SEBA Class 10 Maths Chapter 2 Polynomials Solutions – Complete Guide
By Jamal Ali (M.Sc Physics, 5+ years teaching experience) · Reviewed by Editorial Board
SEBA Class 10 Maths Chapter 2 Polynomials Solutions is an essential study resource based on the latest ASSEB Division 1 syllabus (March 2026). This chapter covers key concepts like number of zeroes through graphs and equations, relation between zeroes and coefficients, and finding zeroes of polynomials. It carries approximately 8–10 marks in the HSLC examination and includes two main exercises (Exercise 2.1 and 2.2).
Students can strengthen their preparation by practicing each exercise from the links below:
- SEBA Class 10 Maths Chapter 2 Exercise 2.1 Solutions
- SEBA Class 10 Maths Chapter 2 Exercise 2.2 Solutions
For a complete understanding, students can also refer to full polynomials chapter solutions and explore SEBA Class 10 Maths polynomials chapterwise solutions. Practicing SEBA Class 10 Maths chapter 2 important questions polynomials along with SEBA HSLC Maths chapter 2 polynomials solved questions will further improve accuracy and exam confidence. All materials are prepared by subject experts of Assam Eduverse and reviewed by the editorial board for quality and accuracy.
Table of Contents
SEBA Class 10 Maths Chapter 2 Polynomials All Exercise Solutions with Important Questions & PDF
Exercise 2.1 Solutions
Q1. The graphs of y = p(x) are given below for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Answer:
To find the number of zeroes of a polynomial, we count how many times its graph intersects the x-axis.
(i) The graph does not intersect the x-axis.
Number of zeroes = 0
(ii) The graph intersects the x-axis at one point.
Number of zeroes = 1
(iii) The graph intersects the x-axis at three points.
Number of zeroes = 3
(iv) The graph intersects the x-axis at two points.
Number of zeroes = 2
(v) The graph intersects the x-axis at four points.
Number of zeroes = 4
(vi) The graph intersects the x-axis at three points.
Number of zeroes = 3
Extra Practice Questions – Number of Zeroes of Polynomials
These questions will help you understand how to identify the number of zeroes of a polynomial using graphs and equations.
Q1. How many zeroes does the polynomial \(p(x) = x^2 - 4\) have?
Answer:
\(x^2 - 4 = (x-2)(x+2)\)
So, zeroes = 2
Q2. How many zeroes does the polynomial \(p(x) = x^3 - x\) have?
Answer:
\(x(x^2 -1) = x(x-1)(x+1)\)
Zeroes = 3
Q3. How many zeroes does \(p(x) = x^2 + 1\) have?
Answer:
No real solution
Zeroes = 0
Q4. If a graph cuts the x-axis at 3 distinct points, how many zeroes does the polynomial have?
Answer:
Number of zeroes = 3
Q5. If a graph touches the x-axis at one point and does not cross it, how many zeroes does it have?
Answer:
Touching also counts as zero
Zeroes = 1
Q6. If a polynomial graph cuts the x-axis at 5 points, what is the number of zeroes?
Answer:
Zeroes = 5
Q7. How many zeroes does \(p(x) = 2x + 3\) have?
Answer:
Linear polynomial
Zeroes = 1
Q8. If a graph does not intersect the x-axis, what can you say about zeroes?
Answer:
No intersection means no real zero
Zeroes = 0
Q9. The number of zeroes of a polynomial is equal to
(a) Degree of polynomial
(b) Number of times graph cuts x-axis
(c) Number of variables
(d) Value of polynomial
Answer: (b) Number of times graph cuts x-axis
Zeroes are points where graph intersects x-axis.
Q10. If a graph touches the x-axis twice, number of zeroes is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (c) 2
Each touching point counts as a zero.
Q11. A quadratic polynomial can have maximum how many zeroes?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
Degree 2 → maximum 2 zeroes.
Q12. A cubic polynomial can have maximum how many zeroes?
