Polynomials for Class 10: The Complete CBSE Guide (2026-27)

A polynomial in one variable x is an algebraic expression of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where all exponents are non-negative integers and all coefficients (a₀, a₁, …, aₙ) are real numbers with aₙ ≠ 0. The highest power of x with a non-zero coefficient is the degree of the polynomial. For example, 5x³ – 2x² + 7x – 9 is a polynomial of degree 3 (cubic), while 2x + 3 is degree 1 (linear). The expression x + 1/x is NOT a polynomial because 1/x = x⁻¹ has a negative exponent. The constant term (a₀) is the coefficient of x⁰. A constant polynomial like 7 has degree 0, but the zero polynomial (p(x) = 0 for all x) has no defined degree — this is a special case you must remember for MCQs. Polynomials are classified by degree: linear (degree 1, general form ax + b), quadratic (degree 2, ax² + bx + c), cubic (degree 3, ax³ + bx² + cx + d), and so on. The CBSE syllabus for polynomials class 10 focuses heavily on quadratic and cubic polynomials because these are where the interesting relationships between zeros and coefficients emerge. Understanding degree is crucial because a polynomial of degree n has at most n real zeros — this fact underpins the entire chapter.

A zero (or root) of a polynomial p(x) is a real number α such that p(α) = 0. Geometrically, zeros are the x-coordinates where the graph of y = p(x) intersects the x-axis. For a linear polynomial ax + b (a ≠ 0), there is exactly one zero: x = –b/a. For a quadratic polynomial ax² + bx + c, there can be zero, one, or two real zeros depending on the discriminant Δ = b² – 4ac (you will study this in detail in Chapter 4 on Quadratic Equations). In polynomials class 10, finding zeros algebraically means solving p(x) = 0. For example, to find the zeros of p(x) = x² – 5x + 6, factor it as (x – 2)(x – 3) and set each factor to zero: x = 2 or x = 3. These are the two zeros. Always verify by substitution: p(2) = 4 – 10 + 6 = 0 ✓ and p(3) = 9 – 15 + 6 = 0 ✓. A common error is to confuse 'zero of the polynomial' with 'the polynomial itself being zero' — remember, a zero is a specific number (or a few numbers), not a function. The relationship between the number of zeros and the degree is strict: a polynomial of degree n has at most n real zeros. This is why a cubic can have 1, 2, or 3 real zeros, but never 4.

This is the most exam-critical concept in polynomials class 10. For a quadratic polynomial p(x) = ax² + bx + c (with a ≠ 0), if α and β are the two zeros, then: Sum of zeros α + β = –b/a, and Product of zeros αβ = c/a. These formulas are derived by comparing p(x) with its factored form a(x – α)(x – β). Expanding a(x – α)(x – β) gives a[x² – (α+β)x + αβ] = ax² – a(α+β)x + a(αβ). Matching coefficients with ax² + bx + c: coefficient of x gives b = –a(α+β) ⟹ α+β = –b/a; constant term gives c = a(αβ) ⟹ αβ = c/a. CBSE loves to ask: 'If α and β are zeros of 2x² – 5x + 3, find α + β and αβ.' Answer: a=2, b=–5, c=3, so α+β = –(–5)/2 = 5/2 and αβ = 3/2. Another common question type: 'Find a quadratic polynomial whose zeros are 3 and –2.' Use the fact that p(x) = k[x² – (α+β)x + αβ]. Here α+β = 3+(–2)=1 and αβ=3(–2)=–6, so p(x) = k(x² – x – 6). Taking k=1 gives x² – x – 6. You can verify by factoring: (x–3)(x+2). This relationship is tested in 2-mark and 3-mark questions every year, and also appears in MCQs. Master it by doing 15–20 problems from NCERT Exercise 2.2.

For a cubic polynomial p(x) = ax³ + bx² + cx + d (a ≠ 0), if α, β, γ are the three zeros, then: α + β + γ = –b/a (sum of zeros), αβ + βγ + γα = c/a (sum of products taken two at a time), and αβγ = –d/a (product of all zeros). These are derived by comparing p(x) = a(x–α)(x–β)(x–γ). Expanding this product gives x³ – (α+β+γ)x² + (αβ+βγ+γα)x – αβγ, then multiply by a and match coefficients. While the CBSE board rarely asks direct cubic coefficient questions in Class 10 (they are more common in Class 11), you should know the formulas for completeness and for solving 'find a cubic polynomial given its zeros' problems. For example, if zeros are 1, 2, 3, then sum = 6, sum of pairwise products = 1·2 + 2·3 + 3·1 = 11, product = 6. So p(x) = k[x³ – 6x² + 11x – 6]. Taking k=1 gives x³ – 6x² + 11x – 6, which factors as (x–1)(x–2)(x–3). This knowledge is useful when you encounter a 3-mark question like 'Verify that 1, 2, 3 are zeros of x³ – 6x² + 11x – 6 and check the relationship between zeros and coefficients.' Such questions blend verification with theory and are high-value.

