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formula-sheet · Mathematics · Chapter 2

Class 9 Mathematics Chapter 2 Polynomials — Formulas & Key Points

Chapter 2 of NCERT Class 9 Mathematics introduces polynomials—algebraic expressions with variables raised to non-negative whole number powers. This formula sheet compiles every definition, theorem, and algebraic identity you need for quick revision. Whether you are solving for zeroes, factorising expressions, or applying the Remainder and Factor Theorems, all key formulas are organised in tables with clear examples. Use this page the night before your exam or during homework to recall concepts instantly.

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Key takeaways

  • ✓A polynomial in one variable has only non-negative whole number exponents; expressions like x + 1/x or √x are not polynomials.
  • ✓Degree of a polynomial is the highest power of the variable with a non-zero coefficient; a linear polynomial has degree 1, quadratic has degree 2, cubic has degree 3.
  • ✓The zero of a polynomial p(x) is any value c where p(c) = 0; a linear polynomial has exactly one zero, a quadratic has at most two, and a cubic has at most three.
  • ✓Remainder Theorem: When p(x) is divided by (x – a), the remainder equals p(a); Factor Theorem: (x – a) is a factor of p(x) if and only if p(a) = 0.
  • ✓Master the seven core algebraic identities—they save time in expansion, factorisation, and fast mental computation during exams.
  • ✓Factorisation of quadratics uses splitting the middle term or the Factor Theorem; for cubics, find one zero first, then divide to get a quadratic.
  • ✓Common mistakes include forgetting the middle term 2xy in (x + y)², confusing (x – y)² with x² – y², and overlooking the constant term when finding zeroes.

Core Definitions and Terminology

Understanding the language of polynomials is the foundation for mastering this chapter. A polynomial in one variable is an algebraic expression where the variable appears only with non-negative integer exponents, and all coefficients are real numbers. For instance, 5x³ – 2x² + 3x – 7 is a polynomial, but x + 1/x is not because 1/x equals x⁻¹, a negative exponent. The degree is the highest exponent with a non-zero coefficient. A constant polynomial like 7 has degree 0, and the zero polynomial has no defined degree. Terms are individual components like 5x³ or –7, and the coefficient is the numerical multiplier. A monomial has one term, a binomial has two, and a trinomial has three. Linear polynomials have degree 1, quadratics degree 2, and cubics degree 3. A zero or root of p(x) is a value c such that p(c) = 0, and a factor is a polynomial that divides another evenly. Memorising these definitions helps you classify and manipulate polynomials confidently in exams.
  • Polynomial: aₙxⁿ + aₙ₋₁xⁿ⁻¹ +... + a₁x + a₀, where all exponents are non-negative integers and coefficients are real.
  • Degree: Highest power of the variable with a non-zero coefficient; determines the polynomial type and maximum number of zeroes.
  • Zero: A value c where p(c) = 0; also called a root of the equation p(x) = 0.
  • Factor: A polynomial that divides another with remainder 0; (x – a) is a factor if p(a) = 0.
  • Monomial, Binomial, Trinomial: Polynomials with 1, 2, or 3 terms respectively.

Major Algebraic Identities — Complete Reference Table

Algebraic identities are equations true for all values of the variables. These seven identities are the backbone of polynomial manipulation in Class 9. Identity I expands the square of a sum: (x + y)² = x² + 2xy + y². Identity II covers the square of a difference: (x – y)² = x² – 2xy + y². Identity III, the difference of squares, factorises instantly: x² – y² = (x + y)(x – y). Identity IV is the square of a trinomial: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx, which has three squared terms and three cross-product terms. Identity V cubes a sum: (x + y)³ = x³ + y³ + 3xy(x + y), or equivalently x³ + 3x²y + 3xy² + y³. Identity VI cubes a difference: (x – y)³ = x³ – y³ – 3xy(x – y). Identity VII factorises a sum of cubes minus 3xyz: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx). Each identity saves calculation time and reduces errors. Practice spotting them in reverse to factorise quickly. For example, recognising x² + 6x + 9 as (x + 3)² using Identity I backwards is faster than trial-and-error.