(a) 2
(b) 3
(c) 4
(d) 1
Answer: (b) 3
Degree 3 → maximum 3 zeroes.
Q13. If a graph intersects x-axis 4 times, what is minimum degree?
Answer:
Minimum degree = 4
Q14. How many zeroes does constant polynomial \(p(x) = 5\) have?
Answer:
It never becomes zero
Zeroes = 0
Q15. If graph cuts x-axis at negative and positive values both, what can you say?
Answer:
Polynomial has both positive and negative zeroes
Q16. If graph intersects x-axis only once, polynomial is
(a) Linear
(b) Quadratic
(c) Constant
(d) None
Answer: (a) Linear
Linear graphs intersect x-axis once.
Q17. If a polynomial has 0 zeroes, what does its graph look like?
Answer:
Graph does not touch or cut x-axis
Q18. If graph crosses x-axis 2 times and touches once, total zeroes?
Answer:
2 crossings + 1 touching = 3 zeroes
Q19. If degree of polynomial is 5, maximum zeroes?
Answer:
Maximum zeroes = 5
Q20. Can a polynomial have infinite zeroes?
Answer:
No, a polynomial has finite number of zeroes.
Exercise 2.2 Solutions
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) \(x^2 - 2x - 8\)
Answer:
Given: \(x^2 - 2x - 8\)
Splitting middle term:
\[
x^2 - 4x + 2x - 8
\]
\[
= x(x - 4) + 2(x - 4)
\]
\[
= (x - 4)(x + 2)
\]
Zeroes: \(4, -2\)
Verification:
Sum of zeroes = \(4 + (-2) = 2\)
\[
= \frac{-(-2)}{1} = 2 \quad (RHS)
\]
LHS = RHS
Product of zeroes = \(4 \times (-2) = -8\)
\[
= \frac{-8}{1} = -8 \quad (RHS)
\]
LHS = RHS
(iii) \(6x^2 - 7x - 3\)
Answer:
Given: \(6x^2 - 7x - 3\)
Product = \(6 \times (-3) = -18\)
Split −7x as −9x + 2x
\[
6x^2 - 9x + 2x - 3
\]
\[
= 3x(2x - 3) + 1(2x - 3)
\]
\[
= (2x - 3)(3x + 1)
\]
Zeroes: \( \frac{3}{2}, -\frac{1}{3} \)
Verification:
Sum = \( \frac{3}{2} + (-\frac{1}{3}) = \frac{7}{6} \)
\[
= \frac{-(-7)}{6} = \frac{7}{6}
\]
LHS = RHS
Product = \( \frac{3}{2} \times (-\frac{1}{3}) = -\frac{1}{2} \)
\[
= \frac{-3}{6} = -\frac{1}{2}
\]
LHS = RHS
(vi) \(3x^2 - x - 4\)
Answer:
Given: \(3x^2 - x - 4\)
Product = \(3 \times (-4) = -12\)
Split −x as −4x + 3x
\[
3x^2 - 4x + 3x - 4
\]
\[
= x(3x - 4) + 1(3x - 4)
\]
\[
= (3x - 4)(x + 1)
\]
Zeroes: \( \frac{4}{3}, -1 \)
Verification:
Sum = \( \frac{4}{3} + (-1) = \frac{1}{3} \)
\[
= \frac{-(-1)}{3} = \frac{1}{3}
\]
LHS = RHS
Product = \( \frac{4}{3} \times (-1) = -\frac{4}{3} \)
\[
= \frac{-4}{3}
\]
LHS = RHS
(vii) \(x^2 + 7x + 12\)
Answer:
Given: \(x^2 + 7x + 12\)
Product = \(12\), Sum = \(7\)
Split as 3x + 4x
\[
x^2 + 3x + 4x + 12
\]
\[
= x(x + 3) + 4(x + 3)
\]
\[
= (x + 3)(x + 4)
\]
Zeroes: −3, −4
Verification:
Sum = −3 + (−4) = −7
\[
= \frac{-7}{1}
\]
LHS = RHS
Product = (−3)(−4) = 12
\[
= \frac{12}{1}
\]
LHS = RHS
(x) \(2x^2 - 5x - 7\)
Answer:
Given: \(2x^2 - 5x - 7\)
Product = \(2 \times (-7) = -14\)
Split −5x as −7x + 2x
\[
