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Updated on 21 Jul 2026, 11:26 IST
These Class 10 Maths Chapter 2 Polynomials Notes PDF explain the complete chapter through short revision notes, properly formatted formulas, graphs, solved examples, important questions, MCQs and previous-year practice. Students can use this page to understand polynomials from the beginning, revise the chapter before an examination or download the complete notes for offline study.
The chapter covers the meaning and types of polynomials, zeroes and their graphical interpretation, the relationship between zeroes and coefficients, formation of quadratic polynomials and the polynomial division algorithm.
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The downloadable Class 10 Maths Chapter 2 Polynomials Notes PDF should contain the complete chapter summary, formulas, diagrams, solved questions and examination practice in one printable file.
A polynomial is an algebraic expression in which the powers of the variable are non-negative integers.
A polynomial in one variable x can be written as:
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₂x² + a₁x + a₀
Here:
p(x) = 3x² − 5x + 7
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The powers of x are 2, 1 and 0. All are non-negative integers.
Example 1:
1/x + 2
Since 1/x = x⁻¹, the expression contains a negative exponent.
Example 2:
√x + 3
Since √x = x¹ᐟ², the expression contains a fractional exponent.
Example 3:
2/(x + 1)
The variable appears in the denominator.
Consider:
p(x) = 4x³ − 7x² + 2x − 9
| Part | Value |
| Variable | x |
| Terms | 4x³, −7x², 2x and −9 |
| Leading term | 4x³ |
| Leading coefficient | 4 |
| Constant term | −9 |
| Degree | 3 |
A number k is a zero of p(x) if:
p(k) = 0
Example:
p(x) = x − 4
p(4) = 4 − 4
p(4) = 0
Therefore, 4 is a zero of p(x).
For a quadratic polynomial:
p(x) = ax² + bx + c
If its zeroes are α and β:
Sum of zeroes:
α + β = −b/a
Product of zeroes:
αβ = c/a
Quadratic polynomial formed from its zeroes:
p(x) = x² − (α + β)x + αβ
Polynomial division identity:
p(x) = g(x)q(x) + r(x)
In words:
Dividend = Divisor × Quotient + Remainder
A polynomial is an expression made from variables and constants using addition, subtraction and multiplication, where every variable has a non-negative integer exponent.
The standard form is:
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₂x² + a₁x + a₀
Term: Each separate part joined by addition or subtraction.
Coefficient: The numerical value multiplying a variable.
Variable: A symbol whose value can change, such as x.
Constant term: A term without a variable.
Leading term: The term containing the highest power.
Leading coefficient: The coefficient of the leading term.
Degree: The highest exponent of the variable with a non-zero coefficient.
| Expression | Polynomial? | Reason |
| 3x² − 2x + 5 | Yes | All powers are non-negative integers |
| 7x³ + x | Yes | Powers are 3 and 1 |
| 6 | Yes | A non-zero constant is a polynomial of degree 0 |
| 1/x + 2 | No | x has exponent −1 |
| √x + 3 | No | x has exponent 1/2 |
| 2/(x + 1) | No | The variable is in the denominator |
| x² + π | Yes | π acts as a constant coefficient |
Consider:
p(x) = 6x⁴ − 3x² + 8x − 11
The polynomial in which every coefficient is zero is called the zero polynomial.
p(x) = 0
The degree of the zero polynomial is generally left undefined in school mathematics because no single highest non-zero power exists.
Polynomials can represent measurable relationships in geometry, science and economics.
Area example:
If the length of a rectangle is x + 3 and its width is x + 2, then:
Area = (x + 3)(x + 2)
Area = x² + 5x + 6
Motion example:
A quadratic polynomial can model the height of an object over time when its motion follows a parabolic path.
Also Check: Algebraic Expressions Revision
Polynomials are classified according to their number of terms and their degree.
Definition: A monomial contains one non-zero term.
Examples:
Definition: A binomial contains two unlike terms.
Examples:
Definition: A trinomial contains three unlike terms.
Examples:
A non-zero constant polynomial has degree 0.
