Polynomials — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Polynomials" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Polynomials" — 9 important questions with detailed answers for CBSE board exam pre…
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NCERT Solutions for Class 10 Mathematics — Polynomials. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Polynomials — All 9 Questions Solved
Q1. State the relationship between zeroes and coefficients of a quadratic polynomial ax² + bx + c.
For a quadratic polynomial ax² + bx + c with zeroes α and β:
Sum of zeroes: α + β = -b/a
Product of zeroes: α × β = c/a
Example: For the polynomial 2x² - 5x + 3, with a = 2, b = -5, c = 3:
If the zeroes are α and β, then:
α + β = -(-5)/2 = 5/2
α × β = 3/2
We can verify: zeroes are 1 and 3/2
1 + 3/2 = 5/2 ✓
1 × 3/2 = 3/2 ✓
Q2. Define a polynomial. What do you mean by the degree of a polynomial?
Polynomial:
A polynomial is an algebraic expression consisting of terms with variables and coefficients, where variables have non-negative integer powers.
General form: p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
where aₙ, aₙ₋₁, ..., a₁, a₀ are constants and n is a non-negative integer.
Degree of a polynomial:
The degree of a polynomial is the highest power of the variable in the polynomial.
Examples:
- p(x) = 3x² + 2x + 5 has degree 2 (quadratic)
- p(x) = x³ - 4x + 1 has degree 3 (cubic)
- p(x) = 7x + 2 has degree 1 (linear)
- p(x) = 5 has degree 0 (constant)
Q3. Divide p(x) = x³ + 3x² + 3x + 1 by g(x) = x + 1 using long division.
Polynomial long division of (x³ + 3x² + 3x + 1) ÷ (x + 1):
Step 1: Divide first term of dividend by first term of divisor
x³ ÷ x = x²
Multiply: x²(x + 1) = x³ + x²
Subtract: (x³ + 3x² + 3x + 1) - (x³ + x²) = 2x² + 3x + 1
Step 2: Repeat with new dividend 2x² + 3x + 1
2x² ÷ x = 2x
Multiply: 2x(x + 1) = 2x² + 2x
Subtract: (2x² + 3x + 1) - (2x² + 2x) = x + 1
Step 3: Repeat with new dividend x + 1
x ÷ x = 1
Multiply: 1(x + 1) = x + 1
Subtract: (x + 1) - (x + 1) = 0
Quotient: x² + 2x + 1
Remainder: 0
Therefore: (x³ + 3x² + 3x + 1) = (x + 1)(x² + 2x + 1)
Note: x² + 2x + 1 = (x + 1)², so the answer is (x + 1)³
Q4. Find the zeroes of the polynomial p(x) = 2x² - 7x + 3.
Using the factorisation method:
We need to find factors of 2 × 3 = 6 that add up to -7.
The factors are -6 and -1 (since -6 + (-1) = -7 and -6 × (-1) = 6)
2x² - 7x + 3 = 2x² - 6x - x + 3
= 2x(x - 3) - 1(x - 3)
= (2x - 1)(x - 3)
Setting each factor to zero:
2x - 1 = 0 → x = 1/2
x - 3 = 0 → x = 3
Verification:
For x = 1/2: 2(1/2)² - 7(1/2) + 3 = 2(1/4) - 7/2 + 3 = 1/2 - 7/2 + 3 = 0 ✓
For x = 3: 2(3)² - 7(3) + 3 = 18 - 21 + 3 = 0 ✓
Therefore, the zeroes are 1/2 and 3
Q5. If α and β are zeroes of x² - 4x + 3, find the polynomial whose zeroes are 2α and 2β.
Step 1: Find the zeroes of x² - 4x + 3
Factorising: x² - 4x + 3 = (x - 1)(x - 3)
Zeroes: α = 1, β = 3
Step 2: Find 2α and 2β
2α = 2(1) = 2
2β = 2(3) = 6
Step 3: Method 1 - Form polynomial with new zeroes
Sum of new zeroes: 2α + 2β = 2 + 6 = 8
Product of new zeroes: (2α)(2β) = 2 × 6 = 12
Polynomial: x² - (sum)x + (product) = x² - 8x + 12
Step 4: Verification
For x = 2: (2)² - 8(2) + 12 = 4 - 16 + 12 = 0 ✓
For x = 6: (6)² - 8(6) + 12 = 36 - 48 + 12 = 0 ✓
Therefore, the required polynomial is x² - 8x + 12
Q6. Use the Remainder Theorem to find the remainder when p(x) = 2x³ - 3x² + 7x - 5 is divided by (x - 2).
