Polynomials — Class 10 Maths practice questions

Question 1

A civil engineer is designing a parabolic arch for a bridge. The shape of the arch is modelled by the polynomial f(x) = –x² + 4x + 5, where x represents the horizontal distance (in metres) from one end of the bridge and f(x) represents the height (in metres) of the arch at that point.

(i) Find the zeroes of the polynomial f(x) and interpret their meaning in the context of the bridge. [1 mark]

(ii) Using the zeroes found in part (i), verify the relationship between zeroes and coefficients of f(x). [1 mark]

(iii) The engineer decides to raise the entire arch by 'k' metres so that the new polynomial becomes g(x) = –x² + 4x + 5 + k. For the new arch to just touch the ground at exactly one point (i.e., the arch has a single point of contact with the ground), find the value of k.

OR

(iii) A second arch of the same bridge is modelled by h(x) = 2x² + (3 – p)x – 6, where p is a constant. If one zero of h(x) is 3/2, find the other zero and the value of p. [2 marks]

Accepted Answer

(i) To find the zeroes of f(x) = –x² + 4x + 5, set f(x) = 0:

⟹ –x² + 4x + 5 = 0
⟹ x² – 4x – 5 = 0
⟹ x² – 5x + x – 5 = 0
⟹ x(x – 5) + 1(x – 5) = 0
⟹ (x + 1)(x – 5) = 0
∴ x = –1 or x = 5

The zeroes of f(x) are –1 and 5.

Contextual interpretation: The zeroes represent the horizontal positions where the height of the arch is zero, i.e., the two points where the arch meets the ground. Since distance cannot be negative, x = 5 m is the meaningful point where the arch touches the ground on the far side; x = –1 is a mathematical zero outside the physical span.

∴ The zeroes are –1 and 5, representing the ground-level points of the arch.

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(ii) For f(x) = –x² + 4x + 5, we have a = –1, b = 4, c = 5.
Let α = –1 and β = 5.

Verification of sum of zeroes:
α + β = –1 + 5 = 4
–b/a = –(4)/(–1) = 4 ✓

Verification of product of zeroes:
αβ = (–1)(5) = –5
c/a = 5/(–1) = –5 ✓

∴ Both relationships are verified. Hence proved.

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(iii) [Main Option]

The new polynomial is g(x) = –x² + 4x + 5 + k.

For the arch to just touch the ground at exactly one point, g(x) = 0 must have exactly one (repeated) real root.

Setting g(x) = 0:
–x² + 4x + (5 + k) = 0
⟹ x² – 4x – (5 + k) = 0

For exactly one real root, the discriminant D = 0.

D = b² – 4ac = (–4)² – 4(1)(–(5 + k))
⟹ D = 16 + 4(5 + k)
⟹ D = 16 + 20 + 4k
⟹ D = 36 + 4k

Setting D = 0:
36 + 4k = 0
⟹ 4k = –36
∴ k = –9

The arch must be lowered by 9 metres (k = –9) so that the new arch just touches the ground at exactly one point.

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(iii) [OR Option]

For h(x) = 2x² + (3 – p)x – 6, let α = 3/2 and β be the other zero.

Using the relationship: product of zeroes = c/a

αβ = c/a = –6/2 = –3

⟹ (3/2) · β = –3
⟹ β = –3 × (2/3)
∴ β = –2

The other zero is –2.

Now, using the relationship: sum of zeroes = –b/a

α + β = –(3 – p)/2

⟹ (3/2) + (–2) = –(3 – p)/2
⟹ 3/2 – 4/2 = –(3 – p)/2
⟹ –1/2 = –(3 – p)/2
⟹ –1 = –(3 – p)
⟹ 1 = 3 – p
⟹ p = 3 – 1
∴ p = 2

∴ The other zero is –2 and the value of p is 2.

Question 2

A civil engineer is designing a parabolic arch for a bridge. The shape of the arch is modelled by the polynomial p(x) = 2x² – 7x + k, where x represents the horizontal distance (in metres) from one end of the arch and p(x) represents the height. The engineer knows that the arch meets the ground (height = 0) at two points whose horizontal distances from one end are α and β (the zeroes of p(x)).