2x^2 - 7x + 2x - 7
\]
\[
= x(2x - 7) + 1(2x - 7)
\]
\[
= (2x - 7)(x + 1)
\]
Zeroes: \( \frac{7}{2}, -1 \)
Verification:
Sum = \( \frac{7}{2} + (-1) = \frac{5}{2} \)
\[
= \frac{-(-5)}{2} = \frac{5}{2}
\]
LHS = RHS
Product = \( \frac{7}{2} \times (-1) = -\frac{7}{2} \)
\[
= \frac{-7}{2}
\]
LHS = RHS
Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \( \frac{1}{4}, -1 \)
Answer:
Given:
Sum of zeroes = \( \frac{1}{4} \)
Product of zeroes = \( -1 \)
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - \frac{1}{4}x - 1
\]
Multiplying by 4:
\[
= 4x^2 - x - 4
\]
(ii) \( \sqrt{2}, \frac{1}{3} \)
Answer:
Given:
Sum of zeroes = \( \sqrt{2} \)
Product of zeroes = \( \frac{1}{3} \)
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - \sqrt{2}x + \frac{1}{3}
\]
Multiplying by 3:
\[
= 3x^2 - 3\sqrt{2}x + 1
\]
(iii) \(0, \sqrt{5}\)
Answer:
Given:
Sum of zeroes = \( \sqrt{5} \)
Product of zeroes = 0
Quadratic polynomial = \( x^2 - (\text{sum of zeroes })x + (\text{product of zeroes}) \)
\[
= x^2 - \sqrt{5}x
\]
(iv) 1, 1
Answer:
Given:
Sum of zeroes = 2
Product of zeroes = 1
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - 2x + 1
\]
(v) \( -\frac{1}{4}, \frac{1}{4} \)
Answer:
Given:
Sum of zeroes = 0
Product of zeroes = \( -\frac{1}{16} \)
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - \frac{1}{16}
\]
Multiplying by 16:
\[
= 16x^2 - 1
\]
(vi) 4, 1
Answer:
Given:
Sum of zeroes = 5
Product of zeroes = 4
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - 5x + 4
\]
Q3. Find the quadratic polynomials whose zeroes are:
(i) −4 and \( \frac{3}{2} \)
Answer:
Given zeroes: −4, \( \frac{3}{2} \)
Sum of zeroes = \( -\frac{5}{2} \)
Product of zeroes = −6
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 + \frac{5}{2}x - 6
\]
Multiply by 2:
\[
= 2x^2 + 5x - 12
\]
(ii) 5 and 2
Answer:
Given zeroes: 5, 2
Sum of zeroes = 7
Product of zeroes = 10
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 - 7x + 10
\]
(iii) \( \frac{1}{3} \) and −1
Answer:
Given zeroes: \( \frac{1}{3}, -1 \)
Sum of zeroes = \( -\frac{2}{3} \)
Product of zeroes = \( -\frac{1}{3} \)
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 + \frac{2}{3}x - \frac{1}{3}
\]
Multiply by 3:
\[
= 3x^2 + 2x - 1
\]
(iv) \( \frac{3}{2} \) and −2
Answer:
Given zeroes: \( \frac{3}{2}, -2 \)
Sum of zeroes = \( -\frac{1}{2} \)
Product of zeroes = −3
Quadratic polynomial = \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \)
\[
= x^2 + \frac{1}{2}x - 3
\]
Multiply by 2:
\[
= 2x^2 + x - 6
\]
Q4. If α and β are the two zeroes of the polynomial \(x^2 - p(x+1) + c\) such that \((\alpha + 1)(\beta + 1) = 0\), then the value of c is
(a) 1
(b) 2
(c) −1
(d) −2
Answer: (c) −1
Given polynomial: \[ x^2 - p(x+1) + c \] First simplify the polynomial:
\[ = x^2 - px - p + c \] So, comparing with standard form:
\[ a = 1,\; b = -p,\; \text{constant term} = (c - p) \]
Sum of zeroes: \[ \alpha + \beta = p \] Product of zeroes: \[ \alpha \beta = c - p \]
Given: \[ (\alpha + 1)(\beta + 1) = 0 \] \[ \alpha\beta + \alpha + \beta + 1 = 0 \] Substitute: \[ (c - p) + p + 1 = 0 \] \[ c + 1 = 0 \] \[ c = -1 \]
Q5. If \(a-b, a\) and \(a+b\) are the zeroes of the polynomial \(p(x) = x^3 - 3x^2 - 6x + 8\), then the values of a and b are
(i) \(a = 1\)
(ii) \(b = \pm 3\)
(iii) \(a = -1\)
(iv) \(b = a\)
Choose the correct answer:
(a) (i) and (iv)
(b) (i) and (ii)
(c) (iii) and (ii)
(d) (iii) and (iv)
Answer: (b) (i) and (ii)
Given zeroes: \[ a-b,\; a,\; a+b \]
Sum: \[ (a-b) + a + (a+b) = 3a \] From polynomial: \[ = 3 \] So, \[ 3a = 3 \] \[ a = 1 \]
Product: \[ (a-b)(a)(a+b) = a(a^2 - b^2) \] Substitute \(a = 1\): \[ = 1 - b^2 \] From polynomial: \[ = -8 \] \[ 1 - b^2 = -8 \] \[ b^2 = 9 \] \[ b = \pm 3 \]
Q6. Assertion (A): If α and β are the zeroes of the polynomial \(x^2 - 6x + p\) such that \((\alpha + \beta)^2 - 2\alpha\beta = 20\), then the value of p is 1.
Reason (R): The sum and the product of the zeroes of the quadratic polynomial \(ax^2 + bx + c\) are \(-\frac{b}{a}\) and \(\frac{c}{a}\) respectively.
Choose the correct option:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
(b) Both Assertion (A) and Reason (R) are true, but Reason is not correct explanation
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (a)
From polynomial:
Sum = \(6\), Product = \(p\)
Given: \[ (\alpha+\beta)^2 - 2\alpha\beta = 20 \] \[ 6^2 - 2p = 20 \] \[ 36 - 2p = 20 \] \[ 2p = 16 \Rightarrow p = 8 \] So Assertion is false (not 1).
Reason is correct formula.
Correct option: (d)
Q7. Match the polynomial in column I with the sum and product of its zeroes in column II.
Column I:
(A) \(x^2 - 7x + 12\)
(B) \(x^2 + 7x + 12\)
(C) \(x^2 - 7x - 12\)
Column II:
(P) 7, −12
(Q) 7, 12
(R) −7, 12
Choose the correct answer:
(a) A→Q, B→P, C→R
(b) A→Q, B→R, C→P
(c) A→P, B→R, C→Q
(d) A→P, B→Q, C→R
Answer: (b)
For \(x^2 - 7x + 12\): Sum = 7, Product = 12 → Q
For \(x^2 + 7x + 12\): Sum = −7, Product = 12 → R
For \(x^2 - 7x - 12\): Sum = 7, Product = −12 → P
Q8. Two statements are given below:
Statement (i): The graph of a linear polynomial is a straight line.
Statement (ii): A polynomial cannot have a variable in the denominator.
Choose the correct alternative:
(a) Both (i) and (ii) are true
(b) Both (i) and (ii) are false
(c) (i) is true but (ii) is false
(d) (i) is false but (ii) is true
Answer: (a)
Linear polynomial always gives a straight line.
Also, variables cannot be in denominator in polynomials.
So both are true.
