Important Questions Class 10 Maths Chapter 2 Polynomials

Crucial questions for Chapter 2: Polynomials in Class 10 Mathematics have been provided here, following the NCERT book. These questions come with solutions to assist students in preparing effectively and achieving good scores in the CBSE Class 10 Mathematics exam. The questions are aligned with the latest syllabus prescribed by the board and are designed based on the current exam pattern. Students can utilize these important questions to practice for all chapters of the 10th standard Mathematics subject, thereby preparing thoroughly for the board exam of 2025-2025. Additionally, detailed solutions are included to aid students in case they encounter difficulties while solving the questions.

Also Check: CBSE Class 10 Maths Important Question Chapter 8 Trignometry

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Short and Long Important Question Answer for Class 10 Polynomials

Q1. If the sum of zeroes of the quadratic polynomial 3x² – kx + 6 is 3, then find the value of k. (2012)
Solution:
Here a = 3, b = -k, c = 6
Sum of the zeroes, (α + β) = −ba = 3 …..(given)
⇒ −(−k)3 = 3
⇒ k = 9

Q2: Find the value of “p” from the polynomial x² + 3x + p, if one of the zeroes of the polynomial is 2.

Solution: As 2 is the zero of the polynomial.

We know that if α is a zero of the polynomial p(x), then p(α) = 0

Substituting x = 2 in x² + 3x + p,

⇒ 2² + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

Also Check: Important Questions for Class 10 Maths Chapter 14 Statistics

Q3. If the sum of the zeroes of the polynomial p(x) = (k² – 14) x² – 2x – 12 is 1, then find the value of k. (2017 D)
Solution:
p(x) = (k² – 14) x² – 2x – 12
Here a = k² – 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 …[Given] ⇒ −ba = 1
⇒ −(−2)k2−14 = 1
⇒ k² – 14 = 2
⇒ k² = 16
⇒ k = ±4

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Q4: Does the polynomial a⁴ + 4a² + 5 have real zeroes?

Solution: In the aforementioned polynomial, let a² = x.

Now, the polynomial becomes,

x² + 4x + 5

Comparing with ax² + bx + c,

Here, b² – 4ac = 4² – 4(1)(5) = 16 – 20 = -4

So, D = b² – 4ac < 0

As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.

Q5. A quadratic polynomial, whose zeroes are -4 and -5, is …. (2016 D)
Solution:
x² + 9x + 20 is the required polynomial.

Also Check: Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13

Q6: Compute the zeroes of the polynomial 4x² – 4x – 8. Also, establish a relationship between the zeroes and coefficients.

Solution:

Let the given polynomial be p(x) = 4x² – 4x – 8

To find the zeroes, take p(x) = 0

Now, factorise the equation 4x² – 4x – 8 = 0

4x² – 4x – 8 = 0

4(x² – x – 2) = 0

x² – x – 2 = 0

x² – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x – 2)(x + 1) = 0

x = 2, x = -1

So, the roots of 4x² – 4x – 8 are -1 and 2.

Relation between the sum of zeroes and coefficients:

-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x²)

Relation between the product of zeroes and coefficients:

(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x²)

Also Check: Polynomials Class 10 Extra Questions Maths Chapter 2

Q7. Find the condition that zeroes of polynomial p(x) = ax² + bx + c are reciprocal of each other. (2017 OD)
Solution:
Let α and 1α be the zeroes of P(x).
P(a) = ax² + bx + c …(given)
Product of zeroes = ca
⇒ α × 1α = ca
⇒ 1 = ca
⇒ a = c (Required condition)
Coefficient of x² = Constant term

Q8.Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2. (2012)
Solution:
Sum of zeroes,
S = (3 + √2) + (3 – √2) = 6
Product of zeroes,
P = (3 + √2) x (3 – √2) = (3)² – (√2)² = 9 – 2 = 7
Quadratic polynomial = x² – Sx + P = x² – 6x + 7

Q.9: Find the quadratic polynomial if its zeroes are 0, √5.

Solution:

A quadratic polynomial can be written using the sum and product of its zeroes as:

x² – (α + β)x + αβ

Where α and β are the roots of the polynomial.

Here, α = 0 and β = √5

So, the polynomial will be:

x² – (0 + √5)x + 0(√5)

= x² – √5x

Q10: Find the value of “x” in the polynomial 2a² + 2xa + 5a + 10 if (a + x) is one of its factors.