Based on this scenario, answer the following:

(i) Find the sum and product of the zeroes α and β of p(x) in terms of k. (1 mark)

(ii) If the difference of the zeroes is 3/2 (i.e., α – β = 3/2), find the value of k. (1 mark)

(iii) Using the value of k found in part (ii), find a quadratic polynomial whose zeroes are (2α + β) and (α + 2β). (2 marks)

OR

(iii) Using the value of k found in part (ii), find the value of α² + β² and hence find (1/α² + 1/β²). (2 marks)

Accepted Answer

For p(x) = 2x² – 7x + k, comparing with ax² + bx + c: a = 2, b = –7, c = k.

**(i) Sum and Product of zeroes:**

By the relationship between zeroes and coefficients of a quadratic polynomial:

α + β = –b/a = –(–7)/2 = **7/2**

αβ = c/a = **k/2**

∴ Sum of zeroes = 7/2 and Product of zeroes = k/2.

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**(ii) Finding the value of k:**

Given: α – β = 3/2

Using the identity (α – β)² = (α + β)² – 4αβ:

⟹ (3/2)² = (7/2)² – 4·(k/2)

⟹ 9/4 = 49/4 – 2k

⟹ 2k = 49/4 – 9/4

⟹ 2k = 40/4 = 10

⟹ k = **5**

∴ The value of k is 5.

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**(iii) [Main Option] Quadratic polynomial with zeroes (2α + β) and (α + 2β):**

From part (i) and (ii): α + β = 7/2 and αβ = k/2 = 5/2.

First, find the sum of the new zeroes:

(2α + β) + (α + 2β) = 3α + 3β = 3(α + β)

⟹ = 3 × 7/2 = **21/2**

Next, find the product of the new zeroes:

(2α + β)(α + 2β) = 2α² + 4αβ + αβ + 2β²

= 2(α² + β²) + 5αβ

Now, α² + β² = (α + β)² – 2αβ = (7/2)² – 2(5/2) = 49/4 – 5 = 49/4 – 20/4 = 29/4

⟹ (2α + β)(α + 2β) = 2(29/4) + 5(5/2) = 29/2 + 25/2 = **54/2 = 27**

Required quadratic polynomial = k[x² – (sum of zeroes)x + (product of zeroes)]

= k[x² – (21/2)x + 27]

Taking k = 2: **2x² – 21x + 54**

∴ The required quadratic polynomial is 2x² – 21x + 54 (or any scalar multiple, e.g. x² – 21x/2 + 27).

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**(iii) [OR Option] Find α² + β² and hence 1/α² + 1/β²:**

From part (i) and (ii): α + β = 7/2 and αβ = 5/2.

**Step 1 — α² + β²:**

α² + β² = (α + β)² – 2αβ

= (7/2)² – 2(5/2)

= 49/4 – 10/2

= 49/4 – 20/4

∴ **α² + β² = 29/4**

**Step 2 — 1/α² + 1/β²:**

1/α² + 1/β² = (α² + β²)/(α²β²) = (α² + β²)/(αβ)²

= (29/4) ÷ (5/2)²

= (29/4) ÷ (25/4)

= 29/4 × 4/25

∴ **1/α² + 1/β² = 29/25**

Question 3

Assertion (A): The sum of the zeroes of the polynomial x² - 5x + 6 is 5.
Reason (R): If α and β are zeroes of ax² + bx + c, then α + β = -b/a.

Accepted Answer

Option (A) is correct.

Explanation: By the relationship between zeroes and coefficients, if α and β are zeroes of ax² + bx + c, then α + β = −b/a. For x² − 5x + 6, a = 1 and b = −5, so α + β = −(−5)/1 = 5. Thus Assertion (A) is true, Reason (R) is true, and R is the correct explanation of A.

Question 4

If α and β are the zeroes of the polynomial p(x) = 5x² − 3√5 x, then the value of α·β is:

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Comparing p(x) = 5x² − 3√5 x + 0 with ax² + bx + c gives a = 5, b = −3√5, c = 0. ∴ α·β = c/a = 0/5 = 0.

Question 5

A polynomial p(x) of degree 2 has its graph crossing the x-axis at exactly two points: x = −2 and x = 5. Which of the following could be p(x)?

Accepted Answer

Option (A) is correct.