Remainder Theorem:
When a polynomial p(x) is divided by (x - a), the remainder is p(a).
Here, p(x) = 2x³ - 3x² + 7x - 5 and we're dividing by (x - 2), so a = 2.
Remainder = p(2)
= 2(2)³ - 3(2)² + 7(2) - 5
= 2(8) - 3(4) + 14 - 5
= 16 - 12 + 14 - 5
= 13
Therefore, the remainder is 13
Q7. Determine if (x - 1) is a factor of p(x) = x³ - 6x² + 11x - 6.
Using the Factor Theorem:
(x - a) is a factor of p(x) if and only if p(a) = 0.
To check if (x - 1) is a factor, calculate p(1):
p(1) = (1)³ - 6(1)² + 11(1) - 6
= 1 - 6 + 11 - 6
= 0
Since p(1) = 0, (x - 1) is a factor of p(x).
Additionally, we can factor completely:
p(x) = x³ - 6x² + 11x - 6
We know (x - 1) is a factor, so we divide:
p(x) = (x - 1)(x² - 5x + 6)
= (x - 1)(x - 2)(x - 3)
Verification: The zeroes are 1, 2, and 3, and p(1) = 0 ✓
Q8. Find a cubic polynomial with zeroes 2, -1, and 3.
If α = 2, β = -1, and γ = 3 are the zeroes of a cubic polynomial, then:
The polynomial can be written as:
p(x) = k(x - α)(x - β)(x - γ)
p(x) = k(x - 2)(x + 1)(x - 3)
where k is any non-zero constant. Taking k = 1:
p(x) = (x - 2)(x + 1)(x - 3)
Expanding step by step:
First, multiply (x - 2)(x + 1):
(x - 2)(x + 1) = x² + x - 2x - 2 = x² - x - 2
Now multiply (x² - x - 2)(x - 3):
= x³ - 3x² - x² + 3x - 2x + 6
= x³ - 4x² + x + 6
Verification:
p(2) = 8 - 16 + 2 + 6 = 0 ✓
p(-1) = -1 - 4 - 1 + 6 = 0 ✓
p(3) = 27 - 36 + 3 + 6 = 0 ✓
Therefore, the polynomial is x³ - 4x² + x + 6
Q9. Simplify (x² - 5x + 6)/(x - 2) by factorisation.
Simplify (x² - 5x + 6)/(x - 2)
Step 1: Factorise the numerator x² - 5x + 6
We need two numbers that multiply to 6 and add to -5.
The factors are -2 and -3 (since (-2) × (-3) = 6 and (-2) + (-3) = -5)
x² - 5x + 6 = (x - 2)(x - 3)
Step 2: Substitute the factorisation into the expression
(x² - 5x + 6)/(x - 2) = ((x - 2)(x - 3))/(x - 2)
Step 3: Cancel the common factor (x - 2)
Note: This is valid for x ≠ 2
= (x - 3)
Therefore, (x² - 5x + 6)/(x - 2) = x - 3 (for x ≠ 2)
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. What is the degree of the polynomial P(x) = 5x^3 - 7x^2 + 2x - √3?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3 — The degree of a polynomial is the highest power of the variable in the polynomial. In this case, the highest power of 'x' is 3.
MCQ 2. Which of the following is a linear polynomial?
- A. x^2 + 2x + 1
- B. 3x - 5
- C. x^3 - 1
- D. x^(1/2) + 3
Answer: B. 3x - 5 — A linear polynomial is a polynomial of degree 1. The polynomial 3x - 5 has the highest power of x as 1.
MCQ 3. How many real zeros does the polynomial P(x) = (x-2)^2 + 4 have?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: A. 0 — P(x) simplifies to x^2 - 4x + 8. The discriminant (b^2 - 4ac) is (-4)^2 - 4(1)(8) = 16 - 32 = -16. Since the discriminant is negative, the quadratic polynomial has no real zeros.