Example:
p(x) = 7
A linear polynomial has degree 1.
General form:
p(x) = ax + b, where a ≠ 0
Example:
p(x) = 3x − 5
A quadratic polynomial has degree 2.
General form:
p(x) = ax² + bx + c, where a ≠ 0
Example:
p(x) = 2x² − 7x + 3
A cubic polynomial has degree 3.
General form:
p(x) = ax³ + bx² + cx + d, where a ≠ 0
Example:
p(x) = x³ − 4x + 2
| Type | Defining feature | General form | Example | Degree |
| Monomial | One term | axⁿ | 5x² | Depends on n |
| Binomial | Two terms | axᵐ + bxⁿ | x + 4 | Highest exponent |
| Trinomial | Three terms | axᵐ + bxⁿ + c | x² + 3x + 2 | Highest exponent |
| Constant | Highest power is 0 | c | 8 | 0 |
| Linear | Highest power is 1 | ax + b | 2x − 5 | 1 |
| Quadratic | Highest power is 2 | ax² + bx + c | x² − 4x + 3 | 2 |
| Cubic | Highest power is 3 | ax³ + bx² + cx + d | x³ − 2x + 1 | 3 |
Classify each polynomial by number of terms and degree.
Answers:
A zero of a polynomial is a value of the variable that makes the polynomial equal to zero.
Consider:
p(x) = x² − 5x + 6
Check whether 2 is a zero.
p(2) = 2² − 5(2) + 6
p(2) = 4 − 10 + 6
p(2) = 0
Therefore, 2 is a zero of p(x).
For:
p(x) = ax + b
Set p(x) equal to zero:
ax + b = 0
ax = −b
x = −b/a
Therefore, the zero of ax + b is:
x = −b/a, where a ≠ 0
Find the zero of:
p(x) = 3x − 12
Set p(x) = 0:
3x − 12 = 0
3x = 12
x = 4
Therefore, the zero is 4.
Verification:
p(4) = 3(4) − 12
p(4) = 12 − 12
p(4) = 0
Find the zeroes of:
p(x) = x² − 5x + 6
Set the polynomial equal to zero:
x² − 5x + 6 = 0
Split the middle term:
x² − 2x − 3x + 6 = 0
Group the terms:
x(x − 2) − 3(x − 2) = 0
Take the common factor:
(x − 2)(x − 3) = 0
Therefore:
x − 2 = 0
or
x − 3 = 0
x = 2
or
x = 3
The zeroes are 2 and 3.
A polynomial of degree n can have at most n zeroes.
| Degree | Polynomial type | Maximum number of zeroes |
| 1 | Linear | 1 |
| 2 | Quadratic | 2 |
| 3 | Cubic | 3 |
| 4 | Quartic | 4 |
The statement gives the maximum number of zeroes. A polynomial may have fewer real zeroes than its degree.
The zeroes of a polynomial are the x-coordinates of the points where its graph meets or touches the x-axis.
A linear polynomial normally produces a straight line.
Example:
p(x) = x − 3
The zero is 3 because:
p(3) = 3 − 3 = 0
The graph crosses the x-axis at:
(3, 0)
A quadratic polynomial produces a parabola.
Consider:
p(x) = ax² + bx + c
The parabola can meet the x-axis in three possible ways.
| Graph behaviour | Number of distinct real zeroes | Meaning |
| Crosses the x-axis at two points | 2 | Two distinct real zeroes |
| Touches the x-axis at one point | 1 | Two equal real zeroes |
| Does not meet the x-axis | 0 | No real zeroes |
p(x) = x² − 5x + 6
Factorisation gives:
p(x) = (x − 2)(x − 3)
The zeroes are 2 and 3.
The graph crosses the x-axis at:
(2, 0) and (3, 0)
p(x) = x² − 4x + 4
Factorise:
p(x) = (x − 2)²
The only distinct zero is 2.
The graph touches the x-axis at:
(2, 0)
p(x) = x² + 4
For every real value of x:
x² ≥ 0
Therefore:
x² + 4 > 0
The graph does not meet the x-axis, so the polynomial has no real zeroes.
