Chapter 2 of 12

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IB Class 9 Mathematics★ 8–10 marks · CBSE Class 9 board (part of Unit II — Algebra)

Polynomials

From counting apples to encoding all of algebra — polynomials are the words of mathematics. Class 9 Polynomials with one-variable polynomials, degree, zeros, the Remainder & Factor Theorems, identities for factorisation, and 25+ exam-style problems with full step-by-step solutions.

⏱ 45 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Identify whether an expression is a polynomial; classify by degree (constant, linear, quadratic, cubic, …)
2Compute the value p(a) of a polynomial by substitution
3Identify zeros of a polynomial and explain their geometric meaning
4Apply the Remainder Theorem to find the remainder when dividing by (x − a) or (ax + b)
5Apply the Factor Theorem to test for factors and find zeros
6Factorise a quadratic by splitting the middle term
7Factorise a cubic completely: find one zero, divide, factorise the resulting quadratic
8Apply the seven core algebraic identities to expand, simplify and factorise expressions
9Recognise and exploit the a + b + c = 0 → a³ + b³ + c³ = 3abc shortcut

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Why this chapter matters

Polynomials are the words of algebra — every later chapter, from linear equations to calculus, manipulates them. This chapter introduces the Remainder Theorem, Factor Theorem and the seven core identities that you'll use through Class 12 and engineering math.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Algebraic expressions and identities (Class 8)

Basic expansion of (a+b)² etc.

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Exponents and powers (Class 8)

Laws of exponents

Polynomials — Class 9 (CBSE)

If algebra is a language, polynomials are its sentences. Linear equations, quadratics, cubic curves, the orbits of satellites, signal processing, machine-learning loss functions — they all start as polynomial expressions. Master this chapter and Class 10's quadratics, Class 11's binomial theorem and Class 12's calculus become natural extensions.

1. The story — why a 'polynomial' deserves its own chapter

A merchant in 9th-century Baghdad tracks the cost of buying spice. Each kilogram costs an unknown number; let's call it $x$ . Five kilograms cost $5 x$ . A bag costs $3$ rupees. Five kilograms in a bag cost $5 x + 3$ .

That little expression — $5 x + 3$ — is a polynomial. Add another variable cost, say transportation of $2 x^{2}$ (it grows quadratically with the distance, ignore why), and the total becomes $2 x^{2} + 5 x + 3$ .

The word polynomial comes from Greek poly (many) + Latin nominalis (name/term) — literally "many-named expression." Each piece like $2 x^{2}$ or $5 x$ is a term. The chapter is about how to operate on these expressions: add, multiply, factorise, and find their "roots" (the input values that make the polynomial equal to zero).

You'll learn techniques here that the same merchant could have used 1200 years ago — and that the engineer building India's Chandrayaan mission uses today.

2. The big picture — three things to take away

Every polynomial has a "shape" determined by its degree. Degree 1 = straight line. Degree 2 = parabola. Degree 3 = cubic curve. Higher degrees get wavier.
Zeros of a polynomial are the heart of algebra. They're the inputs that make the output zero — and they tell you everything geometrically (where the curve crosses the x-axis) and physically (when something happens, equilibrium points, etc.).
Factorisation = decomposition. Just like 12 = 2 × 2 × 3, a polynomial like $x^{2} - 5 x + 6$ factorises to $(x - 2) (x - 3)$ . The factors give you the zeros for free.

3. What exactly is a polynomial?

A polynomial in one variable $x$ is an expression of the form

$p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0},$

where:

$a_{0}, a_{1}, \dots, a_{n}$ are real-number coefficients,
$a_{n} \neq = 0$ (so the highest-power term actually appears),
$n$ is a non-negative integer (so no $x^{1/2}$ , no $1/ x$ , no $x^{- 3}$ ).

Examples — all polynomials:

$7$ (constant — degree 0)
$3 x - 5$ (linear — degree 1)
$2 x^{2} + 5 x + 3$ (quadratic — degree 2)
$x^{3} - 7 x + 6$ (cubic — degree 3)
$- 4 x^{4} + x^{2} + 2$ (quartic — degree 4)

Non-examples — NOT polynomials:

$x^{1/2} + 1$ (fractional power)
$\frac{1}{x} + 3$ (negative power: $x^{- 1}$ )
$x^{2} + x$ (square root of variable)
$sin (x) + x$ (trigonometric, not polynomial)

Tester's rule. If, after expansion, every term is a constant times $x^{k}$ for non-negative integer $k$ , it's a polynomial.

