📘 SCERT Assam Class 10 Maths Chapter 2 Polynomials Exercise 2.2 Solutions (2026–2027) English Medium
✅ SEBA Class 10 Maths Exercise 2.2 Solutions – Digital Pipal Academy
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📚 Chapter 2: Polynomials – Quick Revision
Before solving Exercise 2.2, remember:
A polynomial is an algebraic expression
Zeros of polynomial = values of x where p(x) = 0
✍️ Exercise 2.2 Solutions (Full Explanation)
Question 1: Find the zeros of the following quadratic polynomials and verify the relationship between zeros and coefficients.
- (i) \(x^2 - 2x - 8\)
- (ii) \(4s^2 - 4s + 1\)
- (iii) \(6x^2 - 7x - 3\)
- (iv) \(4u^2 + 8u\)
- (v) \(t^2 - 15\)
- (vi) \(3x^2 - x - 4\)
- (vii) \(x^2 + 7x + 12\)
- (viii) \(x^2 - 4x + 3\)
- (ix) \(x^2 - 6x - 7\)
- (x) \(2x^2 - 5x - 7\)
\[ x^2 - 2x - 8 = x^2 - (4 - 2)x - 8 \] \[ = x^2 - 4x + 2x - 8 \] \[ = x(x - 4) + 2(x - 4) \] \[ = (x - 4)(x + 2) \]
The value of the polynomial becomes zero when: \[ x - 4 = 0 \quad \text{and} \quad x + 2 = 0 \]
\[ \therefore x = 4 \quad \text{and} \quad x = -2 \]
Verification:
Sum of zeroes: \[ = 4 + (-2) = 2 = -\frac{-2}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 4 \times (-2) = -8 = \frac{-8}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 4s^2 - 4s + 1 = 4s^2 - (2 + 2)s + 1 \] \[ = 4s^2 - 2s - 2s + 1 \] \[ = 2s(2s - 1) - 1(2s - 1) \] \[ = (2s - 1)(2s - 1) \]
The value of the polynomial becomes zero when: \[ 2s - 1 = 0 \]
\[ \therefore s = \frac{1}{2}, \frac{1}{2} \]
Verification:
Sum of zeroes: \[ = \frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4} = -\frac{\text{Coefficient of } s}{\text{Coefficient of } s^2} \]
Product of zeroes: \[ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^2} \]
\[ 6x^2 - 3 - 7x = 6x^2 - 7x - 3 \] \[ = 6x^2 - (9 - 2)x - 3 \] \[ = 6x^2 - 9x + 2x - 3 \] \[ = 3x(2x - 3) + 1(2x - 3) \] \[ = (2x - 3)(3x + 1) \]
The value of the polynomial becomes zero when: \[ 2x - 3 = 0 \quad \text{and} \quad 3x + 1 = 0 \]
\[ \therefore x = \frac{3}{2} \quad \text{and} \quad x = -\frac{1}{3} \]
Verification:
Sum of zeroes: \[ = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6} = -\frac{-7}{6} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{1}{2} = \frac{-3}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 4u^2 + 8u = 4u^2 + 8u + 0 \] \[ = 4u(u + 2) \]
The value of the polynomial becomes zero when: \[ 4u = 0 \quad \text{and} \quad u + 2 = 0 \]
\[ \therefore u = 0 \quad \text{and} \quad u = -2 \]
Verification:
Sum of zeroes: \[ = 0 + (-2) = -2 = -\frac{8}{4} = -\frac{\text{Coefficient of } u}{\text{Coefficient of } u^2} \]
Product of zeroes: \[ = 0 \times (-2) = 0 = \frac{0}{4} = \frac{\text{Constant term}}{\text{Coefficient of } u^2} \]
\[ t^2 - 15 = t^2 - (\sqrt{15})^2 \] \[ = (t - \sqrt{15})(t + \sqrt{15}) \]
The value of the polynomial becomes zero when: \[ t - \sqrt{15} = 0 \quad \text{and} \quad t + \sqrt{15} = 0 \]
\[ \therefore t = \sqrt{15} \quad \text{and} \quad t = -\sqrt{15} \]
Verification:
Sum of zeroes: \[ = \sqrt{15} + (-\sqrt{15}) = 0 = -\frac{0}{1} = -\frac{\text{Coefficient of } t}{\text{Coefficient of } t^2} \]
Product of zeroes: \[ = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{\text{Constant term}}{\text{Coefficient of } t^2} \]
