Important questions of Chapter 2 Polynomials for Class 10 Maths are provided here as per the NCERT book. These questions with answers will help the students prepare and score well in the CBSE Class 10 Maths exam. The questions here cover the latest syllabus as prescribed by the board. Also, they are based on the latest exam pattern. Students can practice the important questions for all the chapters of 10th standard Maths subject and prepare well for the board exam. Along with the important questions, students will also find detailed solutions. In case they get stuck while practising the questions, then they can refer to the solutions.

We have also provided some additional questions so that students can revise the chapter by solving these problems in an instant. This chapter deals with expressions carrying unknown variables with varying degrees. The solutions of these expressions are termed as zeros of the polynomial. The polynomial with degree 1 is the linear polynomial, with degree 2 is a quadratic polynomial, with degree 3 is a cubic polynomial.

**Also Check:**

- Important 2 Marks Questions for CBSE 10th Maths
- Important 3 Marks Questions for CBSE 10th Maths
- Important 4 Marks Questions for CBSE 10th Maths

## Class 10 Maths Chapter 2 Important Questions With Solutions

A few important Class 10 polynomials questions are provided below with solutions. These questions include both short and long answer questions to let the students get acquainted with the in-depth concepts.

**1. Find the value of “p” from the polynomial x ^{2} + 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 x^{2} + 3x + p,

⇒ 2^{2} + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

**2. Does the polynomial a ^{4} + 4a^{2} + 5 have real zeroes?**

**Solution:**

In the aforementioned polynomial, let a^{2} = x.

Now, the polynomial becomes,

x^{2} + 4x + 5

Comparing with ax^{2} + bx + c,

Here, b^{2} – 4ac = 4^{2 }– 4(1)(5) = 16 – 20 = -4

So, D = b^{2} – 4ac < 0

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

**3. Compute the zeroes of the polynomial 4x ^{2} – 4x – 8. Also, establish a relationship between the zeroes and coefficients.**

**Solution:**

Let the given polynomial be p(x) = 4x^{2} – 4x – 8

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

Now, factorise the equation 4x^{2} – 4x – 8 = 0

4x^{2} – 4x – 8 = 0

4(x^{2} – x – 2) = 0

x^{2} – x – 2 = 0

x^{2} – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x – 2)(x + 1) = 0

x = 2, x = -1

So, the roots of 4x^{2} – 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 x^{2})

Relation between the product of zeroes and coefficients:

(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x^{2})

**4. 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:

x^{2} – (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:

x^{2} – (0 + √5)x + 0(√5)

= x^{2} – √5x

**5. Find the value of “x” in the polynomial 2a ^{2} + 2xa + 5a + 10 if (a + x) is one of its factors.**

**Solution:**

Let f(a) = 2a^{2} + 2xa + 5a + 10

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

So, f(-x) = 2x^{2} – 2x^{2} – 5x + 10 = 0

-5x + 10 = 0

5x = 10

x = 10/5

Therefore, x = 2

**6. 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.

=> x^{2} + 9 – 6x – 4

= x^{2} – 6x + 5

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

Now, solve x^{2} – 6x + 5 = 0 to get the roots.

So, x^{2} – 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.

**7. α and β are zeroes of the quadratic polynomial x ^{2} – 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

**8. If the zeroes of the polynomial x^{3 }– 3x^{2 }+ x + 1 are a – b, a, a + b, **

**then find the value of a and b.**

**Solution:**

Let the given polynomial be:

p(x) = x^{3 }– 3x^{2 }+ 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.

px^{3 }+ qx^{2} + rx + s = x^{3 }– 3x^{2} + 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 – 𝑏^{2}

=> -1/1 = 1 – 𝑏^{2}

Or, 𝑏^{2} = 1 + 1 =2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥^{3} − 3𝑥^{2} + 𝑥 + 1.

**9. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.**

**(i) 1/4, -1**

**(ii) 1, 1**

**(iii) 4, 1**

**Solution:**

(i) From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α + β

Product of zeroes = αβ

Given,

Sum of zeroes = 1/4

Product of zeroes = -1

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x^{2} – (α + β)x + αβ

= x^{2} – (1/4)x + (-1)

= 4x^{2} – x – 4

Thus, 4x^{2} – x – 4 is the required quadratic polynomial.

(ii) Given,

Sum of zeroes = 1 = α + β

Product of zeroes = 1 = αβ

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x^{2} – (α + β)x + αβ

= x^{2} – x + 1

Thus, x^{2} – x + 1 is the quadratic polynomial.

(iii) Given,

Sum of zeroes, α + β = 4

Product of zeroes, αβ = 1

Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x^{2} – (α + β)x + αβ

= x^{2} – 4x + 1

Thus, x^{2} – 4x +1 is the quadratic polynomial.

**10. Obtain all other zeroes of 3x ^{4 }+ 6x^{3 }– 2x^{2 }– 10x – 5, if two of its zeroes are √(5/3) and-√(5/3).**

**Solution:** Since this is a polynomial of degree 4, hence there will be a total of 4 roots.

√(5/3) and-√(5/3) are zeroes of polynomial f(x).

∴ [x-√(5/3)] [x+√(5/3)] = x^{2}-(5/3)

Therefore, 3x^{2 }+ 6x + 3 = 3x(x + 1) +3 (x + 1)

= (3x + 3)(x + 1)

= 3(x + 1)(x + 1)

= 3(x + 1)(x + 1)

Hence, x + 1 = 0 i.e. x = – 1 , – 1 is a zero of p(x).

So, its zeroes are given by: x = −1 and x = −1.

Therefore, all four zeroes of the given polynomial are:

√(5/3) and-√(5/3), −1 and −1.

nice and helpful questions

my doubt is of relationship between zeros ans coffieients

Please visit: https://byjus.com/maths/relation-between-zeros-and-coefficients/

relationship between zeroes and coefficients means that you have to show that alpha + beta = -b/a and alpha x beta = c/a.

good questions

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Please tell me some tough questions???

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happy with these questions .. will be more happy if got hard questions and i need some difficult questions on relationship between zeroes and coefficient

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I need some hard question from this chapter

Thank you for the questions, these really helped me clear my doubts.

Very useful questions for revision

Very easy but good question thanks byjus