A cubic polynomial can have up to three real zeroes.
Example:
p(x) = x³ − x
Factorise:
p(x) = x(x² − 1)
p(x) = x(x − 1)(x + 1)
The zeroes are:
−1, 0 and 1
The graph meets the x-axis at three points.
For a quadratic polynomial ax² + bx + c, the sum of its zeroes is −b/a and their product is c/a.
Consider:
p(x) = ax² + bx + c
Let its zeroes be α and β.
Then:
α + β = −b/a
and:
αβ = c/a
If α and β are the zeroes, then:
p(x) = a(x − α)(x − β)
Expand the factors:
p(x) = a[x² − αx − βx + αβ]
p(x) = a[x² − (α + β)x + αβ]
Multiply by a:
p(x) = ax² − a(α + β)x + aαβ
Compare this with:
p(x) = ax² + bx + c
The coefficient of x gives:
−a(α + β) = b
Therefore:
α + β = −b/a
The constant term gives:
aαβ = c
Therefore:
αβ = c/a
For:
p(x) = 2x² − 7x + 3
Here:
a = 2
b = −7
c = 3
Sum of zeroes:
α + β = −b/a
α + β = −(−7)/2
α + β = 7/2
Product of zeroes:
αβ = c/a
αβ = 3/2
Therefore:
Sum of zeroes = 7/2
Product of zeroes = 3/2
Consider:
p(x) = x² − 5x + 6
Factorise:
p(x) = (x − 2)(x − 3)
Therefore:
α = 2
β = 3
Sum of zeroes:
α + β = 2 + 3
α + β = 5
Using the coefficients:
−b/a = −(−5)/1
−b/a = 5
Therefore:
α + β = −b/a
Product of zeroes:
αβ = 2 × 3
αβ = 6
Using the coefficients:
c/a = 6/1
c/a = 6
Therefore:
αβ = c/a
The relationship is verified.
One zero of the polynomial 2x² + kx − 6 is 2. Find k.
Since 2 is a zero:
p(2) = 0
Substitute x = 2:
2(2²) + k(2) − 6 = 0
2(4) + 2k − 6 = 0
8 + 2k − 6 = 0
2 + 2k = 0
2k = −2
k = −1
Therefore, k = −1.
| Mistake | Correct rule |
| Writing α + β = b/a | α + β = −b/a |
| Ignoring a | Divide both b and c by a |
| Writing αβ = −c/a | αβ = c/a |
| Using coefficients before arranging the polynomial | First write the polynomial in descending powers |
A monic quadratic polynomial with zeroes α and β is x² − (α + β)x + αβ.
If the zeroes are α and β:
p(x) = (x − α)(x − β)
After expansion:
p(x) = x² − (α + β)x + αβ
A non-zero multiple of this polynomial has the same zeroes.
Form a quadratic polynomial whose zeroes are 3 and 5.
Here:
α = 3
β = 5
Sum of zeroes:
α + β = 3 + 5
α + β = 8
Product of zeroes:
αβ = 3 × 5
αβ = 15
Required polynomial:
p(x) = x² − (α + β)x + αβ
p(x) = x² − 8x + 15
Therefore, the required polynomial is:
x² − 8x + 15
Form a quadratic polynomial whose zeroes are 4 and −2.
Sum:
α + β = 4 + (−2)
α + β = 2
Product:
αβ = 4 × (−2)
αβ = −8
Required polynomial:
p(x) = x² − 2x − 8
Verification:
p(x) = (x − 4)(x + 2)
p(x) = x² − 2x − 8
The sum of two zeroes is 7 and their product is 10. Form the quadratic polynomial.
Use:
p(x) = x² − (sum of zeroes)x + product of zeroes
p(x) = x² − 7x + 10
Therefore, the required polynomial is:
x² − 7x + 10
Form a quadratic polynomial whose zeroes are 1/2 and 2/3.