MCQ 4. If α and β are the zeros of the quadratic polynomial x^2 + 5x + 6, then what is the value of α + β?
- A. 5
- B. -5
- C. 6
- D. -6
Answer: B. -5 — For a quadratic polynomial ax^2 + bx + c, the sum of the zeros is given by α + β = -b/a. Here, a=1, b=5, c=6, so α + β = -5/1 = -5.
MCQ 5. Find the product of the zeros of the polynomial P(x) = 2x^2 - 7x + 3.
- A. 7/2
- B. -7/2
- C. 3/2
- D. -3/2
Answer: C. 3/2 — For a quadratic polynomial ax^2 + bx + c, the product of the zeros is given by αβ = c/a. Here, a=2, b=-7, c=3, so αβ = 3/2.
MCQ 6. If x=2 is a zero of the polynomial P(x) = 2x^2 - 3x + k, what is the value of k?
- A. -2
- B. 2
- C. -4
- D. 4
Answer: A. -2 — If x=2 is a zero, then P(2)=0. So, 2(2)^2 - 3(2) + k = 0 which simplifies to 8 - 6 + k = 0, giving 2 + k = 0, hence k = -2.
MCQ 7. A quadratic polynomial whose sum and product of zeros are -3 and 2 respectively, is:
- A. x^2 + 3x + 2
- B. x^2 - 3x + 2
- C. x^2 + 3x - 2
- D. x^2 - 3x - 2
Answer: A. x^2 + 3x + 2 — A quadratic polynomial can be formed using the formula x^2 - (sum of zeros)x + (product of zeros). Given sum = -3 and product = 2, the polynomial is x^2 - (-3)x + 2 = x^2 + 3x + 2.
MCQ 8. If α and β are the zeros of the polynomial f(x) = x^2 - p(x+1) - c, then (α+1)(β+1) is equal to:
- A. c-1
- B. 1-c
- C. c
- D. 1+c
Answer: B. 1-c — The polynomial is f(x) = x^2 - px - p - c = x^2 - px - (p+c). Here, α+β = p and αβ = -(p+c). Thus, (α+1)(β+1) = αβ + α + β + 1 = -(p+c) + p + 1 = 1-c.
MCQ 9. If the zeros of the quadratic polynomial x^2 + (a+1)x + b are 2 and -3, then:
- A. a=-7, b=-1
- B. a=5, b=-1
- C. a=2, b=-6
- D. a=0, b=-6
Answer: D. a=0, b=-6 — Sum of zeros = 2 + (-3) = -1. Also, sum of zeros = -(a+1)/1 = -(a+1). So, -1 = -(a+1) which implies a+1=1, so a=0. Product of zeros = 2 × (-3) = -6. Also, product of zeros = b/1 = b. So, b=-6. Thus, a=0, b=-6.
MCQ 10. If one zero of the polynomial P(x) = (a^2+9)x^2 + 13x + 6a is the reciprocal of the other, then the value of a is:
- A. 2
- B. 3
- C. 1
- D. -3
Answer: B. 3 — Let the zeros be α and 1/α. Their product is α × (1/α) = 1. For a quadratic polynomial Ax^2 + Bx + C, the product of zeros is C/A. So, 6a / (a^2+9) = 1, which means 6a = a^2+9. Rearranging gives a^2 - 6a + 9 = 0, which is (a-3)^2 = 0, so a=3.
MCQ 11. If α, β are the zeros of x^2 - 6x + 8, then (1/α) + (1/β) is:
- A. 3/4
- B. -3/4
- C. 6/8
- D. 4/3
Answer: A. 3/4 — For x^2 - 6x + 8, α+β = 6 and αβ = 8. Then (1/α) + (1/β) = (α+β)/(αβ) = 6/8 = 3/4.
MCQ 12. If one zero of the quadratic polynomial kx^2 + 3x + k is 2, then the value of k is:
- A. 6/7
- B. -6/7
- C. 7/6
- D. -7/6
Answer: B. -6/7 — If 2 is a zero, then P(2)=0. So, k(2)^2 + 3(2) + k = 0 which simplifies to 4k + 6 + k = 0, giving 5k + 6 = 0, hence k = -6/7.
