Polynomials — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Polynomials" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Polynomials" — 8 important questions with detailed answers for CBSE board exam prep…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

If p(x) = 2x³ - 3x² + 4x - 5, find p(2).
Verify that (x - 1) is a factor of x³ - 3x² + 3x - 1 using the Factor Theorem.
Expand (2x + 3)³ using the binomial expansion identity.
Factorise 6x² + 17x + 5 completely.
Show that x² - 2x + 3 has no real zeros.
Divide 2x³ + 3x² - 5x - 6 by (x + 2) using polynomial long division.
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
If p(x) = 2x³ - 3x² + 4x - 5, find p(2).	✓ Solved
Verify that (x - 1) is a factor of x³ - 3x² + 3x - 1 usin…	✓ Solved
Expand (2x + 3)³ using the binomial expansion identity.	✓ Solved
Factorise 6x² + 17x + 5 completely.	✓ Solved
Show that x² - 2x + 3 has no real zeros.	✓ Solved
Divide 2x³ + 3x² - 5x - 6 by (x + 2) using polynomial lon…	✓ Solved

Showing 6 of 8 questions

Q1: If p(x) = 2x³ - 3x² + 4x - 5, find p(2).

Given: p(x) = 2x³ - 3x² + 4x - 5 Find: p(2) Step 1: Substitute x = 2 p(2) = 2(2)³ - 3(2)² + 4(2) - 5 Step 2: Calculate powers = 2(8) - 3(4) + 4(2) - 5 Step 3: Multiply = 16 - 12 + 8 - 5 Step 4: Add and subtract = 16 + 8 - 12 - 5 = 24 - 17 = 7 Answer: p(2) = 7

Q2: Verify that (x - 1) is a factor of x³ - 3x² + 3x - 1 using the Factor Theorem.

Polynomial: f(x) = x³ - 3x² + 3x - 1 To check if (x - 1) is a factor, apply Factor Theorem: (x - a) is a factor ⟺ f(a) = 0 Step 1: Find the zero of (x - 1) x - 1 = 0 ⟹ x = 1 Step 2: Calculate f(1) f(1) = (1)³ - 3(1)² + 3(1) - 1 = 1 - 3 + 3 - 1 = 0 Step 3: Conclusion Since f(1) = 0, by the Factor Theorem, (x - 1) is a factor of f(x). Note: This polynomial is actually f(x) = (x - 1)³

Q3: Expand (2x + 3)³ using the binomial expansion identity.

Expression: (2x + 3)³ Use identity: (a + b)³ = a³ + 3a²b + 3ab² + b³ Where a = 2x and b = 3 Step 1: Identify a³ a³ = (2x)³ = 8x³ Step 2: Calculate 3a²b 3a²b = 3(2x)²(3) = 3(4x²)(3) = 36x² Step 3: Calculate 3ab² 3ab² = 3(2x)(3)² = 3(2x)(9) = 54x Step 4: Calculate b³ b³ = 3³ = 27 Step 5: Combine all terms (2x + 3)³ = 8x³ + 36x² + 54x + 27 Answer: 8x³ + 36x² + 54x + 27

Q4: Factorise 6x² + 17x + 5 completely.

Polynomial: 6x² + 17x + 5 Step 1: Find two numbers whose product = 6 × 5 = 30 and sum = 17 The numbers are 15 and 2 (since 15 × 2 = 30 and 15 + 2 = 17) Step 2: Split the middle term 6x² + 17x + 5 = 6x² + 15x + 2x + 5 Step 3: Factor by grouping = 3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5) Step 4: Verify (3x + 1)(2x + 5) = 6x² + 15x + 2x + 5 = 6x² + 17x + 5 ✓ Answer: (3x + 1)(2x + 5)

Q5: Show that x² - 2x + 3 has no real zeros.

Polynomial: f(x) = x² - 2x + 3 Method 1: Complete the square f(x) = x² - 2x + 3 = (x² - 2x + 1) - 1 + 3 = (x - 1)² + 2 Since (x - 1)² ≥ 0 for all real x, f(x) = (x - 1)² + 2 ≥ 2 > 0 for all real x Therefore, f(x) never equals zero for any real value of x. Method 2: Discriminant For f(x) = x² - 2x + 3, where a = 1, b = -2, c = 3 Discriminant Δ = b² - 4ac = (-2)² - 4(1)(3) = 4 - 12 = -8 < 0 Since Δ < 0, the quadratic has no real roots. Conclusion: x² - 2x + 3 has no real zeros.

Q6: Divide 2x³ + 3x² - 5x - 6 by (x + 2) using polynomial long division.

Dividend: 2x³ + 3x² - 5x - 6 Divisor: x + 2 Step 1: Divide leading terms 2x³ ÷ x = 2x² Multiply: 2x²(x + 2) = 2x³ + 4x² Subtract: (2x³ + 3x² - 5x - 6) - (2x³ + 4x²) = -x² - 5x - 6 Step 2: Divide leading terms of remainder -x² ÷ x = -x Multiply: -x(x + 2) = -x² - 2x Subtract: (-x² - 5x - 6) - (-x² - 2x) = -3x - 6 Step 3: Divide leading terms -3x ÷ x = -3 Multiply: -3(x + 2) = -3x - 6 Subtract: (-3x - 6) - (-3x - 6) = 0 Answer: 2x³ + 3x² - 5x - 6 = (x + 2)(2x² - x - 3) with remainder 0

Showing 6 of 8 questions. Visit the full page for complete solutions.