Solution:

Let f(a) = 2a² + 2xa + 5a + 10

Since, (a + x) is a factor of 2a² + 2xa + 5a + 10, f(-x) = 0

So, f(-x) = 2x² – 2x² – 5x + 10 = 0

-5x + 10 = 0

5x = 10

x = 10/5

Therefore, x = 2

Q11: How many zeros does the polynomial (x – 3)² – 4 have? Also, find its zeroes.

Solution:

Given polynomial is (x – 3)² – 4

Now, expand this expression.

=> x² + 9 – 6x – 4

= x² – 6x + 5

As the polynomial has a degree of 2, the number of zeroes will be 2.

Now, solve x² – 6x + 5 = 0 to get the roots.

So, x² – x – 5x + 5 = 0

=> x(x – 1) -5(x – 1) = 0

=> (x – 1)(x – 5) = 0

x = 1, x = 5

So, the roots are 1 and 5.

Also Check: Important Questions for Class 10 Maths Chapter 1 – Real Numbers

Q12. Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively. (2015)
Solution:
Quadratic polynomial is
x² – (Sum of zeroes) x + (Product of zeroes)
= x² – (0)x + (-√2)
= x² – √2

Q13. Can (x – 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer. (2016 OD)
Solution:
In case of division of a polynomial by another polynomial, the degree of the remainder (polynomial) is always less than that of the divisor. (x – 2) can not be the remainder when p(x) is divided by (2x + 3) as the degree is the same.

Q14: α and β are zeroes of the quadratic polynomial x² – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.

Solution:

Let, f(x) = x² – 6x + y

From the given,

3α + 2β = 20———————(i)

From f(x),

α + β = 6———————(ii)

And,

αβ = y———————(iii)

Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),

=> α = 20 – 12 = 8

Now, substitute this value in equation (ii),

=> β = 6 – 8 = -2

Substitute the values of α and β in equation (iii) to get the value of y, such as;

y = αβ = (8)(-2) = -16

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Q15: If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, then find the value of a and b.

Solution:

Let the given polynomial be:

p(x) = x³ – 3x² + x + 1

Given,

The zeroes of the p(x) are a – b, a, and a + b.

Now, compare the given polynomial equation with general expression.

px³ + qx² + rx + s = x³ – 3x² + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

-q/p = 3a

Substitute the values q and p.

-(-3)/1 = 3a

a = 1

So, the zeroes are 1 – b, 1, 1 + b.

For the product of zeroes:

Product of zeroes = 1(1 – b)(1 + b)

-s/p = 1 – 𝑏²

=> -1/1 = 1 – 𝑏²

Or, 𝑏² = 1 + 1 =2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥³ − 3𝑥² + 𝑥 + 1.

Also Check: Important Questions for Class 10 Maths Chapter 4 Quadratic Equations

Question 20. Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes. (2014)
Solution: Let Sum of zeroes (α + β) = S = -8 …[Given] Product of zeroes (αβ) = P = 12 …[Given] Quadratic polynomial is x² – Sx + P
= x² – (-8)x + 12

= x² + 8x + 12
= x² + 6x + 2x + 12
= x(x + 6) + 2(x + 6)
= (x + 2)(x + 6)
Zeroes are:
x + 2 = 0 or x + 6 = 0
x = -2 or x = -6

Also Check: Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13

Q21: Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax² + bx + c, a ≠ 0, c ≠ 0.

Solution:

Let α and β be the zeroes of the polynomial f(x) = ax² + bx + c.

So, α + β = -b/a

αβ = c/a

According to the given, 1/α and 1/β are the zeroes of the required quadratic polynomial.

Now, the sum of zeroes = (1/α) + (1/β)

= (α + β)/αβ

= (-b/a)/ (c/a)

= -b/c

Product of two zeroes = (1/α) (1/β)

= 1/αβ

= 1/(c/a)

= a/c

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

= k[x² – (-b/c)x + (a/c)]

= k[x² + (b/c) + (a/c)]

Q22: If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, then find the value of a and b.

Solution:

Let the given polynomial be:

p(x) = x³ – 3x² + x + 1

Given,

The zeroes of the p(x) are a – b, a, and a + b.

Now, compare the given polynomial equation with general expression.

px³ + qx² + rx + s = x³ – 3x² + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

-q/p = 3a

Substitute the values q and p.

-(-3)/1 = 3a

a = 1

So, the zeroes are 1 – b, 1, 1 + b.

For the product of zeroes:

Product of zeroes = 1(1 – b)(1 + b)

-s/p = 1 – 𝑏²

=> -1/1 = 1 – 𝑏²

Or, 𝑏² = 1 + 1 =2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥³ − 3𝑥² + 𝑥 + 1.

Class 10 Maths Chapter 2 Polynomials FAQs