4. Terminology — the four labels you'll see in every exam

For the polynomial $p (x) = 2 x^{3} + 7 x^{2} - 5 x + 1$ :

Label	Meaning	This example
Term	Each individual piece separated by + / −	$2 x^{3}$ , $7 x^{2}$ , $- 5 x$ , $1$
Coefficient	The numerical multiplier of each term	$2, 7, - 5, 1$
Degree	Highest power of $x$ present	$3$
Constant term	The term with no $x$	$1$
Leading coeff.	Coefficient of the highest-power term	$2$

A polynomial is named by its degree:

Degree 0 → constant polynomial (e.g. $7$ ).
Degree 1 → linear polynomial (e.g. $3 x - 5$ ).
Degree 2 → quadratic polynomial (e.g. $x^{2} - 4 x + 4$ ).
Degree 3 → cubic polynomial (e.g. $x^{3} + 2 x - 1$ ).
Degree 4 → quartic (also called biquadratic if no $x^{3}$ or $x$ terms).
Degree $n$ → polynomial of degree $n$ .

Watch this special case. The polynomial $p (x) = 0$ is called the zero polynomial. Its degree is undefined (some books say $- \infty$ ). Don't confuse it with the constant polynomial $p (x) = 7$ , whose degree is $0$ .

5. Polynomials in more than one variable (a brief preview)

You can also have polynomials in two or more variables: $p (x, y) = x^{2} y - 3 x y + 7$ . The degree of such a polynomial is the highest sum of exponents in any single term. For $x^{2} y$ , the sum is $2 + 1 = 3$ ; for $3 x y$ , it's $2$ ; for $7$ , it's $0$ . So the degree is $3$ .

Class 9 focuses on one-variable polynomials. Multi-variable polynomials come back in Class 10 (algebra of expressions) and Class 11 (binomial theorem).

6. Value and zero of a polynomial

The value of $p (x)$ at $x = a$ is just what you get by substituting $a$ for $x$ and simplifying. We write it as $p (a)$ .

Worked example. Find $p (2)$ if $p (x) = x^{3} - 4 x + 1$ .

$p (2) = 2^{3} - 4 (2) + 1 = 8 - 8 + 1 = 1$ .

A zero (or root) of $p (x)$ is a value $a$ such that $p (a) = 0$ .

Worked example. Is $x = 1$ a zero of $p (x) = x^{2} - 3 x + 2$ ?

$p (1) = 1^{2} - 3 (1) + 2 = 1 - 3 + 2 = 0$ . Yes.

Why zeros matter. If you plot $y = p (x)$ , the zeros are exactly the x-coordinates where the curve crosses the x-axis. Geometrically, zeros are where the polynomial vanishes.

Key fact (proved in Class 10): A polynomial of degree $n$ has at most $n$ zeros (counting multiplicities). A linear polynomial has at most one zero, a quadratic at most two, a cubic at most three, and so on.

7. The Remainder Theorem — division shortcut

When you divide a polynomial $p (x)$ by a linear divisor $(x - a)$ , long division gives you a quotient $q (x)$ and a remainder $r$ (which is a constant). The Remainder Theorem says:

Remainder Theorem. When $p (x)$ is divided by $(x - a)$ , the remainder is $p (a)$ .

In other words, the remainder isn't even a new number — it's just the value of $p$ at $a$ , which is one substitution away.

Worked example. Find the remainder when $p (x) = x^{3} - 4 x^{2} + 5 x - 3$ is divided by $(x - 2)$ .

By the Remainder Theorem, the remainder is $p (2)$ : $p (2) = 8 - 16 + 10 - 3 = - 1$ . Remainder = −1.

Verification by long division would give the same answer — but the theorem saves you all the working.

More general form. When $p (x)$ is divided by $(a x + b)$ , the remainder is $p (- b / a)$ .

(Reason: $a x + b = 0 \Rightarrow x = - b / a$ .)

Worked example. Find the remainder when $p (x) = 2 x^{3} - x + 5$ is divided by $(2 x + 3)$ .

$2 x + 3 = 0 \Rightarrow x = - 3/2$ . $p (- 3/2) = 2 \cdot (- 27/8) - (- 3/2) + 5 = - 27/4 + 3/2 + 5 = - 27/4 + 6/4 + 20/4 = - 1/4$ . Remainder = −1/4.

8. The Factor Theorem — the most powerful tool in Class 9 algebra

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

This is just the Remainder Theorem with remainder $= 0$ . "Zero remainder" means $(x - a)$ divides $p (x)$ exactly, which is the definition of being a factor.

Why it's powerful. To check whether $(x - 3)$ is a factor of $p (x) = x^{3} - 7 x + 6$ :

Substitute $x = 3$ : $p (3) = 27 - 21 + 6 = 12 \neq = 0$ . So no, NOT a factor.
Try $x = 1$ : $p (1) = 1 - 7 + 6 = 0$ . So $(x - 1)$ IS a factor.

You now know one factor for free. Polynomial division gives you the rest.

Step-by-step factorisation using the Factor Theorem

Worked example (a CBSE classic). Factorise $p (x) = x^{3} - 23 x^{2} + 142 x - 120$ completely.

Step 1 — Find one zero by trial. Test small integers that divide the constant term $- 120$ . (This is the Rational Root Theorem shortcut.)

$p (1) = 1 - 23 + 142 - 120 = 0$ . ✓ So $(x - 1)$ is a factor.