**Explanation:** For a degree-2 polynomial with zeroes α = −2 and β = 5, the polynomial is k[x² − (α + β)x + αβ]. Here α + β = −2 + 5 = 3 and αβ = (−2)(5) = −10, giving k[x² − 3x − 10]. For k = 1, p(x) = x² − 3x − 10, which matches Option (A). Option (B) gives x² + 3x − 10 = (x − 2)(x + 5), with zeroes at x = 2 and x = −5, not the required values.

Question 6

Assertion (A): The product of the zeroes of the polynomial 2x² - 8x + 6 is 3.
Reason (R): For ax² + bx + c, the product of zeroes is c/a.

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Here a = 2, c = 6, so product of zeroes = 6/2 = 3, confirming Assertion A is true. Reason R states exactly this standard result and is therefore true. Since applying R directly yields the value 3 stated in A, R is the correct explanation of A.

Question 7

Assertion (A): The product of the zeroes of the polynomial 2x² - 8x + 6 is 3.
Reason (R): For ax² + bx + c, the product of zeroes is c/a.

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Here, a = 2, c = 6, so product of zeroes = 6/2 = 3, confirming A is true. R states this exact formula and is true. Since R directly explains why the product equals 3, R is the correct explanation of A.

Question 8

If −3 is a zero of the polynomial p(x) = x² + 5x + (2k + 4), then the value of k is:

Accepted Answer

Option (A) is correct.

**Explanation:** If −3 is a zero of p(x), then p(−3) = 0. Substituting x = −3: (−3)² + 5(−3) + (2k + 4) = 0 ⟹ 9 − 15 + 2k + 4 = 0 ⟹ 2k − 2 = 0 ⟹ k = 1.

Question 9

The zeroes of the quadratic polynomial 5x² + 6x − 8 are:

Accepted Answer

Option (B) is correct.

**Explanation:** Factorising 5x² + 6x − 8: product = 5 × (−8) = −40, sum = 6 ⟹ split as 10 and −4 ⟹ 5x² + 10x − 4x − 8 = 5x(x + 2) − 4(x + 2) = (5x − 4)(x + 2). Setting each factor to zero: 5x − 4 = 0 ⟹ x = 4/5 and x + 2 = 0 ⟹ x = −2. ∴ the zeroes are **−2 and 4/5**.

Question 10

Assertion (A): The product of the zeroes of the polynomial 2x² - 8x + 6 is 3.
Reason (R): For ax² + bx + c, the product of zeroes is c/a.

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Here a = 2, b = −8, c = 6 ⟹ product of zeroes = 6/2 = 3, so Assertion (A) is true. Reason (R) states the correct formula (product of zeroes = c/a), which is true, and it directly explains why the product equals 3 in (A). ∴ Both A and R are true and R is the correct explanation of A.

Question 11

The zeroes of the quadratic polynomial p(x) = 4x² − 4x − 3 are:

Accepted Answer

Option (A) is correct.

**Explanation:** The zeroes of p(x) = 4x² − 4x − 3 are found by factorisation. Two numbers with product 4×(−3) = −12 and sum −4 are −6 and 2, giving 4x² − 6x + 2x − 3 = 2x(2x − 3) + 1(2x − 3) = (2x + 1)(2x − 3). Setting each factor to zero: 2x + 1 = 0 ⟹ x = −1/2, and 2x − 3 = 0 ⟹ x = 3/2. ∴ the zeroes are 3/2 and −1/2.

Question 12

If p(x) = x² − 7x + 10, then the value of p(2) + p(5) is:

Accepted Answer

Option (A) is correct.

Explanation: Substituting x = 2: p(2) = (2)² − 7(2) + 10 = 4 − 14 + 10 = 0. Substituting x = 5: p(5) = (5)² − 7(5) + 10 = 25 − 35 + 10 = 0. Since x = 2 and x = 5 are the zeroes of p(x), both values equal zero. ∴ p(2) + p(5) = 0 + 0 = 0.

Question 13

Assertion (A): The sum of the zeroes of the polynomial x² - 5x + 6 is 5.
Reason (R): If α and β are zeroes of ax² + bx + c, then α + β = -b/a.

Accepted Answer

Option (A) is correct.

**Explanation:** By the relationship between zeroes and coefficients, if α and β are zeroes of ax² + bx + c, then α + β = −b/a. This confirms R is true. Applying R to x² − 5x + 6 (where a = 1, b = −5): α + β = −(−5)/1 = 5, confirming A is true. Since R directly explains why A holds, both A and R are true and R is the correct explanation of A.