The Division Algorithm is the formal statement of polynomial long division. It says: if p(x) and g(x) are polynomials with g(x) ≠ 0, then there exist unique polynomials q(x) (quotient) and r(x) (remainder) such that p(x) = g(x)·q(x) + r(x), where either r(x) = 0 or degree of r(x) < degree of g(x). This is identical in structure to integer division (17 = 5·3 + 2, where 5 is divisor, 3 is quotient, 2 is remainder). In polynomials class 10, you perform long division to find q(x) and r(x), then verify the division algorithm identity. For example, divide p(x) = x³ + 2x² – x + 3 by g(x) = x + 1. Using long division: quotient q(x) = x² + x – 2, remainder r(x) = 5. Verification: (x+1)(x²+x–2) + 5 = x³ + x² – 2x + x² + x – 2 + 5 = x³ + 2x² – x + 3 = p(x) ✓. The degree condition is crucial: degree of remainder (0, since 5 is constant) is less than degree of divisor (1). A 3-mark question typically gives p(x), g(x) and asks you to find q(x), r(x) and verify the division algorithm. Practice the long division procedure until you can do it without errors — this is a guaranteed 3 marks if you show all steps clearly.

The Remainder Theorem states: when a polynomial p(x) is divided by a linear polynomial (x – a), the remainder is p(a). This allows you to find the remainder without performing long division — simply substitute x = a into p(x). For example, to find the remainder when p(x) = x³ – 2x² + x – 1 is divided by (x – 2), compute p(2) = 8 – 8 + 2 – 1 = 1. So remainder = 1. If the divisor is (x + 3) = (x – (–3)), then remainder = p(–3). The Remainder Theorem is a time-saver in exams and is often tested in 1-mark or 2-mark questions. It is also the foundation of the Factor Theorem. Proof: By the division algorithm, p(x) = (x–a)q(x) + r, where r is a constant (since divisor has degree 1, remainder has degree 0). Substitute x=a: p(a) = (a–a)q(a) + r = 0 + r = r. Hence remainder = p(a). Make sure to handle signs correctly: if divisor is (2x–1), rewrite as 2(x – 1/2), so remainder = p(1/2). In polynomials class 10 notes, the Remainder Theorem appears in both Exercise 2.3 and in board exam MCQs.

The Factor Theorem says: (x – a) is a factor of p(x) if and only if p(a) = 0. In other words, a is a zero of p(x) ⟺ (x–a) divides p(x) with remainder 0. This is the key tool for factorising cubic and higher-degree polynomials. Strategy: (1) Test small integer values (factors of the constant term) to find one zero. (2) Use the Factor Theorem to write one linear factor. (3) Divide p(x) by that factor to get a quotient of lower degree. (4) Factor the quotient (usually quadratic) by splitting middle term or formula. For example, factorise p(x) = x³ – 6x² + 11x – 6. Test x=1: p(1)=1–6+11–6=0 ✓, so (x–1) is a factor. Divide p(x) by (x–1) to get x²–5x+6. Factor this: (x–2)(x–3). So p(x) = (x–1)(x–2)(x–3). The Factor Theorem is tested heavily in 3-mark questions: 'Using the Factor Theorem, factorise x³+2x²–x–2.' Answer: Test x=1: 1+2–1–2=0 ✓, so (x–1) is a factor. Long division gives quotient x²+3x+2=(x+1)(x+2). Final answer: (x–1)(x+1)(x+2). This technique is central to polynomials class 10 and appears in at least one board question every year.

Algebraic identities are equations true for all values of the variables. They are powerful shortcuts for expanding and factorising expressions in polynomials class 10. The CBSE syllabus requires you to know these identities cold: (a+b)² = a²+2ab+b², (a–b)² = a²–2ab+b², a²–b² = (a+b)(a–b), (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, (a+b)³ = a³+3a²b+3ab²+b³ = a³+b³+3ab(a+b), (a–b)³ = a³–3a²b+3ab²–b³ = a³–b³–3ab(a–b), and a³+b³+c³–3abc = (a+b+c)(a²+b²+c²–ab–bc–ca). These identities save time in two ways: (1) Expansion — use the identity directly instead of multiplying term-by-term. (2) Factorisation — recognise patterns and apply the identity in reverse. For example, to factorise x²+6x+9, spot that it is (x+3)² from the first identity. To expand (2p–3q)³, use identity 6 with a=2p, b=3q: (2p)³ – 3(2p)²(3q) + 3(2p)(3q)² – (3q)³ = 8p³ – 36p²q + 54pq² – 27q³. Identity 7 (sum of cubes) is tested in clever ways: if a+b+c=0, then a³+b³+c³=3abc (since the factor (a+b+c) becomes zero). Practice all identities with 10+ problems each from NCERT Exercise 2.5 to build instant recognition.