Sum:
α + β = 1/2 + 2/3
Take the common denominator 6:
α + β = 3/6 + 4/6
α + β = 7/6
Product:
αβ = (1/2) × (2/3)
αβ = 1/3
The monic polynomial is:
p(x) = x² − (7/6)x + 1/3
Multiply every term by 6 to remove fractions:
6p(x) = 6x² − 7x + 2
Therefore, one required polynomial is:
6x² − 7x + 2
The polynomial division algorithm states that the dividend equals the divisor multiplied by the quotient, plus the remainder.
For polynomials p(x) and g(x), where g(x) ≠ 0:
p(x) = g(x)q(x) + r(x)
Here:
The degree of r(x) must be less than the degree of g(x).
Divide:
x² + 3x + 2
by:
x + 1
Factorise the dividend:
x² + 3x + 2 = (x + 1)(x + 2)
Therefore:
(x² + 3x + 2) ÷ (x + 1) = x + 2
Quotient:
q(x) = x + 2
Remainder:
r(x) = 0
Verification:
Dividend = Divisor × Quotient + Remainder
x² + 3x + 2 = (x + 1)(x + 2) + 0
x² + 3x + 2 = x² + 3x + 2
The result is correct.
Divide:
p(x) = x² + 2x + 5
by:
g(x) = x + 1
Step 1: Divide the first terms.
x² ÷ x = x
The first term of the quotient is x.
Step 2: Multiply the divisor by x.
x(x + 1) = x² + x
Step 3: Subtract.
(x² + 2x + 5) − (x² + x) = x + 5
Step 4: Divide the new first term.
x ÷ x = 1
The next quotient term is 1.
Step 5: Multiply the divisor by 1.
1(x + 1) = x + 1
Step 6: Subtract.
(x + 5) − (x + 1) = 4
Therefore:
Quotient = x + 1
Remainder = 4
Verification:
p(x) = g(x)q(x) + r(x)
x² + 2x + 5 = (x + 1)(x + 1) + 4
x² + 2x + 5 = x² + 2x + 1 + 4
x² + 2x + 5 = x² + 2x + 5
The result is correct.
[INTERNAL LINK: polynomial division worksheet → /polynomial-division-worksheet/]
The most important Class 10 Polynomials formulas cover zeroes, coefficients, polynomial formation and division.
| Concept | Formula | When to use it | Common mistake |
| General polynomial | p(x) = aₙxⁿ + … + a₁x + a₀ | Identifying degree and coefficients | Treating negative exponents as valid |
| Linear polynomial | p(x) = ax + b | Working with a degree-1 polynomial | Taking a = 0 |
| Zero of a linear polynomial | x = −b/a | Finding the zero directly | Forgetting the negative sign |
| Quadratic polynomial | p(x) = ax² + bx + c | Working with a degree-2 polynomial | Ignoring a |
| Sum of zeroes | α + β = −b/a | Finding or verifying the sum | Writing b/a |
| Product of zeroes | αβ = c/a | Finding or verifying the product | Writing −c/a |
| Polynomial from zeroes | x² − (α + β)x + αβ | Constructing a quadratic polynomial | Reversing the sign of the middle term |
| Division algorithm | p(x) = g(x)q(x) + r(x) | Verifying polynomial division | Allowing degree of r(x) to equal degree of g(x) |
Solved examples show how polynomial definitions and formulas are applied step by step in examination questions.
Identify the type and degree of:
p(x) = 5x³ − 2x + 7
Solution:
The polynomial contains three terms, so it is a trinomial.
The highest power is 3, so its degree is 3.
Therefore, it is a cubic trinomial.
Check whether 3 is a zero of:
p(x) = x² − 4x + 3
Solution:
p(3) = 3² − 4(3) + 3
p(3) = 9 − 12 + 3
p(3) = 0
Therefore, 3 is a zero.
Find the zeroes of:
p(x) = x² − 7x + 12
Solution:
x² − 7x + 12 = 0
x² − 3x − 4x + 12 = 0
x(x − 3) − 4(x − 3) = 0
(x − 3)(x − 4) = 0
Therefore:
x = 3 or x = 4
The zeroes are 3 and 4.
A quadratic graph crosses the x-axis at x = −2 and x = 5. Find the zeroes.