Step 2 — Divide $p (x)$ by $(x - 1)$ (synthetic or long division): $x^{3} - 23 x^{2} + 142 x - 120 = (x - 1) (x^{2} - 22 x + 120)$ .

Step 3 — Factorise the resulting quadratic using the "splitting the middle term" method (you'll see this in Section 10): $x^{2} - 22 x + 120 = x^{2} - 12 x - 10 x + 120 = x (x - 12) - 10 (x - 12) = (x - 10) (x - 12)$ .

Step 4 — Combine. $p (x) = (x - 1) (x - 10) (x - 12) .$

The three zeros are $1, 10, 12$ . The polynomial is now completely factorised.

9. The seven essential algebraic identities

These are the bread-and-butter of Class 9 algebra. Memorise them; appearance in every paper is guaranteed.

$(1) (a + b)^{2} = a^{2} + 2 ab + b^{2}$ $(2) (a - b)^{2} = a^{2} - 2 ab + b^{2}$ $(3) (a + b) (a - b) = a^{2} - b^{2}$ $(4) (x + a) (x + b) = x^{2} + (a + b) x + ab$ $(5) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ $(6) (a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = a^{3} + b^{3} + 3 ab (a + b)$ $(7) (a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} = a^{3} - b^{3} - 3 ab (a - b)$

Two derived identities used constantly:

$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}),$

$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2}) .$

And a famous one used in Olympiads (you'll see it in Section 13):

$a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a) .$

Special case (when $a + b + c = 0$ ): if the three numbers sum to zero, the famous identity collapses to $a^{3} + b^{3} + c^{3} = 3 ab c$ . This is a 4-mark HOTS question almost every year.

10. Factorising a quadratic — splitting the middle term

To factorise $a x^{2} + b x + c$ :

Compute the product $a \cdot c$ .
Find two numbers $p, q$ such that $p + q = b$ and $p \cdot q = a c$ .
Split the middle term: $a x^{2} + p x + q x + c$ .
Group and take out common factors.

Worked example. Factorise $6 x^{2} + 11 x + 3$ .

Step 1: $a \cdot c = 6 \cdot 3 = 18$ . Step 2: Find two numbers that multiply to $18$ and add to $11$ . → $9$ and $2$ . Step 3: $6 x^{2} + 9 x + 2 x + 3$ . Step 4: $3 x (2 x + 3) + 1 (2 x + 3) = (3 x + 1) (2 x + 3)$ .

✦ Answer: $(3 x + 1) (2 x + 3)$ .

Worked example. Factorise $x^{2} - 5 x + 6$ .

Two numbers with product $6$ and sum $- 5$ : $- 2$ and $- 3$ . $x^{2} - 2 x - 3 x + 6 = x (x - 2) - 3 (x - 2) = (x - 2) (x - 3)$ .

11. Eight worked exam examples

Example 1 — Identify (1 mark)

Is $2 x^{3} - 5 x + 7$ a polynomial? Yes. The coefficient $2$ is just a real number; coefficients are allowed to be irrational. (What's not allowed is a non-integer exponent on $x$ .) Degree = 3.

Example 2 — Find the value (1 mark)

Find $p (- 1)$ if $p (x) = 2 x^{3} - 3 x^{2} + 5 x + 1$ . $p (- 1) = 2 (- 1) - 3 (1) + 5 (- 1) + 1 = - 2 - 3 - 5 + 1 = - 9$ .

Example 3 — Remainder Theorem (2 marks)

Find the remainder when $p (x) = x^{4} - 3 x^{2} + 2 x + 1$ is divided by $(x + 1)$ . $x + 1 = 0 \Rightarrow x = - 1$ . $p (- 1) = 1 - 3 - 2 + 1 = - 3$ . Remainder = $- 3$ .

Example 4 — Factor Theorem (2 marks)

Is $(x - 2)$ a factor of $p (x) = x^{3} - 2 x^{2} + 5 x - 10$ ? $p (2) = 8 - 8 + 10 - 10 = 0$ . ✓ So yes, $(x - 2)$ is a factor.

Example 5 — Factorisation (3 marks)

Factorise $4 x^{2} - 9$ . $4 x^{2} - 9 = (2 x)^{2} - 3^{2} = (2 x + 3) (2 x - 3)$ (using identity 3).

Example 6 — Splitting middle term (3 marks)

Factorise $x^{2} - 7 x + 12$ . Numbers with product 12 and sum −7: −3 and −4. $x^{2} - 3 x - 4 x + 12 = x (x - 3) - 4 (x - 3) = (x - 3) (x - 4)$ .

Example 7 — Identity application (3 marks)

Without actually computing, find the value of $98 \times 102$ . $98 \times 102 = (100 - 2) (100 + 2) = 10 0^{2} - 2^{2} = 10000 - 4 = 9996$ (using identity 3).

