Polynomials PYQs

 Previous Year Questions 2026

Q1. The graph of a quadratic polynomial is shown in the figure. The sum and product of zeroes of the polynomial respectively are :  [1 Mark]

(A) 2 and 3 
(B) 2 and -3
(C) -2 and 3 
(D) -2 and -3 

Solution:

Answer: (B) 2 and -3

Chapter: Polynomials

Q2. In which of the following graphs of the polynomials is one of the zeroes twice the other?    [1 Mark]
(A) (i) only 
(B) (i) and (ii) 
(C) (ii) and (iii) 
(D) (ii) only

Solution:

Answer: (D) (ii) only

Chapter: Polynomials

Q3. The graph of polynomial p(x) = k is shown here. Number of zeroes of polynomial p(x) is :  [1 Mark]
(A) 0
(B) 1
(C) 2
(D) infinitely many

Solution:

Answer: (A) 0

Chapter:Polynomials

Q4. The graph of a polynomial $p(x)$ is shown here. The number of zeroes of the polynomial $p(x)$ is [1 Mark]
(A) 5
(B) 1
(C) 0
(D) 4

Solution:

Answer: (D) 4

Chapter:Polynomials

Q5. The value of k for which sum of the zeroes of the polynomial $p(x)=3x^2-kx+6$ is 2, is [1 Mark]
(A) 2
(B) $-6$
(C) $-2$
(D) 6

Solution:

Answer: (D) 6
Chapter: Polynomials

Q6. Verify the relation between the zeroes and the coefficients of the quadratic polynomial $4x^2-9$.  [2 Marks]

Solution:

Answer:
Zeroes of $4x^2-9$ are ${\displaystyle \frac{3}{2}},{\displaystyle \frac{-3}{2}}$
Sum of zeroes $={\displaystyle \frac{3}{2}}+\left({\displaystyle \frac{-3}{2}}\right)={\displaystyle \frac{0}{4}}=-{\displaystyle \frac{\text{Coefficient of }x}{\text{Coefficient of }{x}^2}}$
Product of zeroes $={\displaystyle \frac{3}{2}}\times \left({\displaystyle \frac{-3}{2}}\right)={\displaystyle \frac{-9}{4}}={\displaystyle \frac{\text{Constant term}}{\text{Coefficient of }{x}^2}}$

Chapter: Polynomials

Q7. Verify the relationship between the zeroes and the coefficients of the polynomial $p(x)=3x^2-5x$.  [2 Marks]

Solution:

Answer: Zeroes are $0,{\displaystyle \frac{5}{3}}$.
Sum of zeroes $=0+{\displaystyle \frac{5}{3}}={\displaystyle \frac{5}{3}}={\displaystyle \frac{-\text{coefficient of }x}{\text{coefficient of }{x}^2}}$
Product of zeroes $=0\times {\displaystyle \frac{5}{3}}={\displaystyle \frac{0}{3}}={\displaystyle \frac{\text{constant term}}{\text{coefficient of }{x}^2}}$

Chapter: Polynomials

Q8 (a). If $\alpha ,\beta$ are the zeroes of polynomial $p(x)=6x^2-5x-3$, then find the value of ${\displaystyle \frac{1}{\alpha }}+{\displaystyle \frac{1}{\beta }}$.  [2 Marks]

Solution:

Answer:
$\alpha +\beta ={\displaystyle \frac{5}{6}},\alpha \beta ={\displaystyle \frac{-3}{6}}$
$$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-5}{3}$$
Chapter: Polynomials

Q8 (b). One zero of the polynomial $4x^2-12x+(2k+1)$ is five times the other. Find the value of k.  [2 Marks]

Solution:

Answer:
Let zeroes of the polynomial be $\alpha$ and $5\alpha$.
$\alpha +5\alpha ={\displaystyle \frac{12}{4}}\Rightarrow \alpha ={\displaystyle \frac{1}{2}}$
$\alpha \times 5\alpha ={\displaystyle \frac{2k+1}{4}}$
$k=2$

Chapter:Polynomials

Q9(a). If $\alpha$, $\beta$ are zeroes of the polynomial $p(x)=5x^2-7x-3$, then form a quadratic polynomial whose zeroes are ${\displaystyle \frac{2}{\alpha }}$ and ${\displaystyle \frac{2}{\beta }}$.    [3 Marks]

Solution:

Answer:
$\alpha +\beta ={\displaystyle \frac{7}{5}}$, $\alpha \beta =-{\displaystyle \frac{3}{5}}$
Sum of zeroes of required polynomial:$$\frac{2}{\alpha }+\frac{2}{\beta }=\frac{2(\alpha +\beta )}{\alpha \beta }=\frac{2\times \frac{7}{5}}{-\frac{3}{5}}=-\frac{14}{3}$$Product of zeroes of required polynomial:$$\frac{2}{\alpha }\times \frac{2}{\beta }=\frac{4}{\alpha \beta }=\frac{4}{-\frac{3}{5}}=-\frac{20}{3}$$Required polynomial:$$x^2+\frac{14}{3}x-\frac{20}{3}\text{or}k(3x^2+14x-20)\text{ where }k\text{ is any non-zero real number.}$$ 

