HOTS Questions on Class 10 Maths Chapter 2 ‘Polynomials’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 2 ‘Polynomials’ for Class 10 is designed to push learners beyond routine exercises and develop deep, flexible algebraic reasoning. Aligned with CBSE and NCERT objectives, the questions emphasise conceptual understanding and creative application of polynomial ideas such as degree and coefficients, zeroes and their multiplicities, factor theorem and remainder theorem, division of polynomials, graphs of simple polynomials, and connections between roots and coefficients. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Polynomials

Question 1: The squared difference of zeroes of the polynomial x² + px + 45 equals 144. Find all possible values of p.

Solution: 

Let zeroes be α and β.

From x² + px + 45: α + β = −p    αβ = 45

Given: (α − β)² = 144

(α + β)² − 4αβ = 144

(−p)² − 4(45) = 144

p² − 180 = 144

p² = 324

p = ±18

∴ p = 18 or p = −18

Question 2: α and β are zeroes of p(x) = x² + 8x + 6. Form a quadratic polynomial whose zeroes are α/β and β/α.

Solution: 

From x² + 8x + 6: α + β = −8    αβ = 6

New sum: α/β + β/α = (α² + β²)/(αβ) = [(α+β)² − 2αβ]/(αβ)

= [64 − 12]/6 = 52/6 = 26/3

New product: (α/β)(β/α) = 1

New polynomial: x² − (26/3)x + 1 = (1/3)(3x² − 26x + 3)

Required polynomial: 3x² − 26x + 3 

Question 3: If α and β are the zeroes of the polynomial 2x² − 4x + 5, find the values of: (a) α² + β², (b) (α − β)², and (c) α³ + β³.

Solution: 

From the polynomial 2x² − 4x + 5:

α + β = 4/2 = 2    αβ = 5/2 = 2.5

(a) α² + β² = (α + β)² − 2αβ

= (2)² − 2(5/2) = 4 − 5 = −1

(b) (α − β)² = (α + β)² − 4αβ

= 4 − 4(5/2) = 4 − 10 = −6

(c) α³ + β³ = (α + β)³ − 3αβ(α + β)

= (2)³ − 3(5/2)(2) = 8 − 15 = −7

Question 4: If one zero of the quadratic polynomial (k² + k)x² + 68x + 6k is the reciprocal of the other, find the value of k.

Solution: If one zero is α, the other is 1/α.

Product = α × (1/α) = 1

From the polynomial: Product of zeroes = 6k/(k² + k)

6k/(k² + k) = 1

⇒ 6k = k² + k

⇒ k² − 5k = 0

⇒ k(k − 5) = 0

k = 0 or k = 5

Check k = 0: Leading coefficient = 0 → not a valid polynomial.

So k = 5.

∴ k = 5

Question 5: Find a cubic polynomial with sum of zeroes = 3/2, sum of product of zeroes taken two at a time = −1, and product of zeroes = −1/4.

Solution: 

Using the cubic template: k[x³ − (S₁)x² + (S₂)x − S₃]

S₁ = 3/2, S₂ = −1, S₃ = −1/4

Polynomial = k[x³ − (3/2)x² + (−1)x − (−1/4)]

= k[x³ − (3/2)x² − x + 1/4]

Multiply by 4 (set k = 4): 4x³ − 6x² − 4x + 1

Cubic polynomial: 4x³ − 6x² − 4x + 1

Question 6: If the zeroes of the polynomial f(x) = x³ − 12x² + 39x + a are in Arithmetic Progression, find the value of a.

Solution: 

Let zeroes in AP be (m−d), m, (m+d)

Sum of zeroes: (m−d) + m + (m+d) = 3m = 12 

⇒ m = 4

Sum of pairwise products: (m−d)m + m(m+d) + (m−d)(m+d) = 39

m² − md + m² + md + m² − d² = 39

⇒ 3m² − d² = 39

⇒ 3(16) − d² = 39

⇒ d² = 48 − 39 = 9 

⇒  d = ±3

Zeroes are: 1, 4, 7 (or 7, 4, 1 — same set)

Product of zeroes: 1 × 4 × 7 = 28 = −a/1 

⇒  a = −28

∴ a = −28

Question 7: On dividing polynomial 4x⁴ − 5x³ − 39x² − 46x − 2 by polynomial g(x), the quotient is x² − 3x − 5 and the remainder is −5x + 8. Find g(x).

Solution: 

From Division Algorithm: p(x) = g(x)·q(x) + r(x)

So: g(x) = [p(x) − r(x)] ÷ q(x)

p(x) − r(x) = 4x⁴ − 5x³ − 39x² − 46x − 2 − (−5x + 8)

= 4x⁴ − 5x³ − 39x² − 41x − 10

Divide by q(x) = x² − 3x − 5:

(4x⁴ − 5x³ − 39x² − 41x − 10) ÷ (x² − 3x − 5) = 4x² + 7x + 2

g(x) = 4x² + 7x + 2

Question 8: Obtain all zeroes of p(x) = 3x⁴ − 15x³ + 17x² + 5x − 6, given that two of its zeroes are −1/√3 and 1/√3.

Solution: 

Step 1: Known zeroes are ±1/√3. Form their factor:

(x − 1/√3)(x + 1/√3) = x² − 1/3

Multiply by 3: 3x² − 1

Step 2: Divide p(x) by (3x² − 1):

(3x⁴ − 15x³ + 17x² + 5x − 6) ÷ (3x² − 1) = x² − 5x + 6

Step 3: Factorise x² − 5x + 6 = (x − 2)(x − 3)

All four zeroes: −1/√3, 1/√3, 2, 3

Question 9: If 1 and −1 are both zeroes of polynomial Lx⁴ + Mx³ + Nx² + Rx + P, show that L + N + P = M + R = 0.

Solution: 

Since 1 is a zero: p(1) = L + M + N + R + P = 0   ...(i)

Since −1 is a zero: p(−1) = L − M + N − R + P = 0   ...(ii)

Adding (i) and (ii): 2L + 2N + 2P = 0 ⇒ L + N + P = 0 

Subtracting (ii) from (i): 2M + 2R = 0 ⇒ M + R = 0 

Therefore: L + N + P = M + R = 0

Question 10: Is the polynomial y⁴ + 4y² + 5 guaranteed to have zeroes? Justify your answer using the discriminant or graphical reasoning.

Solution: 

Substitute t = y²: p = t² + 4t + 5. 

For real zeroes of t: discriminant = 16 − 20 = −4 < 0. So t has no real values.

Since the discriminant is negative, there are no real values of t satisfying the equation. Therefore, there is no real y such that y⁴ + 4y² + 5 = 0. Hence, the polynomial has no real zeroes.

Graph reasoning: y⁴ ≥ 0, 4y² ≥ 0, and adding 5 means the minimum value is 5 (at y = 0). The graph never touches the x-axis. 

 

Questions PDF with worked-out examples for Class 10 Chapter 2: Polynomials, perfect for last-minute CBSE exam revision.

Class 10 Chapter 2: Polynomials HOTS PDF