Example 8 — HOTS factorisation (4 marks)

Factorise $x^{3} - 6 x^{2} + 11 x - 6$ . Try integer divisors of $- 6$ : $\pm 1, \pm 2, \pm 3, \pm 6$ . $p (1) = 1 - 6 + 11 - 6 = 0$ . ✓ So $(x - 1)$ is a factor. Divide: $x^{3} - 6 x^{2} + 11 x - 6 \div (x - 1) = x^{2} - 5 x + 6$ . Factorise the quadratic: $(x - 2) (x - 3)$ . $p (x) = (x - 1) (x - 2) (x - 3)$ . Zeros are $1, 2, 3$ .

12. Common pitfalls

Calling $1/ x + 1$ a polynomial. It's not — negative exponents disallowed.
Confusing the constant polynomial with the zero polynomial. $p (x) = 7$ has degree 0; $p (x) = 0$ has undefined degree.
Mis-signing the remainder. Dividing by $(x + 2)$ — substitute $x = - 2$ (not $+ 2$ ).
Forgetting to fully factorise. $x^{3} - x = x (x^{2} - 1)$ is INCOMPLETE — keep going: $= x (x + 1) (x - 1)$ .
Mixing up identities. $(a + b)^{2} \neq = a^{2} + b^{2}$ . There's a cross-term $2 ab$ .
Wrong sign on $a^{3} - b^{3}$ . $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ — the second factor has a $+ ab$ , not $- ab$ .
Stopping after finding one factor. Use the Factor Theorem → divide → continue factorising until linear factors are left.

13. Beyond NCERT — stretch problems

Stretch 1 — The $a + b + c = 0$ trick (olympiad classic)

If $a + b + c = 0$ , prove that $a^{3} + b^{3} + c^{3} = 3 ab c$ .

Use the identity $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ . Given $a + b + c = 0$ , the RHS equals 0, so $a^{3} + b^{3} + c^{3} - 3 ab c = 0$ , giving $a^{3} + b^{3} + c^{3} = 3 ab c$ . ∎

Stretch 2 — Symmetric polynomial trick

If $a + b = 5$ and $ab = 6$ , find $a^{3} + b^{3}$ without solving for $a, b$ .

$a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b) = 5^{3} - 3 \cdot 6 \cdot 5 = 125 - 90 = 35$ .

Stretch 3 — Counting integer zeros (JEE Foundation)

How many integer zeros does $p (x) = x^{4} - 5 x^{2} + 4$ have?

Factorise: $p (x) = (x^{2} - 1) (x^{2} - 4) = (x - 1) (x + 1) (x - 2) (x + 2)$ . Integer zeros: $\pm 1, \pm 2$ . Four integer zeros.

14. Real-world polynomials

Projectile motion. A ball thrown up follows $h (t) = - 5 t^{2} + u t + h_{0}$ — a quadratic polynomial. Solving $h (t) = 0$ gives when it hits the ground.
Compound interest growth. Money compounded for $n$ years grows as $P (1 + r)^{n}$ — expand and you get a polynomial in $n$ (sort of — actually an exponential, but the binomial expansion uses polynomial identities).
Computer graphics. Smooth curves (Bezier curves) used in fonts, animation, and CAD are cubic polynomials.
Loss functions in ML. A linear regression's "loss" is a quadratic polynomial in the model's weights.
Signal processing. Polynomial roots tell engineers whether an audio filter will oscillate or die out.
Encryption. Modern public-key cryptography uses polynomials over finite fields (ECC, RSA-like systems).
Physics — moment of inertia. Often appears as a quadratic in distance from the axis.

15. CBSE exam blueprint

Type	Marks	Typical question	Time
VSA	1	Degree, identify polynomial, value $p (a)$	30 sec
SA-I	2	Remainder/Factor Theorem; simple factorisation	2 min
SA-II	3	Split-the-middle-term; identity-based simplification	4–5 min
LA	4	Cubic factorisation; $a + b + c = 0$ → $a^{3} + b^{3} + c^{3}$	6–8 min

Total marks: 8–10 / 80 in Class 9 finals. This is the highest-yield chapter in algebra at this level. Pure-skill chapter — drill the identities and the factorisation algorithms.

Three exam-day strategies:

Always write the identity you're using before substituting numbers. 1 mark for the rule, 1 mark for the calculation.
When asked "find the remainder when divided by (ax + b)", compute $p (- b / a)$ directly — never long-divide unless explicitly asked.
For cubic factorisation, try small integers ( $\pm 1, \pm 2$ etc.) first; only switch to fractions if integers fail.

16. NCERT exercise walkthrough

Exercise 2.1: 5 questions — what is a polynomial, identifying degree, classifying as linear/quadratic/cubic.
Exercise 2.2: 4 questions — finding value $p (a)$ , zeros of given polynomials.
Exercise 2.3: 3 questions — Remainder Theorem applications.
Exercise 2.4: 5 questions — Factor Theorem; factorising cubic polynomials.
Exercise 2.5: 16 questions — applying identities, big factorisation problems, conjugate / sum-of-cubes tricks.