Solution:
The zeroes are the x-coordinates of the points where the graph crosses the x-axis.
Therefore, the zeroes are:
−2 and 5
For:
p(x) = 2x² − 5x + 2
Factorise:
2x² − 5x + 2 = 0
2x² − 4x − x + 2 = 0
2x(x − 2) − 1(x − 2) = 0
(2x − 1)(x − 2) = 0
Therefore:
x = 1/2 or x = 2
So:
α = 1/2
β = 2
Sum:
α + β = 1/2 + 2
α + β = 5/2
Using coefficients:
−b/a = −(−5)/2
−b/a = 5/2
Product:
αβ = (1/2) × 2
αβ = 1
Using coefficients:
c/a = 2/2
c/a = 1
The relationship is verified.
Form a quadratic polynomial whose zeroes are −3 and 4.
Sum:
α + β = −3 + 4
α + β = 1
Product:
αβ = −3 × 4
αβ = −12
Required polynomial:
p(x) = x² − (α + β)x + αβ
p(x) = x² − x − 12
If the sum of zeroes of 3x² + kx + 4 is 5, find k.
For ax² + bx + c:
α + β = −b/a
Here:
a = 3
b = k
Therefore:
5 = −k/3
Multiply by 3:
15 = −k
k = −15
Divide:
2x² + 7x + 3
by:
2x + 1
Factorise the dividend:
2x² + 7x + 3
2x² + 6x + x + 3
2x(x + 3) + 1(x + 3)
(2x + 1)(x + 3)
Therefore:
Quotient = x + 3
Remainder = 0
Important Polynomials questions test definitions, graphs, zeroes, coefficients, polynomial formation and the division algorithm.
Find the degree of:
7x⁴ − 3x² + 1
Answer: 4
Find the zero of:
p(x) = 5x − 10
Solution:
5x − 10 = 0
5x = 10
x = 2
How many zeroes can a quadratic polynomial have at most?
Answer: 2
Find the sum and product of the zeroes of:
3x² − 8x + 5
Solution:
a = 3
b = −8
c = 5
Sum:
α + β = −b/a
α + β = 8/3
Product:
αβ = c/a
αβ = 5/3
A graph touches the x-axis at x = 4. How many distinct real zeroes does the polynomial have?
Answer: It has one distinct real zero, x = 4.
Form a quadratic polynomial whose zeroes are 2 and −5.
Solution:
Sum:
α + β = 2 + (−5)
α + β = −3
Product:
αβ = 2 × (−5)
αβ = −10
Polynomial:
p(x) = x² − (−3)x − 10
p(x) = x² + 3x − 10
If one zero of 2x² + 5x + k is −2, find k.
Solution:
Since −2 is a zero:
p(−2) = 0
2(−2)² + 5(−2) + k = 0
2(4) − 10 + k = 0
8 − 10 + k = 0
k − 2 = 0
k = 2
A rectangular garden has length x + 5 metres and width x + 2 metres.
Solution:
Area = Length × Width
Area = (x + 5)(x + 2)
Area = x² + 2x + 5x + 10
Area = x² + 7x + 10
When x = 3:
Area = 3² + 7(3) + 10
Area = 9 + 21 + 10
Area = 40 square metres
Polynomials Class 10 MCQs assess quick understanding of degree, zeroes, coefficients, graphs and polynomial operations.
Which expression is a polynomial?
Answer: C
The degree of 5x³ − 2x + 4 is:
Answer: C
The zero of 2x − 8 is:
Answer: C
The sum of the zeroes of x² − 6x + 8 is:
Answer: B
The product of the zeroes of 2x² + 3x − 5 is:
Answer: A
A quadratic graph that does not meet the x-axis has:
Answer: C
A quadratic polynomial whose zeroes are 2 and 3 is:
Answer: B
In p(x) = g(x)q(x) + r(x), r(x) represents:
Answer: D
Assertion: The graph of a quadratic polynomial can intersect the x-axis at most twice.
Reason: A quadratic polynomial has degree 2.
