Previous Year Questions 2025
Q1: Zeroes of the polynomial p(y) (2025)
(a)
(b)
(c)
(d)
Ans: (d)
The given polynomial is p(y) 
For zeroes of the polynomial, put p(y) = 0
Q2: Zeroes of the polynomial p(x) = x2 – 3√2x + 4 are: (2025)
(a) 2,√2
(b) 2√2, √2
(c) 4√2, -√2
(d) √2, 2
Ans: (b)
We have, p(x) = x2 – 3√2x + 4
⇒ p(x) = x2 – 2√2x -√2x + 4
= x(x – 2√2) – √2(x – 2√2)
= (x – 2√2)(x – √2)
For zeroes of the polynomial, put p(x) = 0
⇒ (x – 2√2)(x -√2) = 0
∴ x = 2√2, √2 are zeroes of p(x).
Q3: Find the zeros of the polynomial (2025)
Ans: 
⇒ 3x – 2 = 0 ⇒ x = 2/3 and x + 2 = 0 ⇒ x = -2
x = -2 and x = 2/3 are zeroes of the given polynomial.
Q4: Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is: (2025)
(a) 3
(b) 5
(c) 2
(d) 4
Ans: (c)
Since both polynomials cut the x-axis at two distinct points each, the total number of distinct zeroes of both the polynomials combined is 2.
Q5: If α and β are the zeroes of the polynomial p(x) = x2 – ax – b, then the value of (α + β + αβ) is equal to: (2025)
(a) a + b
(c) a- b
(b) -a – b
(d) -a + b
Ans: (c)
Given polynomial is p(x) = x2 – ax – b, and α and β are zeroes of p(x)
∴ Sum of zeroes = α + β =a
Product of zeroes = αβ = – b
Now, α + β + αβ = a – b
Concept Applied
If α and β are the zeroes of quadratic polynomial p(x) = ax2 + bx + c, then α + β = -b/a, αβ = c/a.
Q6: If α and β are zeroes of the polynomial p(x) = kx2 – 30x + 45k and α +β = αβ, then the value of ‘k’ is: (2025)
(a)
(b)
(c) 3/2
(d) 2/3
Ans: (d)
Given, α and β are zeroes of the polynomial p(x) = kx2 – 30x + 45k
Here, a = k, b = -30, c = 45 k 
∴ α + β = αβ [Given]
Q7: If α and β are the zeroes of the polynomial 3x2 + 6x + k such that then the value of k is: (2025)
(a) -8
(b) 8
(c) -4
(d) 4
Ans: (d)
Compare 3x2 + 6x + k with ax2 + bx + c, we get a = 3, b = 6 and c = k 
Q8: If the zeroes of the polynomialare reciprocals of each other, then the value of bis (2025)
(a) 2
(b) 1/2
(c) -2
(d)
Ans: (a)
Let α and β be the zeroes of the given polynomial 
Q9: If the sum of the zeroes of the polynomial p(x) = (p + 1)x2 + (2p + 3) x + (3p + 4) is – 1, then find the value of ‘p’. (2025)
Ans: The given polynomial is p(x) = (p + 1)x2 + (2p + 3)x + (3p + 4)
Let α and β are zeroes of given polynomial 
⇒ p + 1 = 2p + 3
⇒ p = -2
Hence, the value of p is – 2.
Q10: If α and β are zeroes of the polynomial p(x) = x2 – 2x – 1, then find the value of (2025)
Ans: The given polynomial is p(x) = x2 – 2x – 1;
If α and β are zeroes of given polynomial 
Q11: If ‘α’ and ‘β’ are the zeroes of the polynomial p(y) = y2 – 5y+ 3, then find the value of α4β3 + α3β4 . (2025)
Ans:
We have p(y) = y2 – 5y + 3
Let α and β be zeroes of p(y).
Given, sum of zeroes= α + β = 5
Product of zeroes = αβ = 3
α⁴β³ + α³β⁴
= α³β³ (α + β)
= (αβ)³ (α + β)
= (3)³ (5) = 27 × 5 = 135
Q12: If the zeroes of the polynomial x2 + ax + b are in the ratio 3 : 4, then prove that 12a2 = 49b. (2025)
Ans: Let the zeroes of the polynomial x2 + ax + b be 3x and 4x. 
⇒ 49b = 12a2. Hence proved.
Q13: Find the zeroes of the polynomial p(x) = 3x2 – 4x – 4. Hence, write a polynomial whose each of the zeroes is 2 more than the zeroes of p(x). (2025)
Ans: The polynomial is p(x) = 3x2 – 4x – 4.