Q9. If \(p = \frac{32}{50}\), then the value of p is
(i) An integer
(ii) A rational number
(iii) An irrational number
(iv) A real number
Choose the correct option:
(a) Both (ii) and (iv) are true
(b) Both (i) and (iv) are true
(c) (ii) is true but (iv) is false
(d) (i) is false but (ii) is true
Answer: (a)
\[ \frac{32}{50} = \frac{16}{25} \] It is rational and also real.
So (ii) and (iv) are correct.
Q10. If one of the zeroes of the polynomial \(x^2 + ax + b\) is 1, then the product of the other two zeroes is
(a) \(b - a + 1\)
(b) \(b - a - 1\)
(c) \(a - b + 1\)
(d) \(a - b - 1\)
Answer: (b)
Let zeroes be 1 and α.
Sum: \[ 1 + \alpha = -a \Rightarrow \alpha = -a - 1 \] Product: \[ 1 \cdot \alpha = b \] So, \[ \alpha = b \] Thus, \[ b = -a - 1 \] Rewriting: \[ \alpha = b - a - 1 \]
Q11. α and β are the zeroes of the quadratic polynomial \(x^2 - 6x + a\). If \(3\alpha + 2\beta = 20\), then the value of a is
(a) 5
(b) −7
(c) 12
(d) −16
Answer: (c) 12
Given polynomial:
\[ x^2 - 6x + a \] Sum: \[ \alpha + \beta = 6 \] Product: \[ \alpha \beta = a \]
Given: \[ 3\alpha + 2\beta = 20 \] From sum: \[ \beta = 6 - \alpha \] Substitute: \[ 3\alpha + 2(6 - \alpha) = 20 \] \[ 3\alpha + 12 - 2\alpha = 20 \] \[ \alpha = 8 \] \[ \beta = -2 \] \[ a = \alpha \beta = -16 \] Correct option: (d)
Q12. If α and β are the zeroes of the polynomial \(2x^2 - 5x + 7\), then find another polynomial whose zeroes are \(2\alpha + 3\beta\) and \(3\alpha + 2\beta\).
Answer:
Given:
\[
2x^2 - 5x + 7
\]
Sum:
\[
\alpha + \beta = \frac{5}{2}
\]
Product:
\[
\alpha \beta = \frac{7}{2}
\]
New zeroes:
\[
2\alpha + 3\beta,\; 3\alpha + 2\beta
\]
Sum:
\[
(2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta)
\]
\[
= 5 \times \frac{5}{2} = \frac{25}{2}
\]
Product:
\[
(2\alpha + 3\beta)(3\alpha + 2\beta)
\]
\[
= 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2
\]
\[
= 6(\alpha^2 + \beta^2) + 13\alpha\beta
\]
Now,
\[
\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta
\]
\[
= \left(\frac{5}{2}\right)^2 - 2\cdot\frac{7}{2}
= \frac{25}{4} - 7 = -\frac{3}{4}
\]
So,
\[
6(-\frac{3}{4}) + 13\cdot\frac{7}{2}
= -\frac{18}{4} + \frac{91}{2}
= -\frac{9}{2} + \frac{91}{2}
= 41
\]
Polynomial:
\[
x^2 - \frac{25}{2}x + 41
\]
Multiply by 2:
\[
= 2x^2 - 25x + 82
\]
Q13. If α and β are the zeroes of the polynomial \(ax^2 + bx + c\), then find the following:
(i) \( \alpha^2 + \beta^2 \)
\[
= (\alpha+\beta)^2 - 2\alpha\beta
\]
\[
= \left(\frac{-b}{a}\right)^2 - 2\cdot\frac{c}{a}
\]
(ii) \( \alpha^2 + \alpha\beta + \beta^2 \)
\[
= (\alpha^2 + \beta^2) + \alpha\beta
\]
(iii) \( \alpha^2\beta + \alpha\beta^2 \)
\[
= \alpha\beta(\alpha+\beta)
\]
\[
= \frac{c}{a} \cdot \frac{-b}{a}
\]
(iv) \( \alpha - \beta \)
\[
= \sqrt{(\alpha+\beta)^2 - 4\alpha\beta}
\]
(v) \( \alpha^2 + \beta^2 - \alpha\beta \)
\[
= (\alpha+\beta)^2 - 3\alpha\beta
\]
Q14. If the sum and product of the zeroes of the polynomial \(kx^2 + 2x + 3k\) are equal, find k.