Answer: Both the assertion and reason are true, and the reason correctly explains the assertion.
[INTERACTIVE QUIZ PLACEHOLDER: 10-question no-login diagnostic quiz]
The quiz result should show:
Previous-year questions show how polynomial concepts are tested in actual examinations and help students identify recurring question formats.
Do not publish unsupported frequency claims. Review official question papers and sample papers before stating which subtopic appears most often.
Find a quadratic polynomial whose sum and product of zeroes are 4 and −5 respectively.
Solution:
p(x) = x² − (sum of zeroes)x + product of zeroes
p(x) = x² − 4x − 5
Therefore, the required polynomial is:
x² − 4x − 5
[INTERNAL LINK: Class 10 Polynomials previous-year questions → /polynomials-pyq/]
The most common errors involve invalid exponents, formula signs, graph interpretation and incomplete polynomial division.
| Mistake | Incorrect approach | Correct approach |
| Calling 1/x + 2 a polynomial | Treating 1/x as a normal term | Rewrite it as x⁻¹ and identify the negative exponent |
| Ignoring the leading coefficient | Using α + β = −b | Use α + β = −b/a |
| Reversing formula signs | Writing α + β = b/a | Use α + β = −b/a |
| Misreading a graph | Counting turning points | Count x-axis intersections |
| Confusing a zero with a coefficient | Calling b or c a zero | Substitute values and check p(x) = 0 |
| Incomplete division | Stopping while remainder degree is too high | Continue until degree of remainder is lower |
| Missing terms in division | Writing x³ + 2 without placeholders | Write x³ + 0x² + 0x + 2 |
| Failing to verify | Accepting the result immediately | Check p(x) = g(x)q(x) + r(x) |
Before submitting an answer:
Students can revise Polynomials efficiently by reviewing definitions and formulas first, followed by graphs, solved examples and targeted practice.
Morning:
Afternoon:
Evening:
A Polynomials mind map should connect definitions, types, zeroes, graphs, coefficients, polynomial formation and division on one page.
The mind map should include:
The NCERT exercise guide should map each current textbook exercise to the concept and solving skill required.
Verify the current textbook edition before adding exercise numbers because rationalisation or textbook revisions may change the structure.
Also Check: Important Questions for Class 10 Polynomials
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You can download the complete notes from the PDF button near the top of this page. The file should include formulas, solved examples, graphs, MCQs and important questions without requiring registration.
A polynomial is an algebraic expression in which each variable has a non-negative integer exponent. For example, 3x² − 5x + 7 is a polynomial, but 1/x + 2 is not.
The main topics are types and degrees of polynomials, zeroes, graphical meaning of zeroes, relationships between zeroes and coefficients, formation of quadratic polynomials and polynomial division.
Set the polynomial equal to zero and solve for x. For quadratic polynomials, factorisation is commonly used when the expression can be factorised easily.
The zeroes are the x-coordinates where the polynomial’s graph meets or touches the x-axis. A quadratic graph can have two, one or no distinct real zeroes.
A quadratic polynomial can have at most two zeroes. Its graph may cross the x-axis twice, touch it once or not meet it at all.
For ax² + bx + c with zeroes α and β, α + β = −b/a and αβ = c/a. These formulas allow students to find or verify the sum and product without fully solving the polynomial.
If the zeroes are α and β, the monic quadratic polynomial is x² − (α + β)x + αβ. Any non-zero multiple of this polynomial has the same zeroes.
The most important formulas are α + β = −b/a, αβ = c/a, x² − (α + β)x + αβ and p(x) = g(x)q(x) + r(x).
Students should practise finding zeroes, interpreting graphs, verifying zeroes and coefficients, forming quadratic polynomials, finding unknown parameters and applying the division algorithm. Any claim about the most frequently tested question must be supported by verified paper analysis.
Notes are useful for understanding and revision, but students should also solve NCERT exercises, official sample papers and previous-year questions.
Review the definition and formulas, study the three graph cases, solve one question from each major topic and finish with an error review. The ten-minute and thirty-minute plans on this page provide a structured sequence.