The zeroes are given by p(x) = 0
⇒ 3x2 – 4x – 4 = 0
⇒ 3x2 – 6x + 2x – 4 = 0
⇒ 3x(x – 2) + 2(x – 2) = 0
⇒ (x – 2)(3x + 2) = 0
⇒ x – 2 = 0 or 3x + 2 = 0
⇒ x = 2 and -2/3
Thus, zeroes of new polynomial are 
Hence, new polynomial is 
Taking a= 3, r(x) = 3x2 – 16x + 16
Thus, r(x) = 3x2 – 16x + 16 is the new polynomial with zeroes x = 4 and x = 4/3.
Previous Year Questions 2024
Q1: What should be added from the polynomial x2 – 5x + 4, so that 3 is the zero of the resulting polynomial? (2024)
(a) 1
(b) 2
(c) 4
(d) 5
Ans: (b)
Let, f(x) = x2 – 5x + 4
Let p should be added to f(x) then 3 becomes zero of polynomial.
So, f(3) + p = 0
⇒ (3)2 – 5 × (3) + 4 + p = 0
⇒ 9 + 4 – 15 + p = 0
⇒ – 2 + p = 0
⇒ p = 2
So, 2 should be added.
Q2: Find the zeroes of the quadratic polynomial x2 – 15 and verify the relationship between the zeroes and the coefficients of the polynomial. (2024)
Ans:
x2 – 15 = 0
x2 = 15
x = ± √15
Zeroes will be α = √15 , β = – √15
Verification: Given polynomial is x2 – 15
On comparing above polynomial with
ax2 + bx + c, we have
a = 1, b = 0, c = –15
sum of zeros = α + β 
Product of zeros = αβ 
Hence, verified.
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Previous Year Questions 2023
Q1: The graph of y = p(x) is given, for a polynomial p(x). The number of zeroes of p(x) from the graph is (2023)
(a) 3
(b) 1
(c) 2
(d) 0
Ans: (b)
Here, y = p(x) touches the x-axis at one point
So, number of zeros is one.
Q2: If α, β are the zeroes of a polynomial p(x) = x2 + x – 1, then 1/α + 1/β equals to (2023)
(a) 1
(b) 2
(c) -1
(d) -1/2
Ans: (a)
The polynomial is p(x) = x2 + x – 1.
Step 1: The relationships between the zeroes and coefficients:
Sum of zeroes (α + β): – ba = – 11 = -1
Product of zeroes (αβ): ca = -11 = -1
Step 2: Simplify 1α + 1β:
1α + 1β = α + βαβ
Substitute the values:
α + βαβ = -1-1 = 1
Final Answer: (a) 1
Q3: If α, β are the zeroes of a polynomial p(x) = x2 – 1, then the value of (α + β) is (2023)
(a) 1
(b) 2
(c) -1
(d) 0
Ans: (d)
The polynomial is p(x) = x2 – 1.
Step 1: Sum of zeroes (α + β): – ba = – 01
Step 2: Simplify:
– 01 = 0
Final Answer: (d) 0
Q4: If α, β are the zeroes of a polynomial p(x) = 4x2 – 3x – 7, then (1/α + 1/β) is equal to (2023)
(a) 7/3
(b) -7/3
(c) 3/7
(d) -3/7
Ans: (d)
The polynomial is p(x) = 4x2 – 3x – 7.
Step 1: calculating sum and product of zeroes
Sum of zeroes (α + β): – ba = – (-3)4 = 34
Product of zeroes (αβ): ca = -74
Step 2: Simplify 1α + 1β:
α + βαβ = 34-74 = -37
Final Answer: (d) – 37
Q5: If one zero of the polynomial p(x) = 6x2 + 37x – (k – 2) is reciprocal of the other, then find the value of k. (CBSE 2023)
Ans: We have,
The polynomial is p(x) = 6x2 + 37x – (k – 2).
Step 1: The relationship between the product of zeroes and coefficients:
Product of zeroes (αβ) = ca = -(k – 2)6
It is given that αβ = 1. Substitute this:
-(k – 2)6 = 1
Step 2: Solve for k:
Multiply both sides by 6:
-(k – 2) = 6
Simplify:
k – 2 = -6
k = -4
Final Answer: k = – 4
| Also read: Important Definitions & Formulas: Polynomials |
Previous Year Questions 2022
Q1: If one of the zeroes of a quadratic polynomial ( k – 1 )x2 + kx + 1 is – 3, then the value of k is (2022)
(a) 4/3
(b) -4/3
(c) 2/3
(d) -2/3
Ans: (a)
Given that -3 is a zero of quadratic polynomial (k – 1)2+ kx + 1.
⇒ Putting x = -3 in above equation, we get
∴ (k – 1) (-3)2 + k(-3) +1 = 0
⇒ 9k – 9 – 3k + 1 = 0 ⇒ 6k – 8 = 0
⇒ k = 8/6
⇒ k = 4/3
Q2: If the path traced by the car has zeroes at -1 and 2, then it is given by (2022)
(a) x2 + x + 2
(b) x2 – x + 2
(c) x2 – x – 2
(d) x2 + x – 2
Ans: (c)
The zeroes of the polynomial are -1 and 2.