All exercises are covered by Sections 4–10 above with multiple worked examples. Try them in the practice quiz.

17. 60-second recap

Polynomial = sum of terms $a_{k} x^{k}$ with non-negative integer $k$ and real $a_{k}$ .
Degree = highest power of $x$ .
Value $p (a)$ = substitute $a$ for $x$ .
Zero = value of $x$ that makes $p (x) = 0$ .
Remainder Theorem: remainder when $p (x) \div (x - a)$ is $p (a)$ .
Factor Theorem: $(x - a)$ is a factor iff $p (a) = 0$ .
Seven core identities — memorise.
Splitting the middle term for quadratics.
Cubic factorisation: find one zero via Factor Theorem → divide → factorise the quadratic.

Take the practice quiz and the flashcard deck before moving on. Next chapter: Coordinate Geometry.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Polynomial

p(x) = aₙxⁿ + … + a₁x + a₀

n ∈ ℕ ∪ {0}; aᵢ ∈ ℝ

Value at x = a

p(a) = substitute a for x

Zero / root

p(a) = 0 ⇒ a is a zero of p(x)

Degree-n polynomial has at most n zeros.

Remainder Theorem

p(x) ÷ (x − a) → remainder = p(a)

Saves long division.

Remainder Theorem (general)

p(x) ÷ (ax + b) → remainder = p(−b/a)

Factor Theorem

(x − a) is a factor of p(x) ⇔ p(a) = 0

Both directions hold.

Identity 1

(a + b)² = a² + 2ab + b²

The cross-term 2ab is the killer.

Identity 2

(a − b)² = a² − 2ab + b²

Identity 3

(a + b)(a − b) = a² − b²

The conjugate / difference-of-squares.

Identity 4

(x + a)(x + b) = x² + (a + b)x + ab

Used in splitting the middle term.

Identity 5

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

Identity 6

(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b)

Identity 7

(a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b)

Sum of cubes

a³ + b³ = (a + b)(a² − ab + b²)

Difference of cubes

a³ − b³ = (a − b)(a² + ab + b²)

Three-variable cube identity

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

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

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Calling 1/x + 3 a polynomial

✓ Negative exponents disallowed. Polynomials require non-negative integer powers of x only.

WATCH OUT

✗ Saying the zero polynomial p(x) = 0 has degree 0

✓ Degree of the zero polynomial is undefined (some texts use −∞). p(x) = 7 has degree 0; p(x) = 0 does not.

WATCH OUT

✗ (a + b)² = a² + b²

✓ Wrong — there's a cross-term 2ab. Always expand or use identity (a + b)² = a² + 2ab + b².

WATCH OUT

✗ Wrong sign when dividing by (x + a)

✓ x + a = 0 ⇒ x = −a. Substitute −a (with the negative sign) into p, not +a.

WATCH OUT

✗ Incomplete factorisation

✓ Keep factorising until only linear and irreducible quadratic factors remain. x³ − x = x(x² − 1) is NOT done; continue → x(x − 1)(x + 1).

WATCH OUT

✗ Wrong sign in difference-of-cubes

✓ a³ − b³ = (a − b)(a² + ab + b²). Note the + in the middle factor, NOT a minus.

WATCH OUT

✗ Not trying small integers first

✓ For cubic factorisation, test ±1, ±2, etc. (divisors of the constant term) before trying fractions.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 2.1

Exercise 2.1

What is a polynomial; identifying degree, classifying as linear/quadratic/cubic

Questions

Ex 2.2

Exercise 2.2

Finding the value p(a); zeros of given polynomials

Questions

Ex 2.3

Exercise 2.3

Remainder Theorem applications

Questions

Ex 2.4

Exercise 2.4

Factor Theorem; factorising cubic polynomials

Questions

Ex 2.5

Exercise 2.5

Applying identities to expand, simplify and factorise

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Identify

Which of the following is a polynomial: (a) √x + 3, (b) 1/x + 5, (c) 2x³ − 7x + 4, (d) sin(x) + 1?

▶ Show solution

Step 1 — Check each candidate against the polynomial rule.
  A polynomial requires every term to be a (constant) × x^k with k a non-negative integer.

Step 2 — Apply.
  (a) √x = x^(1/2). Fractional exponent → NOT a polynomial.
  (b) 1/x = x^(−1). Negative exponent → NOT a polynomial.
  (c) 2x³, −7x, 4 all have non-negative integer exponents → polynomial.
  (d) sin(x) is a transcendental function → NOT a polynomial.

✦ Answer: Only (c) is a polynomial.

Degree of (c) = 3 (highest power).

Q2EASY· Degree

Find the degree of the polynomial 4x⁵ − 7x² + 6.

▶ Show solution

Step 1 — Identify the highest power of x present.
  Terms: 4x⁵, −7x², 6. The highest power is 5.

✦ Answer: Degree = 5.

Rule: degree = highest exponent of the variable, regardless of the coefficient's sign or value.

Q3EASY· Value

Find p(3) if p(x) = 2x² − 5x + 1.