All Seven Identities in One Table

Below is a quick-reference table listing all seven core identities, their formulas, and typical use cases. Keep this table handy during revision. Identity I and II are used for expanding squares, Identity III for factorising differences of squares, Identity IV for expanding trinomial squares (often in geometry problems), Identities V and VI for cubing binomials, and Identity VII for factorising symmetric expressions. When you see a problem, match the pattern to one of these identities. For instance, if you need to expand (2a + 3b)², use Identity I with x=2a and y=3b. If you must factorise 49p² – 64q², use Identity III with x=7p and y=8q. Practising pattern recognition with these identities is the fastest way to improve speed and accuracy in Class 9 Mathematics exams.
  • Identity I: (x + y)² = x² + 2xy + y² — Expands square of a sum; use when squaring binomials like (3x + 4)².
  • Identity II: (x – y)² = x² – 2xy + y² — Expands square of a difference; use for (5a – 2b)².
  • Identity III: x² – y² = (x + y)(x – y) — Factorises difference of squares; instant factorisation for a² – b².
  • Identity IV: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx — Expands trinomial square; geometry and three-variable problems.
  • Identity V: (x + y)³ = x³ + y³ + 3xy(x + y) = x³ + 3x²y + 3xy² + y³ — Cubes a sum; use for expanding (a + b)³.
  • Identity VI: (x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³ — Cubes a difference; use for (a – b)³.
  • Identity VII: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx) — Factorises sum of cubes; three-variable symmetric expressions.

Remainder Theorem and Factor Theorem

The Remainder Theorem states: when a polynomial p(x) is divided by (x – a), the remainder is p(a). This means you do not need long division to find the remainder—simply substitute a into the polynomial. For example, dividing x³ + 2x² – 5x + 3 by (x – 2) gives remainder p(2) = 8 + 8 – 10 + 3 = 9. The Factor Theorem extends this: (x – a) is a factor of p(x) if and only if p(a) = 0. In other words, if substituting a gives zero, then (x – a) divides p(x) evenly. This theorem is the key to factorising higher-degree polynomials: test small integer factors of the constant term until you find a zero, use that to write one linear factor, then divide the polynomial to reduce the degree. For cubic polynomials, finding one zero converts the problem into factorising a quadratic, which is straightforward using splitting the middle term or the quadratic formula (covered in later chapters).
  • Remainder Theorem: Divide p(x) by (x – a) → remainder = p(a). Quick check without long division.
  • Factor Theorem: (x – a) is a factor of p(x) ⟺ p(a) = 0. Use to find factors by testing zeroes.
  • To factorise a cubic: find one zero c, write factor (x – c), divide p(x) by (x – c) to get a quadratic, then factorise the quadratic.
  • Test candidates for zeroes: factors of the constant term. For p(x) = x³ – 6x² + 11x – 6, test ±1, ±2, ±3, ±6.
  • Once you have all factors, write p(x) as a product: p(x) = a(x – r₁)(x – r₂)..., where r₁, r₂,... are zeroes.

Factorisation Techniques for Quadratic and Cubic Polynomials

Factorisation converts polynomials into products of simpler factors, making equations easier to solve. For quadratic polynomials ax² + bx + c, the primary method is splitting the middle term: find two numbers that add to b and multiply to ac, then rewrite bx as their sum, and factor by grouping. For example, factorise 6x² + 17x + 5. You need two numbers adding to 17 and multiplying to 30 (6×5). These are 2 and 15. Rewrite: 6x² + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1)(2x + 5). For cubic polynomials, use the Factor Theorem: test small integers (factors of the constant term) until you find a zero. Once you have one factor (x – c), perform polynomial division to obtain a quadratic, then factorise the quadratic. This step-by-step reduction is essential for solving cubic equations. If no rational zero exists, the polynomial may be prime or require advanced techniques beyond Class 9. Always verify your factorisation by expanding the product to check it equals the original polynomial.
  • Quadratic (ax² + bx + c): Split middle term into two parts that add to b and multiply to ac, then group and factor.
  • Cubic: Find one zero using Factor Theorem, write factor (x – zero), divide to get quadratic, factorise quadratic.
  • Test candidates: ±(factors of constant term)/(factors of leading coefficient). For monic cubics (leading coefficient 1), test ±factors of constant.
  • Polynomial division: Use long division or synthetic division to divide p(x) by (x – c) after confirming c is a zero.
  • Check your work: Expand the factors to verify you get the original polynomial.