Step 1: The polynomial with given zeroes is:
p(x) = a(x – α)(x – β)
Substitute the zeroes α = -1 and β = 2:
p(x) = a(x – (-1))(x – 2) = p(x) = a(x + 1)(x – 2)
Step 2: Expand the polynomial:
p(x) = a[(x)(x) + (x)(-2) + (1)(x) + (1)(-2)]
p(x) = a[x2 – x – 2]
Step 3: Assuming a = 1:
p(x) = x2 – x – 2
Final Answer: (c) x2 – x – 2
Q3: The number of zeroes of the polynomial representing the whole curve, is (2022)
(a) 4
(b) 3
(c) 2
(d) 1
Ans: (a)
Given curve cuts the x-axis at four distinct points.
So, number of zeroes will be 4 .
Q4: The distance between C and G is (2022)
(a) 4 units
(b) 6 units
(c) 8 units
(d) 7 units
Ans: (b)
The distance between point C and G is 6 units.
Q5: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6. (2022)
(a) x2 + 5x + 6
(b) x2 – 5x + 6
(c) x2 – 5 x – 6
(d) – x2 + 5x + 6
Ans: (a)
Let α, β be the zeroes of required polynomial p(x).
Given, α + β=-5 and α.β=6
p(x) = x2 – (Sum of zeros)x + (Product of zeros)
∴ p(x)=k[x2 – (-5)x + 6] = k[x2 + 5x + 6]
Thus, one of the polynomial which satisfy the given condition is x2+ 5x + 6
Previous Year Questions 2021
Q1: If one zero of the quadratic polynomial x2 + 3x + k is 2 then find the value of k. (2021)
Ans: Given, polynomial is f(x) =x2 + 3x + k
Since, 2 is zero of the polynomial f(x).
∴ f(2) = 0
⇒ f(2) =(2)2 + 3 x 2 + k
⇒ 4 + 6 + k = 0
⇒ k = -10
Previous Year Questions 2020
Q1: The degree of polynomial having zeroes -3 and 4 only is (2020)
(a) 2
(b) 1
(c) more than 3
(d) 3
Ans: (a)
Since, the polynomial has two zeroes only. So. the degree of the polynomial is 2.
Q2: If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2. then the value of k is (2020)
(a) 10
(b) – 10
(c) -7
(d) -2
Ans: (b)
Given, 2 is a zero of the polynomial
p(x) = x2 + 3x + k
∴ p (2) = 0
⇒ (2)2 + 3(2) + k = 0
⇒ 4 + 6 + k = 0
⇒ 10 + k = 0
⇒ k= -10
Q3: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6 is ________. (2020)
(a) x2 + 5x + 6
(b) x2 – 5x + 6
(c) x2– 5x – 6
(d) -x2 + 5x + 6
Ans: (a)
Let α, β be the zeroes of required polynomial p(x)
Given, α+ β = -5 and αβ = 6
p(x) = k[x2 – (- 5)x + 6]
= k[x2 + 5x + 6]
Thus, one of the polynomial which satisfy the given condition is x2 + 5x + 6.
Q4: Form a quadratic polynomial, the sum and product of whose zeroes are (-3) and 2 respectively. (CBSE 2020)
Ans: Let α, β be the zeroes of required polynomial Given, α + β = -3 and αβ = 2
∴ p(x) = k[x2 – (-3)x + 2] = k(x2 + 3x + 2)
For k = 1 , p (x) = x2 + 3x + 2
Hence, one of the polynomial which satisfy the given condition is x2 + 3x + 2.
Q5: The zeroes of the polynomial x2 – 3x – m(m + 3) are:
(a) m, m + 3
(b) –m, m + 3
(c) m, – (m + 3)
(d) –m, – (m + 3) (CBSE 2020)
Ans: (b)
Given:
x2 − 3x − m(m + 3) = 0
Let’s find the zeroes by applying the quadratic formula:
Substitute into the formula:
Simplify under the square root:
Taking the square root:
So, the zeroes are –m and m + 3.
Thus, the correct answer is (b) –m, m + 3.
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Previous Year Questions 2019
Q1: Find the value of k such that the polynomial x2 – (k + 6)x + 2(2k – 1) has the sum of its zeroes equal to half of its product. [Year 2019, 3 Marks]
Ans: 7
The given polynomial is x2 -(k + 6)x + 2(2k – 1)
According to the question
Sum of zeroes = 1/2(Product of Zeroes ):
⇒ k + 6 = 1/2 x 2 (2k – 1)
⇒ k + 6 = 2k – 1
⇒ k = 7