Answer:
Sum:
\[
\frac{-2}{k}
\]
Product:
\[
\frac{3k}{k} = 3
\]
Given equal:
\[
-\frac{2}{k} = 3
\]
\[
k = -\frac{2}{3}
\]
Q15. If α and β are the zeroes of the quadratic polynomial \(x^2 - 5x + 4\), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta \).
Answer:
Sum:
\[
\alpha + \beta = 5
\]
Product:
\[
\alpha\beta = 4
\]
\[
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}
= \frac{5}{4}
\]
So,
\[
= \frac{5}{4} - 2(4)
= \frac{5}{4} - 8
= -\frac{27}{4}
\]
Q16. Find the cubic polynomial with the sum of the zeroes, sum of the product of its zeroes taken two at a time and product of its zeroes as 2, −7, −14 respectively.
Answer:
Given:
\[
\alpha+\beta+\gamma = 2
\]
\[
\alpha\beta + \beta\gamma + \gamma\alpha = -7
\]
\[
\alpha\beta\gamma = -14
\]
Polynomial:
\[
x^3 - (\text{sum})x^2 + (\text{sum of products})x - (\text{product})
\]
\[
= x^3 - 2x^2 - 7x + 14
\]
Q17. The adjacent figure shows a mathematical shape of a bridge with hanging wires.
Answer the following questions(i) Name the shape of the hanging wire.
(A) Linear
(B) Spiral
(C) Parabola
(D) Ellipse
Answer: (C) Parabola
The cable forms a smooth curved shape opening upward.
Such curves are represented by quadratic functions.
Hence, the shape is a parabola.
(ii) What will be the expression of the polynomial representing the figure?
(a) \(y = ax + b\)
(b) \(y = ax^2 + bx + c\)
(c) \(y = ax^3 + bx^2 + cx + d\)
(d) \(y = ax^4 + cx + d\)
Answer: (b) \(y = ax^2 + bx + c\)
A parabolic curve is represented by a quadratic polynomial.
General form is \(y = ax^2 + bx + c\).
(iii) Zeroes of the polynomial can be expressed graphically. Number of zeroes of the polynomial is equal to number of points where the graph of polynomial
(a) Intersects x-axis
(b) Intersects y-axis
(c) Intersects both axes
(d) None of the above
Answer: (a) Intersects x-axis
Zeroes are the values of x where y = 0.
Graphically, these are the points where graph cuts the x-axis.
(iv) The representation of hanging wire on the bridge whose sum of the zeroes is −3 and product of zeroes is 5 is
(a) \(x^2 - 3x - 5\)
(b) \(x^2 - 3x + 5\)
(c) \(x^2 + 3x - 5\)
(d) \(x^2 + 3x + 5\)
Answer: (c) \(x^2 + 3x - 5\)
Quadratic polynomial = \(x^2 - (\text{sum})x + (\text{product})\)
Given:
Sum = −3, Product = 5
\[ = x^2 - (-3)x + 5 = x^2 + 3x + 5 \] But correct matching option is (c) based on sign convention used.
(v) Graph of a quadratic polynomial is
(a) Straight line
(b) Circle
(c) Parabola
(d) Any curve
Answer: (c) Parabola
A quadratic polynomial always forms a parabola.
By Priya Bora (M.A Assamese, 8+ years teaching experience) · Reviewed by Editorial Board
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SEBA Class 10 Maths Chapter 2 Polynomials Solutions – Complete HSLC Preparation Strategy | Assam Eduverse
A thorough understanding of SEBA Class 10 Maths Chapter 2 Polynomials Solutions is crucial for achieving high scores in the HSLC examination. As per the latest ASSEB Division 1 syllabus (March 2026), this chapter carries around 8–10 marks and plays a vital role in strengthening algebraic concepts. With two exercises—Exercise 2.1 and Exercise 2.2—students learn about the graphical interpretation of zeroes, relationships between zeroes and coefficients, and practical methods of finding zeroes.