▶ Show solution

Step 1 — Substitute x = 3.
  p(3) = 2·(3)² − 5·(3) + 1.

Step 2 — Evaluate.
  = 2·9 − 15 + 1
  = 18 − 15 + 1
  = 4.

✦ Answer: p(3) = 4.

Q4EASY· Zero check

Is x = −2 a zero of p(x) = x³ + 2x² − 5x − 6?

▶ Show solution

Step 1 — Definition of zero: a is a zero iff p(a) = 0.

Step 2 — Compute p(−2).
  p(−2) = (−2)³ + 2·(−2)² − 5·(−2) − 6
        = −8 + 2·4 + 10 − 6
        = −8 + 8 + 10 − 6
        = 4.

Step 3 — Conclude.
  p(−2) = 4 ≠ 0.

✦ Answer: NO, x = −2 is not a zero (p(−2) = 4).

Q5EASY· Identity

Expand (3x + 4)² using an identity.

▶ Show solution

Step 1 — Use identity (a + b)² = a² + 2ab + b² with a = 3x, b = 4.

Step 2 — Substitute.
  (3x)² + 2·(3x)·(4) + (4)²
  = 9x² + 24x + 16.

✦ Answer: 9x² + 24x + 16.

Memo: square the first term, square the second, sum cross-product doubled.

Q6EASY· Identity

Without computing, find 102 × 98 using an identity.

▶ Show solution

Step 1 — Notice the pattern: 102 = 100 + 2 and 98 = 100 − 2.
  This is exactly the (a + b)(a − b) form with a = 100, b = 2.

Step 2 — Apply identity 3: (a + b)(a − b) = a² − b².
  102 × 98 = 100² − 2² = 10000 − 4 = 9996.

✦ Answer: 102 × 98 = 9996.

Trick: any 'nearly-equal' product can be done mentally with this identity.

Q7MEDIUM· Remainder Theorem

Find the remainder when p(x) = x³ + 3x² − 5x + 7 is divided by (x − 2).

▶ Show solution

Step 1 — Remainder Theorem: p(x) ÷ (x − a) gives remainder p(a).
  Here a = 2.

Step 2 — Compute p(2).
  p(2) = (2)³ + 3·(2)² − 5·(2) + 7
       = 8 + 12 − 10 + 7
       = 17.

✦ Answer: Remainder = 17.

Far faster than long division. Always use the theorem when the divisor is linear.

Q8MEDIUM· Remainder Theorem (general)

Find the remainder when p(x) = 3x³ − 2x² + 4x − 1 is divided by (3x − 1).

▶ Show solution

Step 1 — Find the root of the divisor.
  3x − 1 = 0 ⇒ x = 1/3.
  By the general Remainder Theorem, the remainder is p(1/3).

Step 2 — Compute p(1/3).
  p(1/3) = 3·(1/27) − 2·(1/9) + 4·(1/3) − 1
         = 1/9 − 2/9 + 4/3 − 1.

Step 3 — Combine over a common denominator of 9.
         = 1/9 − 2/9 + 12/9 − 9/9
         = (1 − 2 + 12 − 9)/9
         = 2/9.

✦ Answer: Remainder = 2/9.

Working with fractions feels slow but is much faster than long-dividing a cubic by (3x − 1).

Q9MEDIUM· Factor Theorem

Determine whether (x + 1) is a factor of p(x) = x⁴ − x³ − 2x + 2.

▶ Show solution

Step 1 — Apply the Factor Theorem.
  (x + 1) is a factor of p(x) iff p(−1) = 0.

Step 2 — Compute p(−1).
  p(−1) = (−1)⁴ − (−1)³ − 2·(−1) + 2
        = 1 − (−1) + 2 + 2
        = 1 + 1 + 2 + 2
        = 6.

Step 3 — Conclude.
  p(−1) = 6 ≠ 0.

✦ Answer: NO, (x + 1) is NOT a factor (p(−1) = 6).

Q10MEDIUM· Quadratic factorisation

Factorise 6x² + 11x − 10 by splitting the middle term.

▶ Show solution

Step 1 — Compute a·c (the product of leading coefficient and constant term).
  a·c = 6 · (−10) = −60.

Step 2 — Find two numbers with product −60 and sum 11 (the middle coefficient).
  Factors of 60: (1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10).
  With opposite signs to give a negative product:
  15 and −4 → 15 × (−4) = −60 ✓, 15 + (−4) = 11 ✓.

Step 3 — Split the middle term.
  6x² + 15x − 4x − 10.

Step 4 — Group and factor.
  3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5).

Step 5 — Verify by expansion.
  (3x − 2)(2x + 5) = 6x² + 15x − 4x − 10 = 6x² + 11x − 10. ✓

✦ Answer: 6x² + 11x − 10 = (3x − 2)(2x + 5).

Q11MEDIUM· Identity factor

Factorise 25x² − 49y² using an identity.

