## RD Sharma Solutions Class 10 Maths Chapter 2 – Free PDF Download

**RD Sharma Solutions for Class 10 Maths Chapter 2 –Polynomials**are provided here for students to study and excel in their board exams. Mathematics is one of the scoring subjects in Class 10. And that’s why we, at BYJU’S, have created RD Sharma Solutions. These are created by our expert faculty in order to provide a clear understanding of important concepts by giving detailed explanations for the problems. These are very important study sources for any student to get high marks in their **Mathematics** examinations.

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) = x ^{2}– 2x – 8**

**Solution: **

Given,

f(x) = x^{2}– 2x – 8

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

⇒ x^{2}– 2x – 8 = 0

⇒ x^{2}– 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 x^{2}

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x^{2}

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

-8 = -8

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

**(ii) g(s) = 4s ^{2}– 4s + 1**

**Solution: **

Given,

g(s) = 4s^{2}– 4s + 1

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

⇒ 4s^{2}– 4s + 1 = 0

⇒ 4s^{2}– 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 s^{2}

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s^{2}

1/2 x 1/2 = 1/4

1/4 = 1/4

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

**(iii) h(t)=t ^{2}– 15**

**Solution: **

Given,

h(t) = t^{2}– 15 = t^{2}+(0)t – 15

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

⇒ t^{2}– 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 t^{2}

√15 + (-√15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t^{2}

√15 x (-√15) = -15/1

-15 = -15

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

**(iv) f(x) = 6x ^{2}– 3 – 7x**

**Solution: **

Given,

f(x) = 6x^{2}– 3 – 7x

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

⇒ 6x^{2}– 3 – 7x = 0

⇒ 6x^{2}– 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 x^{2}

3/2 + (-1/3) = – (-7) / 6

7/6 = 7/6

Product of roots = constant / coefficient of x^{2}

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

-1/2 = -1/2

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

**(v) p(x) = x ^{2}+ 2√2x – 6**

**Solution: **

Given,

p(x) = x^{2}+ 2√2x – 6

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

⇒ x^{2}+ 2√2x – 6 = 0

⇒ x^{2}+ 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 x^{2}

√2 + (-3√2) = – (2√2) / 1

-2√2 = -2√2

Product of roots = constant / coefficient of x^{2}

√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)=√3x ^{2}+ 10x + 7√3**

**Solution: **

Given,

q(x) = √3x^{2}+ 10x + 7√3

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

⇒ √3x^{2}+ 10x + 7√3 = 0

⇒ √3x^{2}+ 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 x^{2}

-√3 + (-7/√3) = – (10) /√3

(-3-7)/ √3 = -10/√3

-10/ √3 = -10/√3

Product of roots = constant / coefficient of x^{2}

(-√3) x (-7/√3) = (7√3) / √3

7 = 7

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

**(vii) f(x) = x ^{2}– (√3 + 1)x + √3**

**Solution: **

Given,

f(x) = x^{2}– (√3 + 1)x + √3

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

⇒ x^{2}– (√3 + 1)x + √3 = 0

⇒ x^{2}– √3x – x + √3 = 0

⇒ x(x – √3) – 1 (x – √3) = 0

⇒ (x – √3)(x – 1) = 0

This gives us 2 zeros, for

x = √3 and x = 1

Hence, the zeros of the quadratic equation are √3 and 1.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x^{2}

√3 + 1 = – (-(√3 +1)) / 1

√3 + 1 = √3 +1

Product of roots = constant / coefficient of x^{2}

1 x √3 = √3 / 1

√3 = √3

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

** (viii) g(x)=a(x ^{2}+1)–x(a^{2}+1)**

**Solution: **

Given,

g(x) = a(x^{2}+1)–x(a^{2}+1)

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

⇒ a(x^{2}+1)–x(a^{2}+1) = 0

⇒ ax^{2}+ a − a^{2}x – x = 0

⇒ ax^{2}− a^{2}x – x + a = 0

⇒ ax(x − a) − 1(x – a) = 0

⇒ (x – a)(ax – 1) = 0

This gives us 2 zeros, for

x = a and x = 1/a

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

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x^{2}

a + 1/a = – (-(a^{2 }+ 1)) / a

(a^{2} + 1)/a = (a^{2} + 1)/a

Product of roots = constant / coefficient of x^{2}

a x 1/a = a / a

1 = 1

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

**(ix) h(s) = 2s ^{2}– (1 + 2√2)s + √2**

**Solution: **

Given,

h(s) = 2s^{2}– (1 + 2√2)s + √2

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

⇒ 2s^{2}– (1 + 2√2)s + √2 = 0

⇒ 2s^{2}– 2√2s – s + √2 = 0

⇒ 2s(s^{}– √2) -1(s – √2) = 0

⇒ (2s – 1)(s – √2) = 0

This gives us 2 zeros, for

x = √2 and x = 1/2

Hence, the zeros of the quadratic equation are √3 and 1.

