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…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 9 Mathematics — Polynomials. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Polynomials — All 8 Questions Solved
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
Q7. Using the identity (a + b)² = a² + 2ab + b², evaluate 102².
Find: 102²
Write 102 = 100 + 2
Use identity: (a + b)² = a² + 2ab + b² where a = 100, b = 2
Step 1: Calculate a²
a² = (100)² = 10,000
Step 2: Calculate 2ab
2ab = 2(100)(2) = 400
Step 3: Calculate b²
b² = 2² = 4
Step 4: Apply identity
102² = (100 + 2)² = 100² + 2(100)(2) + 2²
= 10,000 + 400 + 4
= 10,404
Answer: 102² = 10,404
Q8. If p(x) = x⁴ + 2x³ - 5x² - 6x + 8, verify that (x + 1) is a factor and find the quotient.
Polynomial: p(x) = x⁴ + 2x³ - 5x² - 6x + 8
Step 1: Check if (x + 1) is a factor using Factor Theorem
For (x + 1), we evaluate p(-1):
p(-1) = (-1)⁴ + 2(-1)³ - 5(-1)² - 6(-1) + 8
= 1 - 2 - 5 + 6 + 8
= 8
Wait, let me recalculate:
p(-1) = 1 - 2 - 5 + 6 + 8 = 8 ≠ 0
Actually, let me verify the polynomial. Using synthetic division with x = -1:
Coefficients: 1, 2, -5, -6, 8
-1 | 1 2 -5 -6 8
| -1 -1 6 0
| 1 1 -6 0 8
Remainder = 8, so (x + 1) is NOT a factor of this polynomial.
For demonstration purposes, if the polynomial were p(x) = x⁴ + 2x³ - 5x² - 6x (remainder 0 case):
Then quotient would be: x³ + x² - 6x
Polynomials — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following expressions is a polynomial?
- A. A) √x + 3
- B. B) x - 1/x
- C. C) x^2 + 5x - 6
- D. D) x^(3/2) + 2
Answer: C. C) x^2 + 5x - 6 — A polynomial must have non-negative integer exponents for its variables. Options A, B, and D contain terms with non-integer or negative exponents, violating this condition.
MCQ 2. What is the degree of the polynomial 5x^3 - 7x^2 + 2x - 1/2?
- A. A) 1
- B. B) 2
- C. C) 3
- D. D) 4
Answer: C. C) 3 — The degree of a polynomial is the highest exponent of the variable in the polynomial. In this expression, the highest exponent of x is 3.
MCQ 3. The expression 4x^2y is an example of a:
- A. A) Monomial
- B. B) Binomial
- C. C) Trinomial
- D. D) Quadrinomial
Answer: A. A) Monomial — A monomial is a polynomial that contains only one term. The given expression 4x^2y consists of a single term.
MCQ 4. What is the coefficient of x in the polynomial 7x^3 - 2x^2 + x - 9?
- A. A) 7
- B. B) -2
- C. C) 1
- D. D) -9
Answer: C. C) 1 — The coefficient of a term is the numerical factor multiplying the variable(s) in that term. For the term 'x', the coefficient is 1 (as x can be written as 1x).
MCQ 5. What is the degree of a non-zero constant polynomial?
- A. A) 0
- B. B) 1
- C. C) Undefined
- D. D) Any integer
Answer: A. A) 0 — A non-zero constant polynomial, such as 5, can be expressed as 5x^0. Therefore, its degree is 0.
MCQ 6. What is the degree of the zero polynomial?
- A. A) 0
- B. B) 1
- C. C) Undefined
- D. D) Any non-negative integer
Answer: C. C) Undefined — The zero polynomial (0) is considered to have an undefined degree because it can be written as 0x^n for any non-negative integer n, making its highest power ambiguous.
MCQ 7. If P(x) = x^100 - x^99 + x^98 - ... + x^2 - x + 1, find P(-1).
- A. A) 0
- B. B) 1
- C. C) 100
- D. D) 101
Answer: D. D) 101 — Substitute x = -1 into the polynomial. Even powers of -1 are 1, and odd powers are -1. The expression becomes 1 - (-1) + 1 - ... + 1 - (-1) + 1 = 1 + 1 + 1 + ... + 1 + 1 + 1. There are 101 terms in total (from x^100 to x^1 plus the constant 1), each evaluating to 1, so P(-1) = 101.
MCQ 8. If P(x, y) = 3x^2y - 5xy + 2y^2, find P(1, 2).
- A. A) 0
- B. B) 4
- C. C) 8
- D. D) 12
Answer: B. B) 4 — Substitute x=1 and y=2 into the polynomial: P(1, 2) = 3(1)^2(2) - 5(1)(2) + 2(2)^2 = 3(1)(2) - 10 + 2(4) = 6 - 10 + 8 = 4.
MCQ 9. Is -2 a zero of the polynomial P(x) = x + 2?
- A. A) Yes
- B. B) No
- C. C) Cannot be determined
- D. D) Only if x is positive
Answer: A. A) Yes — A value 'a' is a zero of P(x) if P(a) = 0. Here, P(-2) = -2 + 2 = 0, so -2 is a zero.
MCQ 10. Find the zero of the polynomial P(x) = 3x - 6.
- A. A) -2
- B. B) 2
- C. C) 3
- D. D) 6
Answer: B. B) 2 — To find the zero of a polynomial, set P(x) = 0. So, 3x - 6 = 0, which gives 3x = 6, and thus x = 2.