\[ 3x^2 - x - 4 = 3x^2 - (4 - 3)x - 4 \] \[ = 3x^2 - 4x + 3x - 4 \] \[ = x(3x - 4) + 1(3x - 4) \] \[ = (3x - 4)(x + 1) \]
The value of the polynomial becomes zero when: \[ 3x - 4 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = \frac{4}{3} \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = \frac{4}{3} + (-1) = \frac{4 - 3}{3} = \frac{1}{3} = -\frac{-1}{3} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{-4}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 + 7x + 12 = x^2 + (3 + 4)x + 12 \] \[ = x^2 + 3x + 4x + 12 \] \[ = x(x + 3) + 4(x + 3) \] \[ = (x + 3)(x + 4) \]
The value of the polynomial becomes zero when: \[ x + 3 = 0 \quad \text{and} \quad x + 4 = 0 \]
\[ \therefore x = -3 \quad \text{and} \quad x = -4 \]
Verification:
Sum of zeroes: \[ = (-3) + (-4) = -7 = -\frac{7}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = (-3) \times (-4) = 12 = \frac{12}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 - 4x + 3 = x^2 - (1 + 3)x + 3 \] \[ = x^2 - x - 3x + 3 \] \[ = x(x - 1) - 3(x - 1) \] \[ = (x - 1)(x - 3) \]
The value of the polynomial becomes zero when: \[ x - 1 = 0 \quad \text{and} \quad x - 3 = 0 \]
\[ \therefore x = 1 \quad \text{and} \quad x = 3 \]
Verification:
Sum of zeroes: \[ = 1 + 3 = 4 = -\frac{-4}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 1 \times 3 = 3 = \frac{3}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 - 6x - 7 = x^2 - (7 - 1)x - 7 \] \[ = x^2 - 7x + x - 7 \] \[ = x(x - 7) + 1(x - 7) \] \[ = (x - 7)(x + 1) \]
The value of the polynomial becomes zero when: \[ x - 7 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = 7 \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = 7 + (-1) = 6 = -\frac{-6}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 7 \times (-1) = -7 = \frac{-7}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 2x^2 - 5x - 7 = 2x^2 - (7 - 2)x - 7 \] \[ = 2x^2 - 7x + 2x - 7 \] \[ = x(2x - 7) + 1(2x - 7) \] \[ = (2x - 7)(x + 1) \]
The value of the polynomial becomes zero when: \[ 2x - 7 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = \frac{7}{2} \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = \frac{7}{2} + (-1) = \frac{7 - 2}{2} = \frac{5}{2} = -\frac{-5}{2} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{7}{2} \times (-1) = -\frac{7}{2} = \frac{-7}{2} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
🎯 Important Exam Questions (Most Repeated)
Find zeros of quadratic polynomial
Verify relationship between zeros and coefficients
Form polynomial from given sum and product
👉 These questions are very important for SEBA board exams
🧠 Pro Tips to Score 90+ in Maths
Practice factorization daily
Always show verification steps
Learn formulas properly
Solve previous year questions
📥 PDF Download (Updated 2026)
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❓ FAQs
Q1. What is Exercise 2.2 in Class 10 Maths Assam?
Exercise 2.2 focuses on finding zeros of polynomials and verifying relationships.
Q2. Is this based on latest SCERT Assam syllabus?
Yes, it is updated for 2026–2027 SEBA/ASSEB syllabus.
Q3. Are these solutions enough for board exam?
Yes, these are exam-oriented and fully sufficient for preparation.
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