Now, for verification,

Sum of zeros = – coefficient of s / coefficient of s^{2}

√2 + 1/2 = – (-(1 + 2√2)) / 2

(2√2 + 1)/2 = (2√2 +1)/2

Product of roots = constant / coefficient of s^{2}

1/2 x √2 = √2 / 2

√2 / 2 = √2 / 2

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

**(x) f(v) = v ^{2}+ 4√3v – 15**

**Solution: **

Given,

f(v) = v^{2}+ 4√3v – 15

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

⇒ v^{2}+ 4√3v – 15 = 0

⇒ v^{2}+ 5√3v – √3v – 15 = 0

⇒ v(v + 5√3) – √3 (v + 5√3) = 0

⇒ (v – √3)(v + 5√3) = 0

This gives us 2 zeros, for

v = √3 and v = -5√3

Hence, the zeros of the quadratic equation are √3 and -5√3.

Now, for verification,

Sum of zeros = – coefficient of v / coefficient of v^{2}

√3 + (-5√3) = – (4√3) / 1

-4√3 = -4√3

Product of roots = constant / coefficient of v^{2}

√3 x (-5√3) = (-15) / 1

-5 x 3 = -15

-15 = -15

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

**(xi) p(y) = y ^{2}+ (3√5/2)y – 5**

**Solution: **

Given,

p(y) = y^{2}+ (3√5/2)y – 5

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

⇒ y^{2}+ (3√5/2)y – 5 = 0

⇒ y^{2}– √5/2 y + 2√5y – 5 = 0

⇒ y(y – √5/2) + 2√5 (y – √5/2) = 0

⇒ (y + 2√5)(y – √5/2) = 0

This gives us 2 zeros, for

y = √5/2 and y = -2√5

Hence, the zeros of the quadratic equation are √5/2 and -2√5.

Now, for verification,

Sum of zeros = – coefficient of y / coefficient of y^{2}

√5/2 + (-2√5) = – (3√5/2) / 1

-3√5/2 = -3√5/2

Product of roots = constant / coefficient of y^{2}

√5/2 x (-2√5) = (-5) / 1

– (√5)^{2} = -5

-5 = -5

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

**(xii) q(y) = 7y ^{2}– (11/3)y – 2/3**

**Solution: **

Given,

q(y) = 7y^{2}– (11/3)y – 2/3

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

⇒ 7y^{2}– (11/3)y – 2/3 = 0

⇒ (21y^{2} – 11y -2)/3 = 0

⇒ 21y^{2} – 11y – 2 = 0

⇒ 21y^{2} – 14y + 3y – 2 = 0

⇒ 7y(3y – 2) – 1(3y + 2) = 0

⇒ (3y – 2)(7y + 1) = 0

This gives us 2 zeros, for

y = 2/3 and y = -1/7

Hence, the zeros of the quadratic equation are 2/3 and -1/7.

Now, for verification,

Sum of zeros = – coefficient of y / coefficient of y^{2}

2/3 + (-1/7) = – (-11/3) / 7

-11/21 = -11/21

Product of roots = constant / coefficient of y^{2}

2/3 x (-1/7) = (-2/3) / 7

– 2/21 = -2/21

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

**2. For each of the following, find a quadratic polynomial whose sum and product, respectively, of the zeros are as given. Also, find the zeros of these polynomials by factorisation.**

**(i) -8/3 , 4/3**

**Solution: **

A quadratic polynomial formed for the given sum and product of zeros is given by

f(x) = x^{2} + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -8/3 and product of zero= 4/3

Thus,

The required polynomial f(x) is,

⇒ x^{2} – (-8/3)x + (4/3)

⇒ x^{2} + 8/3x + (4/3)

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

⇒ x^{2} + 8/3x + (4/3) = 0

⇒ 3x^{2} + 8x + 4 = 0

⇒ 3x^{2} + 6x + 2x + 4 = 0

⇒ 3x(x + 2) + 2(x + 2) = 0

⇒ (x + 2) (3x + 2) = 0

⇒ (x + 2) = 0 and, or (3x + 2) = 0

Therefore, the two zeros are -2 and -2/3.