For quadratic polynomials ax²+bx+c, the most common CBSE technique is splitting the middle term. The idea: find two numbers p and q such that p+q=b and pq=ac. Rewrite bx as px+qx, then factor by grouping. For example, factorise 6x²+17x+5. Here a=6, b=17, c=5, so ac=30. Find two numbers that add to 17 and multiply to 30: 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 to get one linear factor, then factorise the resulting quadratic. Another technique is grouping: for x³+x²+x+1, group as (x³+x²)+(x+1) = x²(x+1)+1(x+1) = (x+1)(x²+1). This works when terms naturally pair up. Polynomials class 10 NCERT Exercise 2.4 and 2.5 drill these methods extensively. Practice until you can spot the right technique in under 10 seconds. In the board exam, factorisation questions are worth 2–3 marks and are straightforward if you have practiced 50+ problems. Common errors: wrong signs when splitting, forgetting to take out the GCD first, or not checking the final answer by expanding it back.

Polynomials class 10 notes are essential for board exam success. The chapter carries 7 marks out of 80 in the CBSE Class 10 Maths paper (2024-25 pattern): typically 2 marks from the MCQ section (4 questions of 1 mark each cover the entire Unit 1 Algebra, so expect 1–2 MCQs on polynomials) and 5 marks from the theory section (one 3-mark question on division algorithm or factorisation, one 2-mark question on zeros and coefficients). The 3-mark question usually asks you to divide two polynomials, verify the division algorithm, or factorise a cubic using the Factor Theorem. The 2-mark question asks you to find a quadratic polynomial given its zeros, or find the sum/product of zeros given a polynomial, or verify the relationship between zeros and coefficients. MCQs test definitions (degree, zero, remainder theorem) and quick calculations. Exam strategy: (1) Memorise all coefficient formulas (–b/a, c/a for quadratic; –b/a, c/a, –d/a for cubic). (2) Practice 10 division algorithm problems so you can do long division in under 3 minutes without errors. (3) Drill 20 Factor Theorem problems to develop zero-finding intuition. (4) Do every NCERT Exercise 2.2, 2.3, 2.4 problem — board questions are often minor variations of these. (5) In the exam, show all steps in division and factorisation; even if your final answer is wrong, you can get partial marks for method.

CBSE board exams for polynomials class 10 follow a predictable pattern. From 2020–2024, these question types appeared repeatedly: (1) 'If α and β are zeros of px²+qx+r, find the value of α²+β² (or 1/α+1/β, or α²β+αβ²).' Strategy: use (α+β)² = α²+β²+2αβ to find α²+β², or factorise the expression in terms of sum and product. (2) 'Find a quadratic polynomial whose zeros are 2+√3 and 2–√3.' Strategy: sum = 4, product = (2+√3)(2–√3)=4–3=1, so polynomial is x²–4x+1. (3) 'Verify that 1, 2, 3 are zeros of x³–6x²+11x–6 and verify the relationship between zeros and coefficients.' Strategy: substitute each value, check p(1)=p(2)=p(3)=0, then compute sum=6=–(–6)/1 ✓, sum of pairs=11=11/1 ✓, product=6=–(–6)/1 ✓. (4) 'Divide 3x⁴+5x³–7x²+2x+2 by x²+3x+1 and verify the division algorithm.' Strategy: perform polynomial long division, write quotient and remainder, then verify p(x)=g(x)q(x)+r(x). (5) 'Factorise x³–3x²–9x–5 completely.' Strategy: test x=–1: –1–3+9–5=0 ✓, so (x+1) is a factor. Divide to get x²–4x–5=(x–5)(x+1). Final answer: (x+1)²(x–5). Practicing these 5 types covers 90% of board questions. CBSE also loves to ask value-based or application questions: 'The sum of zeros of a quadratic polynomial is 6 and product is 8; if the coefficient of x² is 1, find the polynomial and use it to model the height of a ball.' Answer: x²–6x+8. Such questions test both polynomials class 10 formulas and your ability to connect math to context.

Understanding where students go wrong helps you avoid losing easy marks. Mistake 1: Confusing zero of polynomial with value of polynomial. 'Zero' means the input that makes p(x)=0, not the output. Mistake 2: Sign error in sum of zeros formula. The formula is –b/a (note the negative sign); students often write b/a. Example: for 2x²–5x+3, sum of zeros is –(–5)/2=5/2, not –5/2. Mistake 3: Forgetting the degree condition in division algorithm. You must state 'deg r(x) < deg g(x) or r(x)=0' to get full marks. Mistake 4: Incomplete factorisation. If you factor x³–x as x(x²–1), you must factor further to x(x+1)(x–1). Stopping early loses marks. Mistake 5: Substitution errors in Factor Theorem. When testing x=–2, remember (–2)³=–8, not +8; sign mistakes here lead to wrong conclusions. Mistake 6: Not verifying answers. Always substitute back: if you claim (x–2) is a factor, check p(2)=0. Mistake 7: Misapplying identities. Using a²+b²=(a+b)² is WRONG; the correct identity is (a+b)²=a²+2ab+b². Mistake 8: In splitting middle term, choosing p and q that satisfy p+q=b but not pq=ac (or vice versa). Both conditions must hold. Mistake 9: Writing polynomial in wrong order (not descending powers) makes errors more likely. Mistake 10: Rushing through MCQs and misreading 'which is NOT a polynomial' as 'which is a polynomial'. In polynomials class 10, avoiding these traps can mean the difference between 5/7 and 7/7.

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