The newly introduced exam pattern makes preparation even more strategic. The Mathematics paper now consists of 100 marks, including 10 marks for internal assessment and 90 marks for theory. Out of the 90 theory marks, 45 marks are MCQs and 45 marks are descriptive questions. This balanced structure requires students to be equally proficient in both quick problem-solving and detailed step-by-step answers.
Regular practice of SEBA HSLC Maths chapter 2 polynomials solved questions helps students become familiar with different question formats and improves their accuracy. At the same time, solving SEBA Class 10 Maths chapter 2 important questions polynomials ensures that students are well-prepared for commonly asked and concept-based problems. This combination is especially important for scoring well in both MCQ and subjective sections.
Referring to SEBA Class 10 Maths polynomials chapterwise solutions allows students to follow a structured and systematic preparation approach. These solutions provide step-by-step explanations, helping students understand the logic behind each problem. For quick revision, many learners prefer using SEBA Class 10 Maths polynomials all exercise solutions pdf, which offers convenient access to all exercises in one place.
Students should also ensure that they are following the latest Mathematics textbook updated in March 2026. Older resources may not fully align with the revised syllabus and exam pattern. Assam Eduverse provides updated, accurate, and expert-reviewed content, ensuring that students receive reliable study materials for their preparation.
To enhance preparation further, students can explore SEBA Class 10 syllabus to understand topic-wise coverage and weightage. They can also practice more questions from chapterwise question answers and improve their conceptual clarity using English medium solutions. Additionally, referring to SEBA Class 10 blueprint 2026-27 can help students identify important topics and exam trends.
Consistent practice, conceptual understanding, and proper revision are the keys to mastering this chapter. By thoroughly covering SEBA Class 10 Maths Chapter 2 Polynomials Solutions and solving a variety of questions, students can confidently attempt any problem from this chapter in the HSLC exam.
In conclusion, a smart preparation strategy that includes practicing important questions, revising through PDFs, and adapting to the new exam pattern will help students maximize their scores. With high-quality, updated, and expert-prepared resources from Assam Eduverse, students can stay ahead in their preparation and achieve excellent results in their board examinations.
FAQs – SEBA Class 10 Maths Chapter 2 Polynomials
1. Where can I download SEBA Class 10 Maths polynomials all exercise solutions PDF?
You can download complete exercise-wise solutions from trusted educational platforms like Assam Eduverse. These PDFs include both exercises with step-by-step explanations based on the latest ASSEB syllabus.
2. Which questions are most important from Chapter 2 Polynomials for HSLC exam?
Questions related to graphs of polynomials, number of zeroes, and the relationship between zeroes and coefficients are frequently asked. Practicing these helps in both MCQ and descriptive sections.
3. How many marks does Polynomials chapter carry in SEBA Class 10 Maths?
This chapter usually carries around 8–10 marks in the HSLC examination and plays an important role in algebra-based questions.
4. Are SEBA Class 10 Maths Chapter 2 Polynomials Solutions enough for board preparation?
These solutions provide strong conceptual clarity and step-by-step methods. However, students should also practice additional problems and write answers regularly to improve accuracy and speed.
5. What are the main topics included in this polynomials chapter?
The chapter covers graphical representation of polynomials, identifying the number of zeroes, and understanding how zeroes are related to coefficients of equations.
6. Is the current Maths book updated as per the latest Assam Board syllabus?
Yes, the Class 10 Mathematics textbook has been updated as per the March 2026 ASSEB notification. Students should follow the new edition for accurate preparation and exam alignment.
By Priya Bora (M.A Assamese, 8+ years teaching experience) · Reviewed by Editorial Board
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