▶ Show solution

Step 1 — Recognise the difference-of-squares pattern.
  25x² = (5x)², 49y² = (7y)². So the expression is a² − b² with a = 5x, b = 7y.

Step 2 — Apply identity 3.
  a² − b² = (a + b)(a − b) = (5x + 7y)(5x − 7y).

✦ Answer: 25x² − 49y² = (5x + 7y)(5x − 7y).

Q12MEDIUM· Sum/diff of cubes

Factorise 8x³ + 27.

▶ Show solution

Step 1 — Recognise as a sum of cubes.
  8x³ = (2x)³, 27 = 3³. So this is a³ + b³ with a = 2x, b = 3.

Step 2 — Apply identity a³ + b³ = (a + b)(a² − ab + b²).
  = (2x + 3)((2x)² − (2x)(3) + 3²)
  = (2x + 3)(4x² − 6x + 9).

Step 3 — Check if the quadratic factor factorises further.
  Discriminant of 4x² − 6x + 9 = (−6)² − 4·4·9 = 36 − 144 = −108 < 0.
  No real roots ⇒ irreducible over ℝ.

✦ Answer: 8x³ + 27 = (2x + 3)(4x² − 6x + 9).

Q13MEDIUM· Cube of binomial

Expand (2x − 3y)³.

▶ Show solution

Step 1 — Apply identity (a − b)³ = a³ − 3a²b + 3ab² − b³.
  Set a = 2x, b = 3y.

Step 2 — Substitute.
  = (2x)³ − 3·(2x)²·(3y) + 3·(2x)·(3y)² − (3y)³
  = 8x³ − 3·4x²·3y + 3·2x·9y² − 27y³
  = 8x³ − 36x²y + 54xy² − 27y³.

✦ Answer: 8x³ − 36x²y + 54xy² − 27y³.

Check by counting terms: a cube of a binomial always has 4 terms — correct here.

Q14HARD· Cubic factorisation

Factorise x³ − 6x² + 11x − 6 completely.

▶ Show solution

Step 1 — Find one zero using the Factor Theorem.
  Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0. ✓
  So (x − 1) is a factor.

Step 2 — Divide p(x) by (x − 1).
  Long division (or synthetic):
  x³ − 6x² + 11x − 6 ÷ (x − 1) = x² − 5x + 6.

Step 3 — Factorise the resulting quadratic.
  x² − 5x + 6: need two numbers with product 6 and sum −5.
  Numbers: −2 and −3.
  x² − 2x − 3x + 6 = x(x − 2) − 3(x − 2) = (x − 2)(x − 3).

Step 4 — Combine.
  p(x) = (x − 1)(x − 2)(x − 3).

Step 5 — Verify by checking each zero.
  p(1) = 0 ✓, p(2) = 8 − 24 + 22 − 6 = 0 ✓, p(3) = 27 − 54 + 33 − 6 = 0 ✓.

✦ Answer: x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3).

Zeros: 1, 2, 3.

Q15HARD· a + b + c = 0 trick

If a + b + c = 0, find the value of a³ + b³ + c³ − 3abc.

▶ Show solution

Step 1 — Recall the three-variable identity.
  a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Step 2 — Substitute the given condition.
  a + b + c = 0 ⇒ the entire RHS is 0 × (anything) = 0.

Step 3 — Therefore a³ + b³ + c³ − 3abc = 0, i.e. a³ + b³ + c³ = 3abc.

✦ Answer: a³ + b³ + c³ − 3abc = 0 (equivalently, a³ + b³ + c³ = 3abc).

Follow-up: Find 5³ + (−2)³ + (−3)³. Since 5 + (−2) + (−3) = 0, the trick says 5³ + (−2)³ + (−3)³ = 3·5·(−2)·(−3) = 90. Verify: 125 − 8 − 27 = 90. ✓

Q16HARD· Symmetric trick

If a + b = 5 and ab = 6, find a³ + b³ without solving for a, b.

▶ Show solution

Step 1 — Use the identity (a + b)³ = a³ + b³ + 3ab(a + b).

Step 2 — Rearrange to isolate a³ + b³.
  a³ + b³ = (a + b)³ − 3ab(a + b).

Step 3 — Substitute the given values.
  = 5³ − 3·6·5
  = 125 − 90
  = 35.

✦ Answer: a³ + b³ = 35.

Why elegant: we never had to solve the quadratic x² − 5x + 6 = 0 (whose roots are a, b = 2, 3). Always use symmetric identities when only sum and product are given.

Q17HARD· Find a, b

If (x − 2) and (x + 1) are both factors of p(x) = x³ + ax² + bx − 6, find a and b.

▶ Show solution

Step 1 — Apply the Factor Theorem twice.
  (x − 2) is a factor ⇒ p(2) = 0.
  (x + 1) is a factor ⇒ p(−1) = 0.

