RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials (2024)

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Our subject specialists formulate these RD Sharma Solutions for Class 10 Maths Chapter 2 to help students with their exam preparation to achieve good marks in Maths. This exercise can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve problems. By practising RD Sharma Solutions for Class 10 Maths, students will be able to grasp the concepts perfectly. It also helps in boosting their confidence, which plays a crucial role in the examinations. Let’s get an insight into this chapter to get a better idea of what it’s about.

Polynomial and its types
Geometrical representation of linear and quadratic polynomials
The geometric meaning of the zeros of a polynomial
Relationship between the zeros and coefficients of a polynomial

Chapter 2 Polynomials

RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers
RD Sharma Solutions Class 10 Maths Chapter 2 Polynomials
RD Sharma Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
RD Sharma Solutions Class 10 Maths Chapter 4 Triangles
RD Sharma Solutions Class 10 Maths Chapter 5 Trigonometric Ratios
RD Sharma Solutions Class 10 Maths Chapter 6 Trigonometric Identities
RD Sharma Solutions Class 10 Maths Chapter 7 Statistics
RD Sharma Solutions Class 10 Maths Chapter 8 Quadratic Equations
RD Sharma Solutions Class 10 Maths Chapter 9 Arithmetic Progressions
RD Sharma Solutions Class 10 Maths Chapter 10 Circles
RD Sharma Solutions Class 10 Maths Chapter 11 Constructions
RD Sharma Solutions Class 10 Maths Chapter 12 Some Applications of Trigonometry
RD Sharma Solutions Class 10 Maths Chapter 13 Probability
RD Sharma Solutions Class 10 Maths Chapter 14 Co-ordinate Geometry
RD Sharma Solutions Class 10 Maths Chapter 15 Areas Related to Circles
RD Sharma Solutions Class 10 Maths Chapter 16 Surface Areas and Volumes

Exercise

Exercise 2.1 Chapter 2 Polynomials
Exercise 2.2 Chapter 2 Polynomials
Exercise 2.3 Chapter 2 Polynomials

RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

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RD Sharma Solutions for Class 10 Maths Chapter 2 Exercise 2.1 Page No: 2.33

1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

(i) f(x) = x²– 2x – 8

Solution:

Given,

f(x) = x²– 2x – 8

To find the zeros, we put f(x) = 0

⇒ x²– 2x – 8 = 0

⇒ x²– 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x – 4)(x + 2) = 0

This gives us 2 zeros, for

x = 4 and x = -2

Hence, the zeros of the quadratic equation are 4 and -2.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x²

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x²

4 x (-2) = (-8) / 1

-8 = -8

Therefore, the relationship between zeros and their coefficients is verified.

(ii) g(s) = 4s²– 4s + 1

Solution:

Given,

g(s) = 4s²– 4s + 1

To find the zeros, we put g(s) = 0

⇒ 4s²– 4s + 1 = 0

⇒ 4s²– 2s – 2s + 1= 0

⇒ 2s(2s – 1) – (2s – 1)= 0

⇒ (2s – 1)(2s – 1) = 0

This gives us 2 zeros, for

s = 1/2 and s = 1/2

Hence, the zeros of the quadratic equation are 1/2 and 1/2.

Now, for verification,

Sum of zeros = – coefficient of s / coefficient of s²

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s²

1/2 x 1/2 = 1/4

1/4 = 1/4

Therefore, the relationship between zeros and their coefficients is verified.

(iii) h(t)=t²– 15

Solution:

Given,

h(t) = t²– 15 = t²+(0)t – 15

To find the zeros, we put h(t) = 0

⇒ t²– 15 = 0

⇒ (t + √15)(t – √15)= 0

This gives us 2 zeros, for

t = √15 and t = -√15

Hence, the zeros of the quadratic equation are √15 and -√15.

Now, for verification,

Sum of zeros = – coefficient of t / coefficient of t²

√15 + (-√15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t²

√15 x (-√15) = -15/1

-15 = -15

Therefore, the relationship between zeros and their coefficients is verified.

