Important Questions Class 10 Maths Chapter 2 Polynomials

Important questions for Class 10 Maths Chapter 2 Polynomials are provided here as per the NCERT book. These questions are provided with answers that 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 2022-2023. Along with the important questions, students will also find detailed solutions. In case they get stuck while solving the questions, then they can refer to the solutions.

We have also provided some additional important questions from the examination point of view for Chapter 2 Polynomials of Class 10 Maths so that students can revise the chapter by solving these problems instantly. This chapter deals with expressions carrying unknown variables with different degrees. The solutions of these expressions are termed 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.

Students can also get access to Class 10 Maths Chapter 2 Polynomials MCQs here.

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.

Q.1: 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

Q.2: 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.

Q.3: 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)

Q.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:

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

Q.5: 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

Q.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.

=> 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.

Q.7: 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

Q.8: 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.

Q.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:-
x² – (α + β)x + αβ

= x² – (1/4)x + (-1)

= 4x² – x – 4

Thus, 4x² – 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:-

x² – (α + β)x + αβ

= x² – x + 1

Thus, x² – 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:-

x² – (α + β)x + αβ

= x² – 4x + 1

Thus, x² – 4x +1 is the quadratic polynomial.

Q.10: Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 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)] = x²-(5/3)

Therefore, 3x² + 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.

Q.11: 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)]

Q.12: Divide the polynomial f(x) = 3x² – x³ – 3x + 5 by the polynomial g(x) = x – 1 – x² and verify the division algorithm.

Solution:

Given,

f(x) = 3x² – x³ – 3x + 5

g(x) = x – 1 – x²

Dividing f(x) = 3x² – x³ – 3x + 5 by g(x) = x – 1 – x²

Here,

Quotient = q(x) = x – 2

Remainder = r(x) = 3

By division algorithm of polynomials,

Dividend = (Quotient × Divisor) + Remainder

So,

[q(x) × g(x)] + r(x) = (x – 2)(x – 1 – x²) + 3

= x² – x – x³ – 2x + 2 + 2x² + 3

= 3x² – x³ – 3x + 5

= f(x)

Hence, the division algorithm is verified.

Q.13: For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?

Solution:

Given,

f(x) = 3x⁴ – 9x³ + x² + 15x + k

g(x) = 3x² – 5

Dividing f(x) by g(x),

Given that f(x) is completely divisible by 3x² – 5.

So, the remainder = 0

k + 10 = 0

k = -10

Q.14: If 4 is a zero of the cubic polynomial x³ – 3x² – 10x + 24, find its other two zeroes.

Solution:

Given cubic polynomial  is p(x) = x³ – 3x² – 10x + 24

4 is a zero of p(x).

So, (x – 4) is the factor of p(x).

Let us divide the given polynomial by (x – 4).

Here, the quotient = x² + x – 6

= x² + 3x – 2x – 6

= x(x + 3) – 2(x + 3)

= (x – 2)(x + 3)

Therefore, the other two zeroes of the given cubic polynomial are 2 and -3.

Practice Questions for Class 10 Maths Chapter 2 Polynomials

1. Find the zeroes of the quadratic polynomial 7y² – (11/3)y – (2/3) and verify the relationship between the zeroes and the coefficients.
2. Find all zeroes of the polynomial (2x⁴ – 9x³ + 5x² + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3).
3. If the remainder on division of x³ + 2x² + kx +3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx – 18.
4. For which values of a and b, are the zeroes of q(x) = x³ + 2x² + a also the zeroes of the polynomial p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + b? Which zeroes of p(x) are not the zeroes of q(x)?
5. Verify that 3, –1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
6. Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2, respectively. Also find its zeroes.