Finding Zeroes of Linear, Quadratic, and Cubic Polynomials

A zero of a polynomial p(x) is a solution to the equation p(x) = 0. For a linear polynomial ax + b (a ≠ 0), there is exactly one zero: x = –b/a. Solve by isolating x. For quadratic polynomials ax² + bx + c, set the polynomial to zero and factorise (or use the quadratic formula in later chapters). A quadratic can have 0, 1, or 2 real zeroes depending on the discriminant Δ = b² – 4ac (positive → 2 zeroes, zero → 1 zero, negative → 0 real zeroes). For cubic polynomials, there can be 1, 2, or 3 real zeroes. Use the Factor Theorem to find the first zero, factor out (x – zero), then solve the resulting quadratic. The fundamental theorem of algebra guarantees that a degree-n polynomial has exactly n roots (counting multiplicities and complex roots), but in Class 9 we focus on real zeroes. Remember: the number of zeroes equals the degree at most. Graphically, zeroes are the x-intercepts where the polynomial curve crosses or touches the x-axis.
  • Linear p(x)=ax+b: Zero is x = –b/a. One zero always exists.
  • Quadratic p(x)=ax²+bx+c: Set to zero, factorise, solve. At most 2 real zeroes.
  • Cubic p(x)=ax³+bx²+cx+d: Find one zero by testing, factor, solve quadratic. At most 3 real zeroes.
  • Geometric meaning: Zeroes are x-coordinates where the graph intersects the x-axis.
  • Multiplicity: If (x–c)² is a factor, c is a repeated zero; the graph touches but does not cross the x-axis at c.

Memory Tricks and Mnemonics for Identities

Memorising the seven algebraic identities is easier with mnemonics and patterns. For Identity I, (x+y)², remember 'square-cross-square': square the first term x², cross-term 2xy, square the second y². For Identity II, (x–y)², the middle term is negative: x² – 2xy + y². Identity III, x²–y², chant 'difference of squares equals sum times difference'. For Identity IV, (x+y+z)², visualise three squares (x², y², z²) plus three cross-terms (2xy, 2yz, 2zx). For Identity V, (x+y)³, expand as x³ + y³ + 3xy(x+y), or remember 'cube-cube-3-cross-sum'. Identity VI is the same with subtraction. Identity VII is symmetric: x³+y³+z³–3xyz factorises neatly when sum of variables is a factor. Practice writing each identity five times, then test yourself by covering the right side and writing the expansion. Use flashcards or the Leitner system. On exam day, write all seven identities on the margin of your answer sheet first thing—this mental cue prevents mid-exam blanks.
  • (x+y)²: Square first, double cross, square second → x² + 2xy + y².
  • (x–y)²: Same as above but middle term negative → x² – 2xy + y².
  • x²–y²: Difference of squares → (x+y)(x–y). Instant factorisation.
  • (x+y+z)²: Three squares plus three double cross-terms → x²+y²+z²+2xy+2yz+2zx.
  • (x+y)³: x³ + y³ + 3xy(x+y) or expand to x³+3x²y+3xy²+y³.
  • (x–y)³: x³ – y³ – 3xy(x–y) or x³–3x²y+3xy²–y³.
  • x³+y³+z³–3xyz: Factorise as (x+y+z)(x²+y²+z²–xy–yz–zx).