**(ii) 21/8 , 5/16**

**Solution: **

A quadratic polynomial formed for the given sum and product of zeros is given by

f(x) = x^{2} + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = 21/8 and product of zero = 5/16

Thus,

The required polynomial f(x) is,

⇒ x^{2} – (21/8)x + (5/16)

⇒ x^{2} – 21/8x + 5/16

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

⇒ x^{2} – 21/8x + 5/16 = 0

⇒ 16x^{2} – 42x + 5 = 0

⇒ 16x^{2} – 40x – 2x + 5 = 0

⇒ 8x(2x – 5) – 1(2x – 5) = 0

⇒ (2x – 5) (8x – 1) = 0

⇒ (2x – 5) = 0 and, or (8x – 1) = 0

Therefore, the two zeros are 5/2 and 1/8.

**(iii) -2√3, -9**

**Solution: **

A quadratic polynomial formed for the given sum and product of zeros is given by

f(x) = x^{2} + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -2√3 and product of zero = -9

Thus,

The required polynomial f(x) is,

⇒ x^{2} – (-2√3)x + (-9)

⇒ x^{2} + 2√3x – 9

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

⇒ x^{2} + 2√3x – 9 = 0

⇒ x^{2} + 3√3x – √3x – 9 = 0

⇒ x(x + 3√3) – √3(x + 3√3) = 0

⇒ (x + 3√3) (x – √3) = 0

⇒ (x + 3√3) = 0 and, or (x – √3) = 0

Therefore, the two zeros are -3√3 and √3.

**(iv) -3/2√5, -1/2**

**Solution: **

A quadratic polynomial formed for the given sum and product of zeros is given by

f(x) = x^{2} + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -3/2√5 and product of zero = -1/2

Thus,

The required polynomial f(x) is,

⇒ x^{2} – (-3/2√5)x + (-1/2)

⇒ x^{2} + 3/2√5x – 1/2

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

⇒ x^{2} + 3/2√5x – 1/2 = 0

⇒ 2√5x^{2} + 3x – √5 = 0

⇒ 2√5x^{2} + 5x – 2x – √5 = 0

⇒ √5x(2x + √5) – 1(2x + √5) = 0

⇒ (2x + √5) (√5x – 1) = 0

⇒ (2x + √5) = 0 and, or (√5x – 1) = 0

Therefore, the two zeros are -√5/2 and 1/√5.

**3.** **Ifα and βare the zeros of the quadratic polynomialf(x) = x ^{2 }– 5x + 4, find the value of1/α + 1/β – 2αβ. **

**Solution: **

From the question, it’s given that

α and βare the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4

So, we can find

Sum of the roots =α+β= -b/a = – (-5)/1 = 5

Product of the roots =αβ= c/a = 4/1 = 4

To find, 1/α +1/β – 2αβ

⇒ [(α +β)/ αβ] – 2αβ

⇒ (5)/ 4 – 2(4) = 5/4 – 8 = -27/ 4

**4.** **Ifα and βare the zeros of the quadratic polynomialp(y) = 5y ^{2 }– 7y + 1, find the value of1/α+1/β.**

**Solution: **

From the question, it’s given that

α and βare the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1

So, we can find

Sum of the roots =α+β= -b/a = – (-7)/5 = 7/5

Product of the roots =αβ= c/a = 1/5

To find, 1/α +1/β

⇒ (α +β)/ αβ

⇒ (7/5)/ (1/5) = 7

**5.** **Ifα and βare the zeros of the quadratic polynomialf(x)=x ^{2} – x – 4, find the value of1/α+1/β–αβ.**

**Solution: **

From the question, it’s given that:

α and βare the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = – 4

So, we can find

Sum of the roots =α+β= -b/a = – (-1)/1 = 1

Product of the roots =αβ= c/a = -4 /1 = – 4

To find, 1/α +1/β – αβ

⇒ [(α +β)/ αβ] – αβ

⇒ [(1)/ (-4)] – (-4) = -1/4 + 4 = 15/ 4

**6**. **Ifα and βare the zeroes of the quadratic polynomialf(x) = x ^{2 }+ x – 2, find the value of1/α – 1/β.**

**Solution: **

From the question, it’s given that:

α and βare the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = – 2

So, we can find

Sum of the roots =α+β= -b/a = – (1)/1 = -1

Product of the roots =αβ= c/a = -2 /1 = – 2

To find 1/α – 1/β

⇒ [(β – α)/ αβ]

⇒

**7.** **If one of the zeros of the quadratic polynomial f(x) = 4x ^{2 }– 8kx – 9is negative of the other, then find the value of k.**

**Solution: **

From the question, it’s given that

The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9

And, for roots to be negative of each other, let the roots be α and – α.

So, we can find

Sum of the roots =α – α= -b/a = – (-8k)/1 = 8k = 0 [∵ α – α= 0]

⇒ k = 0

**8.** **If the sum of the zeroes of the quadratic polynomialf(t)=kt ^{2 }+ 2t + 3kis equal to their product, then find the value of k.**

**Solution: **

Given,

The quadratic polynomial f(t)=kt^{2 }+ 2t + 3k,where a = k, b = 2 and c = 3k.

And,

Sum of the roots = Product of the roots

⇒ (-b/a) = (c/a)

⇒ (-2/k) = (3k/k)

⇒ (-2/k) = 3

∴ k = -2/3

**9.** **Ifα and βare the zeros of the quadratic polynomialp(x) = 4x ^{2 }– 5x – 1, find the value ofα^{2}β+αβ^{2}.**

**Solution: **

From the question, it’s given that:

α and βare the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1

So, we can find

Sum of the roots =α+β= -b/a = – (-5)/4 = 5/4

Product of the roots =αβ= c/a = -1/4

To find, α^{2}β+αβ^{2}

⇒ αβ(α +β)

⇒ (-1/4)(5/4) = -5/16

**10**. **Ifα and βare the zeros of the quadratic polynomialf(t)=t ^{2}– 4t + 3, find the value ofα^{4}β^{3}+α^{3}β^{4}.**

**Solution:**

From the question, it’s given that:

α and βare the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3

So, we can find

Sum of the roots =α+β= -b/a = – (-4)/1 = 4

Product of the roots =αβ= c/a = 3/1 = 3

To find, α^{4}β^{3}+α^{3}β^{4}

⇒ α^{3}β^{3 }(α +β)

⇒ (αβ)^{3 }(α +β)

⇒ (3)^{3 }(4) = 27 x 4 = 108

### 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) = 2x ^{3 }+ x^{2 }– 5x + 2; 1/2, 1, -2**

**Solution: **

Given, f(x) = 2x^{3 }+ x^{2 }– 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) = x ^{3 }– 4x^{2 }+ 5x – 2; 2, 1, 1**

**Solution: **

Given, g(x) = x^{3 }– 4x^{2 }+ 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 ax^{3 }+ bx^{2 }+ cx + d.

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

⇒ f(x) = k [x^{3} – (sum of roots)x^{2} + (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 [x^{3} – (3)x^{2} + (-1)x – (-3)]

∴ f(x) = k [x^{3} – 3x^{2} – x + 3)]

Where k is any non-zero real number.

**3.** **If the zeros of the polynomial f(x) = 2x ^{3}– 15x^{2}+ 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(a^{2} –d^{2}) = -d/a = -(30)/2 = 15

⇒ a(a^{2} –d^{2}) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)^{2} –d^{2}] = 15

⇒ 5[(25/4) –d^{2}] = 30

⇒ (25/4) – d^{2} = 6

⇒ 25 – 4d^{2} = 24

⇒ 1 = 4d^{2}

∴ 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) = x ^{3}– 6x^{2 }+ 11x – 6, g(x) = x^{2 }+ x +1**

