CBSE Class 10 Maths Chapter 2 Polynomials Important Questions

Mathematics is one of the most important subjects in Class 10, and every chapter plays a big role in scoring good marks. Among them, Chapter 2 Polynomials is very important for both board exams and practice. To prepare well, students must go through class 10 maths chapter 2 extra questions because these questions help in understanding the concepts deeply. Practicing polynomials class 10 extra questions will not only strengthen your basics but also give you the confidence to solve any type of problem in exams.

In this chapter, students learn about the definition of polynomials, their types, and different methods to solve them. Questions related to factorization, finding zeros of polynomials, and verifying relationships between coefficients and zeros are very common in exams. That is why class 10 polynomials extra questions are designed in such a way that they cover all important topics of the chapter. By solving ch 2 maths class 10 extra questions, students can revise every important concept before the exam.

Practicing from chapter 2 maths class 10 extra questions also helps in improving speed and accuracy, which are very important during board exams. These extra questions give students an idea about the variety of questions that may appear in the exam.

The polynomial class 10 extra questions provided here are prepared by keeping in mind the exam pattern and previous year question papers. Students who practice class 10 maths ch 2 extra questions regularly can easily score better marks in the exam. Solving polynomials extra questions class 10 helps in removing doubts and clears concepts in a simple way.

In short, practicing polynomial extra questions class 10 is the best way to master this chapter. These class 10 maths chapter 2 extra questions not only improve problem-solving skills but also boost confidence. Whether you are preparing for school exams or the CBSE board exams, going through ch 2 maths class 10 extra questions is a must. By using these resources, students can make their preparation strong and score high in Mathematics.

Do Check: CBSE Class 10 Chapter 1 Important Questions

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

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

Do Check: CBSE Class 10 Maths Important Questions

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.

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)

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

Do Check: NCERT Solutions for Class 10 Maths

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.

Do Check: CBSE Class 10 Maths Sample Papers

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

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.

Do Check: CBSE Class 10 Maths Syllabus

Question 16. 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

Question 17: 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)]

Do Check: CBSE Class 10 Maths Notes

Question 18: 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.