MCQ 11. If x = -1 is a zero of the polynomial P(x) = kx^2 + 2x, find the value of k.
- A. A) 1
- B. B) -2
- C. C) 2
- D. D) 0
Answer: C. C) 2 — Since x = -1 is a zero, P(-1) must be 0. Substitute x = -1 into the polynomial: k(-1)^2 + 2(-1) = 0, which simplifies to k - 2 = 0, therefore k = 2.
MCQ 12. When the polynomial x^3 - ax^2 + 6x - a is divided by (x - a), the remainder is:
- A. A) 0
- B. B) a
- C. C) 5a
- D. D) -a
Answer: C. C) 5a — By the Remainder Theorem, when a polynomial P(x) is divided by (x - a), the remainder is P(a). So, P(a) = a^3 - a(a^2) + 6a - a = a^3 - a^3 + 6a - a = 5a.
MCQ 13. If the polynomials ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by (x - 3), find the value of a.
- A. A) 1
- B. B) -1
- C. C) 0
- D. D) 2
Answer: B. B) -1 — Let P(x) = ax^3 + 4x^2 + 3x - 4 and Q(x) = x^3 - 4x + a. By the Remainder Theorem, P(3) = Q(3). P(3) = 27a + 4(9) + 3(3) - 4 = 27a + 36 + 9 - 4 = 27a + 41. Q(3) = (3)^3 - 4(3) + a = 27 - 12 + a = 15 + a. Equating them: 27a + 41 = 15 + a => 26a = 15 - 41 => 26a = -26 => a = -1.
MCQ 14. Is (x - 1) a factor of x^3 - x^2 + x - 1?
- A. A) Yes
- B. B) No
- C. C) Cannot be determined
- D. D) Only if x is even
Answer: A. A) Yes — By the Factor Theorem, (x - 1) is a factor of P(x) if P(1) = 0. Substituting x=1: P(1) = (1)^3 - (1)^2 + (1) - 1 = 1 - 1 + 1 - 1 = 0. Since P(1)=0, (x-1) is a factor.
MCQ 15. For what value of k is (2x + 1) a factor of 4x^3 + 2x^2 - kx - 1?
- A. A) 1
- B. B) 0
- C. C) -1
- D. D) 2
Answer: D. D) 2 — By the Factor Theorem, if (2x + 1) is a factor, then P(-1/2) must be 0. P(-1/2) = 4(-1/2)^3 + 2(-1/2)^2 - k(-1/2) - 1 = 4(-1/8) + 2(1/4) + k/2 - 1 = -1/2 + 1/2 + k/2 - 1 = 0. This simplifies to k/2 - 1 = 0, so k/2 = 1, and k = 2.
MCQ 16. Factorize x^2 - 4y^2.
- A. A) (x - 2y)^2
- B. B) (x + 2y)^2
- C. C) (x - 2y)(x + 2y)
- D. D) (x - 4y)(x + 4y)
Answer: C. C) (x - 2y)(x + 2y) — This expression is in the form of the algebraic identity a^2 - b^2 = (a - b)(a + b). Here, a = x and b = 2y, so x^2 - (2y)^2 = (x - 2y)(x + 2y).
MCQ 17. Expand (x + 3)(x + 5).
- A. A) x^2 + 8x + 15
- B. B) x^2 + 15x + 8
- C. C) x^2 + 8x + 8
- D. D) x^2 + 15
Answer: A. A) x^2 + 8x + 15 — Using the identity (x + a)(x + b) = x^2 + (a + b)x + ab. Here a = 3 and b = 5, so (x + 3)(x + 5) = x^2 + (3 + 5)x + (3)(5) = x^2 + 8x + 15.
MCQ 18. Factorize x^2 + 7x + 12.
- A. A) (x + 3)(x + 4)
- B. B) (x + 2)(x + 6)
- C. C) (x + 1)(x + 12)
- D. D) (x - 3)(x - 4)
Answer: A. A) (x + 3)(x + 4) — To factorize by middle term splitting, find two numbers whose product is 12 and whose sum is 7. These numbers are 3 and 4. So, x^2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4).
MCQ 19. Simplify (2x - 3y)^2.
- A. A) 4x^2 - 9y^2
- B. B) 4x^2 + 9y^2
- C. C) 4x^2 - 12xy + 9y^2
- D. D) 4x^2 + 12xy + 9y^2
Answer: C. C) 4x^2 - 12xy + 9y^2 — Using the identity (a - b)^2 = a^2 - 2ab + b^2. Here a = 2x and b = 3y. So, (2x - 3y)^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = 4x^2 - 12xy + 9y^2.
MCQ 20. Factorize 8x^3 + 27y^3.
- A. A) (2x + 3y)(4x^2 - 6xy + 9y^2)
- B. B) (2x + 3y)(4x^2 + 6xy + 9y^2)
- C. C) (2x - 3y)(4x^2 + 6xy + 9y^2)
- D. D) (2x + 3y)^3
Answer: A. A) (2x + 3y)(4x^2 - 6xy + 9y^2) — This expression is in the form of the identity a^3 + b^3 = (a + b)(a^2 - ab + b^2). Here, a = 2x and b = 3y. So, (2x)^3 + (3y)^3 = (2x + 3y)((2x)^2 - (2x)(3y) + (3y)^2) = (2x + 3y)(4x^2 - 6xy + 9y^2).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.