Common Mistakes, Sign Errors, and Notation Pitfalls

Students frequently make avoidable errors in polynomial problems. One classic mistake is forgetting the middle term in (x+y)²: writing x²+y² instead of x²+2xy+y². Another is confusing (x–y)² with x²–y²; the former expands to x²–2xy+y², the latter factorises to (x+y)(x–y). Sign errors are rampant: when substituting a negative value into p(x), always use parentheses. For example, if x=–2, then x³=(–2)³=–8, not 8. In factorisation, students sometimes split the middle term incorrectly or forget to verify the product equals the original polynomial. When applying the Factor Theorem, test p(a) carefully—arithmetic slips here invalidate the entire solution. Notation errors include writing x^2 instead of x² in handwritten work, or omitting the multiplication sign in 2x(3x+1), leading to ambiguity. Always double-check your final answer by substitution: if you claim (x–1) is a factor, verify p(1)=0. Write neatly, use parentheses liberally, and show all steps—partial marks depend on clear working.
  • Do not forget the 2xy term in (x+y)²; x²+y² is wrong, x²+2xy+y² is correct.
  • (x–y)² ≠ x²–y². The former is x²–2xy+y², the latter is (x+y)(x–y).
  • Use parentheses for negative substitutions: p(–2) means replace x with (–2), not –2.
  • Check your factorisation by expanding: (x+3)(x–2) should equal x²+x–6, not something else.
  • When testing zeroes, compute p(a) accurately; arithmetic errors here break Factor Theorem logic.
  • Write exponents clearly: x² not x^2 in exams; ambiguous notation loses marks.

Solved Mini-Examples Applying Key Formulas

Working through concrete examples cements understanding. Example 1: Expand (3a+4b)² using Identity I. Let x=3a, y=4b. Then (3a+4b)²=(3a)²+2(3a)(4b)+(4b)²=9a²+24ab+16b². Example 2: Factorise 25p²–49q² using Identity III. Recognise it as (5p)²–(7q)², so apply x²–y²=(x+y)(x–y): (5p+7q)(5p–7q). Example 3: Use the Remainder Theorem to find the remainder when x⁴–3x²+2x–5 is divided by (x–1). Compute p(1)=1–3+2–5=–5. Remainder is –5. Example 4: Check if (x+2) is a factor of p(x)=x³+3x²–4. Compute p(–2)=(–2)³+3(–2)²–4=–8+12–4=0. Yes, (x+2) is a factor. Example 5: Factorise x³–6x²+11x–6. Test x=1: p(1)=1–6+11–6=0, so (x–1) is a factor. Divide: (x³–6x²+11x–6)÷(x–1)=x²–5x+6=(x–2)(x–3). Full factorisation: (x–1)(x–2)(x–3). These examples illustrate the workflow: recognise the pattern, apply the correct identity or theorem, verify the result.