Question 14

The zeroes of the quadratic polynomial p(x) = 3x² − 5x − 2 are:

Accepted Answer

Option (A) is correct.

Explanation: Factorising p(x) = 3x² − 5x − 2 by splitting the middle term (product = 3×(−2) = −6, sum = −5 ⟹ split as −6 and +1): 3x² − 6x + x − 2 = 3x(x − 2) + 1(x − 2) = (3x + 1)(x − 2). Setting each factor to zero gives x = 2 and x = −1/3, which matches Option (A). Verification: sum of zeroes = 2 + (−1/3) = 5/3 = −(−5)/3 ✓; product = 2×(−1/3) = −2/3 = −2/3 ✓.

Question 15

Assertion (A): The product of the zeroes of the polynomial 2x² - 8x + 6 is 3.
Reason (R): For ax² + bx + c, the product of zeroes is c/a.

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Applying this to 2x² − 8x + 6 (a = 2, c = 6): product of zeroes = 6/2 = 3, so Assertion (A) is true. Reason (R) correctly states the standard result "product of zeroes = c/a," which is true, and it is precisely the formula used to verify (A). ∴ Both A and R are true and R is the correct explanation of A.

Question 16

Assertion (A): The product of the zeroes of the polynomial 2x² - 8x + 6 is 3.
Reason (R): For ax² + bx + c, the product of zeroes is c/a.

Accepted Answer

Option (A) is correct.

Explanation: For a quadratic polynomial ax² + bx + c, the product of zeroes = c/a. Here a = 2, b = −8, c = 6 ⟹ product of zeroes = 6/2 = 3, so Assertion (A) is true. Reason (R) states the correct formula (product of zeroes = c/a), so R is also true, and R directly explains why the product equals 3 in Assertion (A).

Question 17

A polynomial p(x) is drawn on a coordinate plane. Its graph cuts the x-axis at x = −3 and x = 4, and touches the x-axis at x = 1. What is the total number of zeroes of p(x)?

Accepted Answer

Option (B) is correct.

Explanation: The total number of zeroes of a polynomial equals the number of distinct x-values where the graph meets the x-axis, whether it cuts or touches. The graph cuts the x-axis at x = −3 and x = 4, giving one zero each, and touches the x-axis at x = 1, giving one zero (of even multiplicity). Thus the total number of zeroes is 3.

Question 18

If a polynomial p(x) = x² − 3x + 2, then the value of p(0) + p(3) is:

Accepted Answer

Option (B) is correct.

**Explanation:** Substituting x = 0: p(0) = 0² − 3(0) + 2 = 2. Substituting x = 3: p(3) = 3² − 3(3) + 2 = 9 − 9 + 2 = 2. ∴ p(0) + p(3) = 2 + 2 = **4**.

Question 19

The zeroes of the polynomial $x^2 + px + q$ are half the zeroes of the polynomial $2x^2 + 8x - 10$. The value of $p$ is:

Accepted Answer

Option (C) is correct.

**Explanation:** For the polynomial 2x² + 8x − 10, dividing throughout by 2 gives x² + 4x − 5 = 0 ⟹ (x + 5)(x − 1) = 0, so the zeroes are −5 and 1. The zeroes of x² + px + q are half of these, i.e., −5/2 and 1/2. Using the relation sum of zeroes = −p/1, we get −5/2 + 1/2 = −p ⟹ −4/2 = −p ⟹ −2 = −p ∴ p = 2.

Question 20

Assertion (A): The sum of the zeroes of the polynomial x² - 5x + 6 is 5.
Reason (R): If α and β are zeroes of ax² + bx + c, then α + β = -b/a.

Accepted Answer

Option (A) is correct.

Explanation: By the relationship between zeroes and coefficients, for ax² + bx + c, α + β = −b/a. For x² − 5x + 6, a = 1, b = −5 ⟹ α + β = −(−5)/1 = 5, so Assertion (A) is true. Reason (R) states exactly this formula, which is correct, and it directly explains why the sum equals 5 in (A). ∴ Both A and R are true and R is the correct explanation of A.

Polynomials: Class 10 Maths Practice Questions

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