(iv) f(x) = 6x²– 3 – 7x

Solution:

Given,

f(x) = 6x²– 3 – 7x

To find the zeros, we put f(x) = 0

⇒ 6x²– 3 – 7x = 0

⇒ 6x²– 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

This gives us 2 zeros, for

x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x²

3/2 + (-1/3) = – (-7) / 6

7/6 = 7/6

Product of roots = constant / coefficient of x²

3/2 x (-1/3) = (-3) / 6

-1/2 = -1/2

Therefore, the relationship between zeros and their coefficients is verified.

(v) p(x) = x²+ 2√2x – 6

Solution:

Given,

p(x) = x²+ 2√2x – 6

To find the zeros, we put p(x) = 0

⇒ x²+ 2√2x – 6 = 0

⇒ x²+ 3√2x – √2x – 6 = 0

⇒ x(x + 3√2) – √2 (x + 3√2) = 0

⇒ (x – √2)(x + 3√2) = 0

This gives us 2 zeros, for

x = √2 and x = -3√2

Hence, the zeros of the quadratic equation are √2 and -3√2.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x²

√2 + (-3√2) = – (2√2) / 1

-2√2 = -2√2

Product of roots = constant / coefficient of x²

√2 x (-3√2) = (-6) / 2√2

-3 x 2 = -6/1

-6 = -6

Therefore, the relationship between zeros and their coefficients is verified.

(vi) q(x)=√3x²+ 10x + 7√3

Solution:

Given,

q(x) = √3x²+ 10x + 7√3

To find the zeros, we put q(x) = 0

⇒ √3x²+ 10x + 7√3 = 0

⇒ √3x²+ 3x +7x + 7√3x = 0

⇒ √3x(x + √3) + 7 (x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

This gives us 2 zeros, for

x = -√3 and x = -7/√3

Hence, the zeros of the quadratic equation are -√3 and -7/√3.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x²

-√3 + (-7/√3) = – (10) /√3

(-3-7)/ √3 = -10/√3

-10/ √3 = -10/√3

Product of roots = constant / coefficient of x²

(-√3) x (-7/√3) = (7√3) / √3

7 = 7

Therefore, the relationship between zeros and their coefficients is verified.

(vii) f(x) = x²– (√3 + 1)x + √3

Solution:

Given,

RD Sharma Solutions for Class 10 Maths Chapter 2 Exercise 2.2 Page No: 2.43

1. Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeros and coefficients in each of the following cases:

(i) f(x) = 2x³+ x²– 5x + 2; 1/2, 1, -2

Solution:

Given, f(x) = 2x³+ x²– 5x + 2, where a= 2, b= 1, c= -5 and d= 2

For x = 1/2

f(1/2) = 2(1/2)³ + (1/2)² – 5(1/2) + 2

= 1/4 + 1/4 – 5/2 + 2 = 0

⇒ f(1/2) = 0, hence, x = 1/2 is a root of the given polynomial.

For x = 1

f(1) = 2(1)³ + (1)² – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

⇒ f(1) = 0, hence, x = 1 is also a root of the given polynomial.

For x = -2

f(-2) = 2(-2)³ + (-2)² – 5(-2) + 2

= -16 + 4 + 10 + 2 = 0

⇒ f(-2) = 0, hence, x = -2 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1/2 + 1 – 2 = – (1)/2

-1/2 = -1/2

Sum of the products of the zeros taken two at a time = c/a

(1/2 x 1) + (1 x -2) + (1/2 x -2) = -5/ 2

1/2 – 2 + (-1) = -5/2

-5/2 = -5/2

Product of zeros = – d/a

1/2 x 1 x (– 2) = -(2)/2

-1 = -1

Hence, the relationship between the zeros and coefficients is verified.

(ii) g(x) = x³– 4x²+ 5x – 2; 2, 1, 1

Solution:

Given, g(x) = x³– 4x²+ 5x – 2, where a= 1, b= -4, c= 5 and d= -2

For x = 2

g(2) = (2)³ – 4(2)² + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

⇒ f(2) = 0, hence x = 2 is a root of the given polynomial.