One-Glance Last-Minute Revision Box

Before your exam, scan this summary box. Polynomial: aₙxⁿ+...+a₀, exponents non-negative integers. Degree: highest power with non-zero coefficient. Zero: p(c)=0. Linear: ax+b, one zero –b/a. Quadratic: ax²+bx+c, at most 2 zeroes. Cubic: ax³+bx²+cx+d, at most 3 zeroes. Identity I: (x+y)²=x²+2xy+y². Identity II: (x–y)²=x²–2xy+y². Identity III: x²–y²=(x+y)(x–y). Identity IV: (x+y+z)²=x²+y²+z²+2xy+2yz+2zx. Identity V: (x+y)³=x³+y³+3xy(x+y). Identity VI: (x–y)³=x³–y³–3xy(x–y). Identity VII: x³+y³+z³–3xyz=(x+y+z)(x²+y²+z²–xy–yz–zx). Remainder Theorem: divide p(x) by (x–a), remainder=p(a). Factor Theorem: (x–a) factor ⟺ p(a)=0. Factorise quadratics by splitting middle term. Factorise cubics by finding one zero, dividing, then factorising quadratic. Always verify by substitution or expansion. Use parentheses for negative numbers. Write identities in exam margin first. Practise pattern recognition. Stay calm, show all steps, and double-check arithmetic. Good luck!
  • Polynomial: aₙxⁿ+...+a₀, n≥0, exponents whole numbers.
  • Degree: highest power with non-zero coefficient.
  • Zero: c where p(c)=0. Linear→1 zero, Quadratic→≤2, Cubic→≤3.
  • Identity I: (x+y)²=x²+2xy+y². Identity II: (x–y)²=x²–2xy+y².
  • Identity III: x²–y²=(x+y)(x–y). Identity IV: (x+y+z)²=x²+y²+z²+2xy+2yz+2zx.
  • Identity V: (x+y)³=x³+y³+3xy(x+y). Identity VI: (x–y)³=x³–y³–3xy(x–y).
  • Identity VII: x³+y³+z³–3xyz=(x+y+z)(x²+y²+z²–xy–yz–zx).
  • Remainder Theorem: p(x)÷(x–a) remainder=p(a). Factor Theorem: (x–a) factor ⟺ p(a)=0.
  • Factorise: split middle term for quadratics, find one zero for cubics then divide.
  • Verify every answer, use parentheses, write identities in exam margin.

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Frequently asked questions

What is the difference between a polynomial and an algebraic expression?+
A polynomial is a special algebraic expression where the variable has only non-negative whole number exponents and all coefficients are real. For example, 3x²+2x–1 is a polynomial, but x+1/x or √x are not because they involve negative or fractional exponents.
How do I remember all seven algebraic identities for the exam?+
Write each identity five times daily and use mnemonics: (x+y)² is 'square-cross-square', (x²–y²) is 'sum times difference'. On exam day, write all seven in the margin first thing—it takes 30 seconds and prevents blanks mid-exam.
What is the Factor Theorem and when do I use it?+
The Factor Theorem states (x–a) is a factor of p(x) if and only if p(a)=0. Use it to check whether a value is a zero, or to find factors of cubic polynomials by testing small integers. It converts guesswork into a systematic search.
How do I factorise a cubic polynomial step by step?+
First, find one zero by testing factors of the constant term. Once you have a zero c, write (x–c) as a factor. Divide the cubic by (x–c) using long division or synthetic division to get a quadratic. Factorise the quadratic using splitting the middle term. Combine all factors.
Why is (x+y)² not equal to x²+y²?+
Because squaring a binomial means multiplying (x+y)(x+y), which gives x²+xy+yx+y²=x²+2xy+y². The cross-term 2xy is essential. Forgetting it is the most common mistake in Class 9 polynomial problems.
Can a quadratic polynomial have more than two zeroes?+
No. A polynomial of degree n has at most n zeroes. A quadratic (degree 2) can have 0, 1, or 2 real zeroes, but never 3 or more. This is guaranteed by the Fundamental Theorem of Algebra.
What if I test all factors of the constant term and none gives p(a)=0?+
Then the polynomial has no rational zeroes. It may be prime (cannot be factorised over integers) or its zeroes may be irrational or complex. For CBSE Class 9, exam questions are designed so at least one rational zero exists.
How do I use the Remainder Theorem to save time in exams?+
Instead of long division, simply substitute the value into the polynomial. If dividing p(x) by (x–3), just compute p(3). The answer is the remainder. This takes 10 seconds versus 2 minutes of division—huge time saver.
What is the degree of the zero polynomial?+
The zero polynomial (0 itself) has no defined degree. Every other constant polynomial like 5 has degree 0, but the zero polynomial is a special case because it equals 0x⁰=0x¹=0x²=... for all powers.
How does CBSETUTOR.ai help with polynomial problems?+
Upload a photo of any polynomial question and get instant step-by-step solutions 24×7. It covers all NCERT exercises, practice problems, and board questions for ₹999/month (all subjects, classes 6-12). Start with a 3-day free trial and see results immediately.

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