CBSE Class 9 Maths — Polynomials

Chapter Overview & Weightage

Polynomials is one of the foundational chapters in Class 9 Maths. It sits in the Algebra unit, which carries about 20 marks in the CBSE Class 9 annual exam. Within that unit, Polynomials typically contributes 6–8 marks.

Questions from this chapter appear in all sections — 1-mark MCQs (identify degree/type), 2-mark short answers (find zeros, verify), and 3-mark questions (factor theorem application, factorise). The 4-mark question sometimes involves combining Polynomials with Number Systems.

Marks weightage area	Expected marks
Definitions, types, degree	1–2 marks
Zeros of polynomials	2–3 marks
Remainder and Factor Theorem	2–3 marks
Factorisation using identities	2–3 marks

Key Concepts You Must Know

What is a polynomial? An algebraic expression with non-negative integer exponents and real coefficients. $3x^2 - 5x + 2$ is a polynomial; $\frac{1}{x} + 3$ is not (negative exponent).

Types by degree:

Constant polynomial — degree 0, e.g., $7$
Linear polynomial — degree 1, e.g., $2x + 3$
Quadratic polynomial — degree 2, e.g., $x^2 - 4$
Cubic polynomial — degree 3, e.g., $x^3 - 2x + 1$

Types by number of terms:

Monomial — 1 term: $5x^2$
Binomial — 2 terms: $x^2 + 3$
Trinomial — 3 terms: $x^2 - 5x + 6$

Zero of a polynomial: The value of $x$ for which $p(x) = 0$ . A linear polynomial has exactly one zero; a quadratic can have up to two zeros.

Remainder Theorem: When $p(x)$ is divided by $(x - a)$ , the remainder is $p(a)$ . No need to actually do the division.

Factor Theorem: $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$ .

Important Formulas

( a + b )^2 = a^2 + 2ab + b^2

( a - b )^2 = a^2 - 2ab + b^2

a^2 - b^2 = (a+b)(a-b)

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

If $p(x)$ is divided by $(x - a)$ , remainder $= p(a)$ .

Special case: divided by $(ax + b)$ , substitute $x = -b/a$ .

Solved Previous Year Questions

PYQ 1 — Find the value of k (2-mark type)

Q: If $x = 2$ is a zero of $p(x) = kx^2 + 2x - 3$ , find $k$ .

Solution: Since $x = 2$ is a zero, $p(2) = 0$ .

k(2)^2 + 2(2) - 3 = 0

4k + 4 - 3 = 0

4k + 1 = 0

k = -\frac{1}{4}

PYQ 2 — Remainder Theorem (2-mark type)

Q: Find the remainder when $p(x) = x^3 - 6x^2 + 14x - 3$ is divided by $(x - 3)$ .

Solution: By Remainder Theorem, substitute $x = 3$ :

p(3) = 27 - 54 + 42 - 3 = 12

Remainder is 12. No long division needed.

PYQ 3 — Factorise using identities (3-mark type)

Q: Factorise $8x^3 + 27y^3$ .

Solution: Recognise this as $a^3 + b^3$ where $a = 2x$ and $b = 3y$ .

8x^3 + 27y^3 = (2x)^3 + (3y)^3

= (2x + 3y)[(2x)^2 - (2x)(3y) + (3y)^2]

= (2x + 3y)(4x^2 - 6xy + 9y^2)

Difficulty Distribution

Level	Types of questions	Marks
Easy	Identify degree/type, find value of polynomial at a point	1–2
Medium	Remainder Theorem, find zero, factorise trinomial	2–3
Hard	Factor Theorem with unknowns, cube identities, sum $a^3+b^3+c^3$ when $a+b+c=0$	3–4

Expert Strategy

Start this chapter by memorising all the algebraic identities thoroughly — these identities are not just for Class 9, they appear in Class 10, 11, 12, and even JEE. Think of them as investments.

For Factor Theorem problems, always try small integers first ( $\pm 1, \pm 2, \pm 3$ ) as trial values. If $p(1) = 0$ , then $(x-1)$ is a factor, and you just reduced a cubic to a quadratic.

When factorising cubics by splitting the middle term approach doesn’t work, always go back to Factor Theorem — it never fails.

For the identity $a^3 + b^3 + c^3 = 3abc$ when $a + b + c = 0$ — this is a very common 1-marker trick question. Spot that $a + b + c = 0$ in the problem and directly write $3abc$ as the answer.

Common Traps

Trap 1: Confusing degree with number of terms. $5x^3$ has degree 3 but only one term (monomial). Never count terms to find degree.

Trap 2: Wrong sign in Factor Theorem. If the factor is $(x + 2)$ , substitute $x = -2$ (not $+2$ ). The zero of $(x + 2)$ is $x = -2$ .

Trap 3: Students apply the identity $a^2 - b^2 = (a+b)(a-b)$ but forget that $a$ and $b$ can be expressions like $2x$ or $3y$ . Always rewrite as $(a)^2 - (b)^2$ first to identify $a$ and $b$ clearly.

Trap 4: Writing $\sqrt{x}$ or $x^{-1}$ inside and calling it a polynomial. A polynomial requires non-negative integer exponents. Watch out for expressions like $x^{1/2} + 3$ — these are NOT polynomials.