For x = 1

g(1) = (1)³ – 4(1)² + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

⇒ g(1) = 0, hence, x = 1 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1 + 1 + 2 = – (-4)/1

4 = 4

Sum of the products of the zeros taken two at a time = c/a

(1 x 1) + (1 x 2) + (2 x 1) = 5/ 1

1 + 2 + 2 = 5

5 = 5

Product of zeros = – d/a

1 x 1 x 2 = -(-2)/1

2 = 2

Hence, the relationship between the zeros and coefficients is verified.

2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeros as 3, -1 and -3, respectively.

Solution:

Generally,

A cubic polynomial, say, f(x), is of the form ax³+ bx²+ cx + d.

And, can be shown w.r.t its relationship between roots as.

⇒ f(x) = k [x³ – (sum of roots)x² + (sum of products of roots taken two at a time)x – (product of roots)]

Where k is any non-zero real number.

Here,

f(x) = k [x³ – (3)x² + (-1)x – (-3)]

∴ f(x) = k [x³ – 3x² – x + 3)]

Where k is any non-zero real number.

3. If the zeros of the polynomial f(x) = 2x³– 15x²+ 37x – 30 are in A.P., find them.

Solution:

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)

And given, the zeros are in A.P.

So, let’s consider the roots as

α = a – d, β = a and γ = a +d

Where a is the first term and d is a common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

⇒ Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 ⇒ a = 5/2

⇒ Product of roots = (a – d) x (a) x (a + d) = a(a² –d²) = -d/a = -(30)/2 = 15

⇒ a(a² –d²) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)² –d²] = 15

⇒ 5[(25/4) –d²] = 30

⇒ (25/4) – d² = 6

⇒ 25 – 4d² = 24

⇒ 1 = 4d²

∴ d = 1/2 or -1/2

Taking d = 1/2 and a = 5/2

We get,

The zeros as 2, 5/2 and 3

Taking d = -1/2 and a = 5/2

We get,

The zeros as 3, 5/2 and 2

RD Sharma Solutions for Class 10 Maths Chapter 2 Exercise 2.3 Page No: 2.57

1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:

(i)f(x) = x³– 6x²+ 11x – 6, g(x) = x²+ x +1

Solution:

Given,

f(x) = x³– 6x²+11x – 6, g(x) = x²+x + 1

Thus,

q(x) = x – 7 and r(x) = 17x +1

(ii) f(x) =10x⁴+ 17x³– 62x²+ 30x – 3, g(x) = 2x²+ 7x + 1

Solution:

Given,

f(x) =10x⁴+ 17x³– 62x²+ 30x – 3 and g(x) = 2x²+ 7x + 1

Thus,

q(x) = 5x² – 9x – 2 and r(x) = 53x – 1

(iii)f(x) = 4x³ + 8x² + 8x + 7, g(x)= 2x²– x + 1

Solution:

Given,

f(x) = 4x³ + 8x² + 8x + 7 and g(x)= 2x²– x + 1

Thus,

q(x) = 2x + 5 and r(x) = 11x + 2

(iv)f(x) = 15x³– 20x² + 13x – 12, g(x) = x²– 2x + 2

Solution:

Given,

f(x) = 15x³– 20x² + 13x – 12 and g(x) = x²– 2x + 2

Thus,

q(x) = 15x + 10 and r(x) = 3x – 32

2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

(i)g(t) = t²–3; f(t)=2t⁴ + 3t³ – 2t² – 9t – 12

Solution:

Given,

g(t) = t²– 3; f(t) =2t⁴ + 3t³ – 2t² – 9t – 12

Since the remainder r(t) = 0, we can say that the first polynomial is a factor of the second polynomial.

(ii)g(x) = x³ – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1

Solution:

Given,

g(x) = x³ – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1

Since the remainder r(x) = 2 and is not equal to zero, we can say that the first polynomial is not a factor of the second polynomial.

(iii) g(x) = 2x²– x + 3; f(x) = 6x⁵− x⁴ + 4x³ – 5x² – x –15

Solution:

Given,

g(x) = 2x²– x + 3; f(x)=6x⁵− x⁴ + 4x³ – 5x² – x –15

Since the remainder r(x) = 0, we can say that the first polynomial is not a factor of the second polynomial.

