorise.
Expand (2a − 3b + c)² using Identity 5.
Factorise: 27x³ − 125y³

Open the Practice tab for graded questions on Polynomials.