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.
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…
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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.
- + 5 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 |
Showing 3 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 ...
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 ...
Showing 3 of 8 questions. Visit the full page for complete solutions.