Chapter:Polynomials

Q9(b). Find the zeroes of the polynomial $p(x)=3x^2+7x-20$ and verify the relationship between its zeroes and the coefficients.    [3 Marks]

Solution:

Answer:
Zeroes are $-4$ and ${\displaystyle \frac{5}{3}}$.
$$\text{Sum of the zeroes}=-4+\frac{5}{3}=-\frac{7}{3}=-\frac{\text{coefficient of }x}{\text{coefficient of }{x}^2}$$$$\text{Product of zeroes}=-4\times \frac{5}{3}=-\frac{20}{3}=\frac{\text{constant term}}{\text{coefficient of }{x}^2}$$ 

Chapter:Polynomials

Q10. Find a quadratic polynomial whose sum and product of zeroes are 0 and $-9$, respectively. Also, find the zeroes of the polynomial so obtained.   [3 Marks]

Solution:

Answer:
Quadratic polynomial $=x^2-(\text{sum of zeroes})x+(\text{product of zeroes})$
$$=x^2-(0)x+(-9)=x^2-9$$For zeroes : $x^2-9=0\Rightarrow (x-3)(x+3)=0$
$$\text{Zeroes}=\mathbf{3}\text{ and }-\mathbf{3}$$

Chapter : Polynomials

Previous Year Questions 2025

Q1: Zeroes of the polynomial p(y)   (1 Mark)
(a) 
(b) 
(c) 
(d) 

Solution:

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:  (1 Mark)
(a) 2,√2 
(b) 2√2, √2 
(c) 4√2, -√2 
(d) √2, 2 

Solution:

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   (2 Marks)

Solution:

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:  (1 Mark)

(a) 3
(b) 5
(c) 2
(d) 4

Solution:

Ans: (c)
Zeroes of a polynomial are the x-coordinates of the points where its graph cuts the x-axis. From the given figure, both polynomial graphs intersect the x-axis at the same two points. Since these intersection points are common to both curves, the total number of distinct zeroes of the two polynomials together 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:  (1 Mark)
(a) a + b 
(c) a- b 
(b) -a - b 
(d) -a + b

Solution:

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:  (1 Mark)
(a) 
(b) 
(c) 3/2
(d) 2/3

Solution:

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:  (1 Mark)
(a) -8 
(b) 8 
(c) -4 
(d) 4

Solution:

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  (1 Mark)
(a) 2 
(b) 1/2
(c) -2
(d) 

Solution:

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'.  (2 Marks)

Solution:

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   (2 Marks)

Solution:

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 .  (2 Marks)

Solution:

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.   (2 Marks)

Solution:

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).  (5 Marks)

Solution:

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.

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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?  (1 Mark)
(a) 1
(b) 2
(c) 4
(d) 5

Solution:

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. (3 Marks)

Solution:

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.

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  (1 Mark)

(a) 3
(b) 1
(c) 2
(d) 0

Solution:

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  (1 Mark)
(a) 1
(b) 2
(c) -1
(d) -1/2

Solution:

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  (1 Mark)
(a) 1
(b) 2
(c) -1
(d) 0 

Solution:

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  (1 Mark)
(a) 7/3
(b) -7/3
(c) 3/7
(d) -3/7

Solution:

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. (3 Marks)

Solution:

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  (1 Mark)
(a) 4/3
(b) -4/3
(c) 2/3
(d) -2/3

Solution:

Ans: (a)
 Given that -3 is a zero of quadratic polynomial (k - 1)x2+ 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  (1 Mark)
(a) x2 + x + 2
(b) x2 - x + 2
(c) x2 - x - 2
(d) x2 + x - 2

Solution:

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  (1 Mark)
(a) 4
(b) 3
(c) 2
(d) 1 

Solution:

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  (1 Mark)

(a) 4 units
(b) 6 units
(c) 8 units
(d) 7 units

Solution:

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.  (1 Mark)
(a) x2 + 5x + 6
(b) x2 - 5x + 6
(c) x2 - 5 x - 6
(d) - x2 + 5x + 6 

Solution:

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

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Previous Year Questions 2021

Q1: If one zero of the quadratic polynomial x2 + 3x + k is 2 then find the value of k. (2 Marks)

Solution:

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  (1 Mark)
(a) 2
(b) 1
(c) more than 3
(d) 3 

Solution:

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  (1 Mark)
(a) 10
(b) - 10
(c) -7
(d) -2

Solution:

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 ________.  (1 Mark)
(a) x2 + 5x + 6
(b) x2 - 5x + 6
(c) x2- 5x - 6
(d) -x2 + 5x + 6

Solution:

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. (2 Marks)

Solution:

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:  (1 Mark) 
(a) m, m + 3 
(b) -m, m + 3 
(c) m, - (m + 3) 
(d) -m, - (m + 3)

Solution:

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.

Also read: Important Definitions & Formulas: Polynomials

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. (2 Marks)

Solution:

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