Step 2 — Write both conditions.
  p(2) = 8 + 4a + 2b − 6 = 0 ⇒ 4a + 2b = −2 ⇒ 2a + b = −1. … (i)
  p(−1) = −1 + a − b − 6 = 0 ⇒ a − b = 7. … (ii)

Step 3 — Solve the simultaneous system.
  Add (i) and (ii): (2a + b) + (a − b) = −1 + 7 ⇒ 3a = 6 ⇒ a = 2.
  Substitute in (ii): 2 − b = 7 ⇒ b = −5.

Step 4 — Verify.
  p(x) = x³ + 2x² − 5x − 6.
  p(2) = 8 + 8 − 10 − 6 = 0 ✓.
  p(−1) = −1 + 2 + 5 − 6 = 0 ✓.

✦ Answer: a = 2, b = −5.

Q18HARD· Integer zeros

Find all integer zeros of p(x) = x⁴ − 5x² + 4.

▶ Show solution

Step 1 — Notice this is a quadratic in x² — a 'biquadratic'.
  Let y = x². Then p becomes q(y) = y² − 5y + 4.

Step 2 — Factorise the quadratic in y.
  q(y) = (y − 1)(y − 4).

Step 3 — Substitute back y = x².
  p(x) = (x² − 1)(x² − 4).

Step 4 — Apply the difference-of-squares identity twice.
  x² − 1 = (x − 1)(x + 1).
  x² − 4 = (x − 2)(x + 2).

Step 5 — Complete factorisation.
  p(x) = (x − 1)(x + 1)(x − 2)(x + 2).

Step 6 — Read off the zeros.
  Zeros: 1, −1, 2, −2.

✦ Answer: Four integer zeros — namely ±1 and ±2.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Polynomial = sum of aₖxᵏ with non-negative integer k and real coefficients.
•Degree = highest power of x.
•Value p(a) — just substitute.
•Zero of p(x) — value of x making p(x) = 0.
•Remainder Theorem: p(x) ÷ (x − a) → remainder p(a).
•Factor Theorem: (x − a) is a factor ⇔ p(a) = 0.
•Seven core identities — memorise.
•Splitting the middle term for quadratic factorisation.
•Cubic factorisation: Factor Theorem → divide → factor the quadratic.
•If a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc — annual HOTS question.

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Every other polynomial has a unique highest power. The zero polynomial has NO terms — so there's no highest power to pick. Convention varies: 'undefined' or '−∞'. CBSE uses 'undefined.'

Yes. The term 5 = 5 · x⁰, so its degree is 0. (Different from the zero polynomial, which has all coefficients zero.)

It's a direct consequence of the Remainder Theorem. If p(a) = 0, the remainder of p(x) ÷ (x − a) is zero, meaning (x − a) divides p(x) exactly — which is the definition of being a factor.

Try simple fractions like ±1/2, ±1/3, ±2/3 — these come from the 'Rational Root Theorem' which says any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. If still nothing works, the cubic has no rational root and you'd need numerical methods (out of Class 9 scope).

Each identity replaces 3–5 lines of algebra with 1 line. Across a paper that has 4–5 identity-related questions, you save 15–20 minutes. Time you can spend on harder problems.

Functionally synonymous. 'Zero' is preferred when discussing the polynomial itself, 'root' is more common when the polynomial = 0 is treated as an equation. Both refer to the same value.

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Polynomials

By the end of this chapter you'll be able to…

Before you start — revise these

Polynomials — Class 9 (CBSE)

1. The story — why a 'polynomial' deserves its own chapter

2. The big picture — three things to take away

3. What exactly is a polynomial?

4. Terminology — the four labels you'll see in every exam

5. Polynomials in more than one variable (a brief preview)

6. Value and zero of a polynomial

7. The Remainder Theorem — division shortcut

8. The Factor Theorem — the most powerful tool in Class 9 algebra

Step-by-step factorisation using the Factor Theorem

9. The seven essential algebraic identities

10. Factorising a quadratic — splitting the middle term

11. Eight worked exam examples

Example 1 — Identify (1 mark)

Example 2 — Find the value (1 mark)

Example 3 — Remainder Theorem (2 marks)

Example 4 — Factor Theorem (2 marks)

Example 5 — Factorisation (3 marks)

Example 6 — Splitting middle term (3 marks)

Example 7 — Identity application (3 marks)

Example 8 — HOTS factorisation (4 marks)

12. Common pitfalls

13. Beyond NCERT — stretch problems

Stretch 1 — The a+b+c=0 trick (olympiad classic)

Stretch 2 — Symmetric polynomial trick

Stretch 3 — Counting integer zeros (JEE Foundation)

14. Real-world polynomials

15. CBSE exam blueprint

16. NCERT exercise walkthrough

17. 60-second recap

Key formulas & results

Common mistakes & fixes

NCERT exercises (with solutions)

Practice problems

5-minute revision

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Why is the zero polynomial's degree undefined?

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Why memorise so many identities?

What's the difference between 'zero' and 'root' of a polynomial?

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Linear Equations in Two Variables

Introduction to Euclid's Geometry

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Stretch 1 — The $a + b + c = 0$ trick (olympiad classic)