Polynomials Exemplar — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Polynomials Exemplar" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Polynomials Exemplar" — 6 important questions with detailed answers for CBSE board…
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Key Questions Covered:
- If p(x) = x^3 - 3x^2 + 2x, find p(-1) and p(2).
- Factorize: x^4 - 1.
- Divide x^4 + x^3 + x^2 + x + 1 by (x - 1) and find quotient and remainder.
- If x = 2 is a zero of p(x) = x^3 - 6x^2 + 11x - 6, factorize p(x).
- Verify that x = -2 is a zero of p(x) = x^3 + 6x^2 + 11x + 6.
- Find all zeros of p(x) = x^3 - x^2 - x + 1.
Solutions Summary:
| Question | Status |
|---|---|
| If p(x) = x^3 - 3x^2 + 2x, find p(-1) and p(2). | ✓ Solved |
| Factorize: x^4 - 1. | ✓ Solved |
| Divide x^4 + x^3 + x^2 + x + 1 by (x - 1) and find quotie… | ✓ Solved |
| If x = 2 is a zero of p(x) = x^3 - 6x^2 + 11x - 6, factor… | ✓ Solved |
| Verify that x = -2 is a zero of p(x) = x^3 + 6x^2 + 11x + 6. | ✓ Solved |
| Find all zeros of p(x) = x^3 - x^2 - x + 1. | ✓ Solved |
Showing 6 of 6 questions
Q1: If p(x) = x^3 - 3x^2 + 2x, find p(-1) and p(2).
p(-1) = (-1)^3 - 3(-1)^2 + 2(-1) = -1 - 3 - 2 = -6. p(2) = (2)^3 - 3(2)^2 + 2(2) = 8 - 12 + 4 = 0.
Q2: Factorize: x^4 - 1.
x^4 - 1 = (x^2)^2 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1).
Q3: Divide x^4 + x^3 + x^2 + x + 1 by (x - 1) and find quotient and remainder.
Using synthetic division or long division: (x^4 + x^3 + x^2 + x + 1) ÷ (x - 1). At x = 1: p(1) = 1 + 1 + 1 + 1 + 1 = 5 (remainder). Quotient: x^3 + 2x^2 + 3x + 4 with remainder 5.
Q4: If x = 2 is a zero of p(x) = x^3 - 6x^2 + 11x - 6, factorize p(x).
Since x = 2 is a zero, (x - 2) is a factor. Dividing: p(x) = (x - 2)(x^2 - 4x + 3) = (x - 2)(x - 1)(x - 3).
Q5: Verify that x = -2 is a zero of p(x) = x^3 + 6x^2 + 11x + 6.
p(-2) = (-2)^3 + 6(-2)^2 + 11(-2) + 6 = -8 + 24 - 22 + 6 = 0. Yes, x = -2 is a zero.
Q6: Find all zeros of p(x) = x^3 - x^2 - x + 1.
p(x) = x^2(x - 1) - (x - 1) = (x - 1)(x^2 - 1) = (x - 1)(x - 1)(x + 1) = (x - 1)^2(x + 1). Zeros are x = 1 (double) and x = -1.
Showing 6 of 6 questions. Visit the full page for complete solutions.
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