**Solution: **

Given,

f(x) = x^{3}– 6x^{2 }+11x – 6, g(x) = x^{2 }+x + 1

Thus,

q(x) = x – 7 and r(x) = 17x +1

**(ii) f(x) =10x ^{4 }+ 17x^{3 }– 62x^{2 }+ 30x – 3, g(x) = 2x^{2}+ 7x + 1**

**Solution:**

**Given,**

**f(x) =**10x^{4 }+ 17x^{3 }– 62x^{2 }+ 30x – 3 and g(x) = 2x^{2}**+ 7x + 1**

**Thus,**

q(x) = 5x^{2} – 9x – 2 and r(x) = 53x – 1

**(iii)f(x) = 4x ^{3} + 8x^{2} + 8x + 7, g(x)= 2x^{2 }– x + 1**

**Solution:**

Given,

f(x) = 4x^{3} + 8x^{2} + 8x + 7 and g(x)= 2x^{2 }– x + 1

**Thus,**

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

**(iv)f(x) = 15x ^{3 }– 20x^{2} + 13x – 12, g(x) = x^{2 }– 2x + 2**

**Solution:**

Given,

f(x) = 15x^{3 }– 20x^{2} + 13x – 12 and g(x) = x^{2 }– 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) = t ^{2}–3; f(t)=2t^{4} + 3t^{3} – 2t^{2} – 9t – 12**

**Solution: **

Given,

g(t) = t^{2 }– 3; f(t) =2t^{4} + 3t^{3} – 2t^{2} – 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) = x ^{3} – 3x + 1; f(x) = x^{5} – 4x^{3} + x^{2} + 3x + 1**

**Solution: **

Given,

g(x) = x^{3} – 3x + 1; f(x) = x^{5} – 4x^{3} + x^{2} + 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) = 2x ^{2}– x + 3; f(x) = 6x^{5 }− x^{4} + 4x^{3} – 5x^{2} – x –15**

**Solution: **

Given,

g(x) = 2x^{2}– x + 3; f(x)=6x^{5 }− x^{4} + 4x^{3} – 5x^{2} – 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)= 2x ^{4 }+ x^{3 }– 14x^{2 }– 19x–6, if two of its zeroes are -2 and -1.**

**Solution:**

Given,

f(x)= 2x^{4 }+ x^{3 }– 14x^{2 }– 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) = x^{2 }+ x + 2x + 2 = x^{2 }+ 3x +2 …… (i)

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

The quotient is 2x^{2} – 5x – 3.

⇒ f(x)= (2x^{2} – 5x – 3)( x^{2 }+ 3x +2)

For obtaining the other 2 zeros of the polynomial,

We put,

2x^{2} – 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) = x ^{3 }+ 13x^{2 }+ 32x + 20, if one of its zeros is -2.**

**Solution: **

Given,

**f(x)= **x^{3 }+ 13x^{2 }+ 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)= (x^{2} + 11x + 10)( x + 2)

So, by putting x^{2} + 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) = x ^{4} – 3x^{3} – x^{2} + 9x – 6, if the two of its zeroes are−√3 and √3.**

**Solution: **

**Given, **

f(x) = x^{4} – 3x^{3} – x^{2} + 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).

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

⇒ f(x)= (x^{2} – 3x + 2)( x^{2 }– 3)

So, by putting x^{2} – 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)= 2x ^{4 }– 2x^{3 }– 7x^{2 }+ 3x + 6, if the two of its zeroes are−√(3/2) and √(3/2).**

**Solution: **

**Given, **

f(x)= 2x^{4 }– 2x^{3 }– 7x^{2 }+ 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).

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

⇒ f(x)= (2x^{2} – 2x – 4)( x^{2 }– 3/2)= 2(x^{2} – x – 2)( x^{2 }– 3/2)

So, by putting x^{2} – 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 polynomialx ^{4 }+ x^{3 }– 34x^{2 }– 4x + 120, if the two of its zeros are 2 and -2.**

**Solution: **

**Let, **

f(x) = x^{4} + x^{3} – 34x^{2} – 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).

⇒ x^{2 }– 4 is a factor of f(x). Hence, performing division algorithm, we get

⇒ f(x)= (x^{2} + x – 30)( x^{2 }– 4)

So, by putting x^{2} + 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.

## Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 2

Q1

### What is the use of practising RD Sharma Solutions for Class 10 Maths Chapter 2?

Practising RD Sharma Solutions for Class 10 Maths Chapter 2 provides students with an idea about the sample of questions that will be asked in the board exam, which would help students prepare competently. These solutions are useful resources which can provide them with all the vital information in the most precise form. These solutions cover all the topics included in the RD Sharma syllabus, prescribed by the current CBSE board.

Q2

### 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.

Q3

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