Polynomials — Mathematics Class 9 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Polynomials (Class 9) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Polynomials (CBSE & HBSE)
Builds the algebraic backbone for Class 10. CBSE asks Remainder/Factor Theorem proofs and identity-based factorisation. HBSE prefers direct factorisation by splitting the middle term.
Polynomial in One Variable — Definitions
Polynomial p(x)
An expression of the form a₀ + a₁x + a₂x² + … + aₙxⁿ with aₙ ≠ 0 and each aᵢ a real number.
- Degree = highest power of x (here n).
- aₙ = leading coefficient, a₀ = constant term.
Classification
| By degree | Name | Example |
|---|---|---|
| 0 | Constant | 7 |
| 1 | Linear | 2x + 3 |
| 2 | Quadratic | x² − 5x + 6 |
| 3 | Cubic | x³ − 8 |
| n | n-th degree | — |
| By number of terms | Name | Example |
|---|---|---|
| 1 | Monomial | 7x³ |
| 2 | Binomial | x + 5 |
| 3 | Trinomial | x² − 4x + 7 |
Value of Polynomial at x = a
p(a) = a₀ + a₁a + a₂a² + … — substitute and evaluate.
Zero of a Polynomial
c is a zero of p(x) iff p(c) = 0.
- Linear ax + b has exactly one zero: x = −b/a.
- A constant polynomial (non-zero) has no zero.
- The zero polynomial has every real number as a zero (degree not defined).
CBSE Tip: Don't confuse "zero of polynomial" with "zero polynomial" — examiners love this trap.
Remainder and Factor Theorems
Remainder Theorem
When p(x) is divided by the linear divisor (x − a), the remainder is p(a).
Why it works: by division algorithm p(x) = (x − a)q(x) + r. Putting x = a kills the (x − a) term ⇒ r = p(a).
Generalisation
If divisor is (ax + b), root is x = −b/a, so remainder = p(−b/a).
Factor Theorem
(x − a) is a factor of p(x) ⇔ p(a) = 0.
If one root is known, divide by the corresponding linear factor to obtain a quotient of lower degree, then factorise the quotient.
Strategy to Factorise a Cubic
- Hit-and-trial integer roots dividing the constant term (±1, ±2, ±3, …).
- Once a root a is found, divide p(x) by (x − a).
- Factorise the quotient (usually quadratic) by splitting middle term.
Example: p(x) = x³ − 23x² + 142x − 120.
- Try x = 1: 1 − 23 + 142 − 120 = 0 ✓ ⇒ (x − 1) is a factor.
- Long-divide: quotient x² − 22x + 120.
- Split: (x − 10)(x − 12).
- p(x) = (x − 1)(x − 10)(x − 12).
HBSE Tip: Always test ±1 first — it works in ~70% of textbook problems.
Algebraic Identities
Standard Identities (Class 9 NCERT)
- (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.
Common Uses
- Expand quickly (avoid binomial multiplication).
- Factorise sums/differences of cubes: x³ ± y³ = (x ± y)(x² ∓ xy + y²).
- Numerical shortcuts: 105² = (100 + 5)² = 10000 + 1000 + 25 = 11025.
Splitting the Middle Term (Quadratic ax² + bx + c)
Find two numbers whose product = ac and sum = b.
Example: 6x² + 17x + 5
- ac = 30, find pair (15, 2): 15 × 2 = 30, 15 + 2 = 17 ✓
- 6x² + 15x + 2x + 5 = 3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5).
CBSE Tip: Identity #8 appears almost every year in some form. Memorise it.
Frequently asked questions
Are these Polynomials notes free?
Yes — the Polynomials notes for Mathematics (Class 9) on Siksha Sarovar are completely free to read, with no account required.
Do these notes follow CBSE and HBSE?
Yes. The Polynomials notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Polynomials chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Polynomials.