MCQ 13. If α and β are the zeros of the polynomial 2x^2 + 5x + k such that α^2 + β^2 + αβ = 21/4, then k is:
- A. 3
- B. -3
- C. 2
- D. -2
Answer: C. 2 — For 2x^2 + 5x + k, α+β = -5/2 and αβ = k/2. Given α^2 + β^2 + αβ = 21/4, we can write (α+β)^2 - 2αβ + αβ = 21/4, which simplifies to (α+β)^2 - αβ = 21/4. Substituting the values: (-5/2)^2 - k/2 = 21/4 => 25/4 - k/2 = 21/4 => k/2 = 25/4 - 21/4 = 4/4 = 1. So, k=2.
MCQ 14. What is the remainder when x^3 - 3x^2 + 5x - 3 is divided by x-1?
- A. 0
- B. 1
- C. 2
- D. 4
Answer: A. 0 — By the Remainder Theorem, the remainder when P(x) is divided by x-a is P(a). Here, a=1. So, P(1) = (1)^3 - 3(1)^2 + 5(1) - 3 = 1 - 3 + 5 - 3 = 0.
MCQ 15. Find all the zeros of x^4 - 5x^3 + 2x^2 + 10x - 8, if two of its zeros are √2 and -√2.
- A. [√2, -√2, 1, 4]
- B. [√2, -√2, -1, -4]
- C. [√2, -√2, 1, -4]
- D. [√2, -√2, -1, 4]
Answer: A. [√2, -√2, 1, 4] — Since √2 and -√2 are zeros, (x-√2)(x+√2) = x^2-2 is a factor. Dividing the polynomial by x^2-2 gives x^2 - 5x + 4. Factoring this quadratic, we get (x-1)(x-4). So the other zeros are 1 and 4.
MCQ 16. If α, β, γ are the zeros of the cubic polynomial x^3 - 3x^2 + x + 1, then αβ + βγ + γα is:
- A. 1
- B. -1
- C. 3
- D. -3
Answer: A. 1 — For a cubic polynomial ax^3+bx^2+cx+d, the sum of the products of zeros taken two at a time is c/a. Here, a=1, c=1. So, αβ + βγ + γα = 1/1 = 1.
MCQ 17. If the zeros of the quadratic polynomial x^2+px+q are twice the zeros of 2x^2-5x-3, then values of p and q are:
- A. p=-5, q=-6
- B. p=5, q=6
- C. p=5, q=-6
- D. p=-5, q=6
Answer: A. p=-5, q=-6 — For 2x^2-5x-3, let zeros be α1, β1. Sum α1+β1 = 5/2, Product α1β1 = -3/2. For x^2+px+q, new zeros are 2α1, 2β1. New sum = 2(α1+β1) = 2(5/2) = 5. Since sum = -p, p=-5. New product = (2α1)(2β1) = 4α1β1 = 4(-3/2) = -6. Since product = q, q=-6.
MCQ 18. If the polynomial x^4 + 2x^3 + 8x^2 + 12x + 18 is divided by another polynomial x^2 + 5, the remainder comes out to be px + q. Find the values of p and q.
- A. p=1, q=-2
- B. p=2, q=-1
- C. p=-1, q=2
- D. p=2, q=1
Answer: B. p=2, q=-1 — Performing polynomial long division of (x^4 + 2x^3 + 8x^2 + 12x + 18) by (x^2 + 5) yields a quotient of x^2 + 2x + 3 and a remainder of 2x - 1. Therefore, px+q = 2x-1, which means p=2 and q=-1.
MCQ 19. Which of the following is NOT a polynomial?
- A. 3x^2 + 5x - 7
- B. √2x + 1
- C. x^(1/2) + 4x
- D. y^3 - 2y + 5
Answer: C. x^(1/2) + 4x — A polynomial must have non-negative integer powers of the variable. In x^(1/2) + 4x, the term x^(1/2) has a fractional power (1/2), so it is not a polynomial.
MCQ 20. If the sum of the zeros of the polynomial 2x^2 - (k+1)x + 5 is 4, then the value of k is:
- A. 5
- B. 6
- C. 7
- D. 8
Answer: C. 7 — The sum of zeros for ax^2+bx+c is -b/a. Here, the sum is -(-(k+1))/2 = (k+1)/2. Given that the sum is 4, we have (k+1)/2 = 4, which means k+1 = 8, so k = 7.
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.