3. Obtain all zeroes of the polynomial f(x)= 2x⁴+ x³– 14x²– 19x–6, if two of its zeroes are -2 and -1.

Solution:

Given,

f(x)= 2x⁴+ x³– 14x²– 19x – 6

If the two zeros of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)

⇒ (x+2)(x+1) = x²+ x + 2x + 2 = x²+ 3x +2 …… (i)

This means that (i) is a factor of f(x). So, by performing the division algorithm, we get

The quotient is 2x² – 5x – 3.

⇒ f(x)= (2x² – 5x – 3)( x²+ 3x +2)

For obtaining the other 2 zeros of the polynomial,

We put,

2x² – 5x – 3 = 0

⇒ (2x + 1)(x – 3) = 0

∴ x = -1/2 or 3

Hence, all the zeros of the polynomial are -2, -1, -1/2 and 3.

4. Obtain all zeroes of f(x) = x³+ 13x²+ 32x + 20, if one of its zeros is -2.

Solution:

Given,

f(x)= x³+ 13x²+ 32x + 20

And -2 is one of the zeros. So, (x + 2) is a factor of f(x),

Performing the division algorithm, we get

⇒ f(x)= (x² + 11x + 10)( x + 2)

So, by putting x² + 11x + 10 = 0, we can get the other 2 zeros.

⇒ (x + 10)(x + 1) = 0

∴ x = -10 or -1

Hence, all the zeros of the polynomial are -10, -2 and -1.

5. Obtain all zeroes of the polynomialf(x) = x⁴ – 3x³ – x² + 9x – 6, if the two of its zeroes are−√3 and √3.

Solution:

Given,

f(x) = x⁴ – 3x³ – x² + 9x – 6

Since two of the zeroes of the polynomial are −√3 and √3 so, (x + √3) and (x–√3) are factors of f(x).

⇒ x²– 3 is a factor of f(x). Hence, performing division algorithm, we get

⇒ f(x)= (x² – 3x + 2)( x²– 3)

So, by putting x² – 3x + 2 = 0, we can get the other 2 zeros.

⇒ (x – 2)(x – 1) = 0

∴ x = 2 or 1

Hence, all the zeros of the polynomial are −√3, 1, √3 and 2.

6. Obtain all zeroes of the polynomialf(x)= 2x⁴– 2x³– 7x²+ 3x + 6, if the two of its zeroes are−√(3/2) and √(3/2).

Solution:

Given,

f(x)= 2x⁴– 2x³– 7x²+ 3x + 6

Since two of the zeroes of the polynomial are −√(3/2) and √(3/2) so, (x + √(3/2)) and (x –√(3/2)) are factors of f(x).

⇒ x²– (3/2) is a factor of f(x). Hence, performing division algorithm, we get

⇒ f(x)= (2x² – 2x – 4)( x²– 3/2)= 2(x² – x – 2)( x²– 3/2)

So, by putting x² – x – 2 = 0, we can get the other 2 zeros.

⇒ (x – 2)(x + 1) = 0

∴ x = 2 or -1

Hence, all the zeros of the polynomial are −√(3/2), -1, √(3/2) and 2.

7. Find all the zeroes of the polynomialx⁴+ x³– 34x²– 4x + 120, if the two of its zeros are 2 and -2.

Solution:

Let,

f(x) = x⁴ + x³ – 34x² – 4x + 120

Since two of the zeroes of the polynomial are −2 and 2 so, (x + 2) and (x – 2) are factors of f(x).

⇒ x²– 4 is a factor of f(x). Hence, performing division algorithm, we get

⇒ f(x)= (x² + x – 30)( x²– 4)

So, by putting x² + x – 30 = 0, we can get the other 2 zeros.

⇒ (x + 6)(x – 5) = 0

∴ x = -6 or 5

Hence, all the zeros of the polynomial are 5, -2, 2 and -6.

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What are Polynomials according to RD Sharma Solutions for Class 10 Maths Chapter 2?

Polynomials are algebraic expressions that consist of variables and coefficients. Variables are also sometimes called indeterminates. We can perform arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variables. By referring to these solutions, students get to clear their doubts instantly and also can exercise additional questions.

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