In this chapter, we provide RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.1 pdf, free RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.2 book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 2 |

Chapter Name | Polynomials |

Exercise | 2.2 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **2 Polynomials Ex 2.2 Download PDF**

**Chapter 2: Polynomials Exercise – 2.2**

**Question: 1**

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

(i) f(x) = 2x^{3}+ x^{2}– 5x + 2; 1/2, 1, – 2

(ii) g(x) = x^{3}– 4x^{2} + 5x – 2; 2, 1, 1

**Solution: **

(i) f(x) = 2x^{3 }+ x^{2 }– 5x + 2; 1/2, 1, – 2

(a) By putting x = 1/2 in the above equation, we will get

(b) By putting x = 1 in the above equation, we will get

f(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

(c) By putting x = -2 in the above equation, we will get

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

= -16 + 4 + 10 + 2 = – 16 + 16 = 0

Now,

Sum of zeroes = α + β + γ = – b/a

Product of the zeroes = αβ + βγ + αγ = c/a

Hence, verified.

(ii) g(x) = x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

(a) By putting x = 2 in the given equation, we will get

g(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 18 – 18 = 0

(b) By putting x = 1 in the given equation, we will get

g(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2

= 0

Now,

Sum of zeroes= α + β + γ =-b/a

⇒ 2 + 1 + 1 = −(−4)

4 = 4

Product of the zeroes = αβ + βγ + αγ = c/a

2 × 1 + 1 × 1 + 1 × 2 = 5

2 + 1 + 2 = 5

5 = 5

αβγ = –(−2)

2 × 1 × 1 = 2

2 = 2

Hence, verified.

**Question: 2**

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

**Solution:**

Any cubic polynomial is of the form ax^{3} + bx^{2} + cx + d:

= x^{3} – (sum of the zeroes) x^{2} + (sum of the products of its zeroes) x – (product of the zeroes)

= x^{3 }– 3x^{2} + (−1)x + (−3)

= k[x^{3} – 3x^{2} – x – 3]

k is any non-zero real numbers.

**Question: 3**

If the zeroes of the polynomial f(x) = 2x^{3 }– 15x^{2} + 37x – 30, find them.

**Solution: **

Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial.

f(x) = 2x^{3 }– 15x^{2 }+ 37x – 30

And, a (a^{2} + d^{2}) = 15

**Question: 4**

Find the condition that the zeroes of the polynomial f(x) = x^{3} + 3px^{2} + 3qx + r may be in A.P.

**Solution: **

f(x) = x^{3 }+ 3px^{2} + 3qx + r

Let, a – d, a, a + d be the zeroes of the polynomial.

Then,

The sum of zeroes = – b/a

a + a – d + a + d = -3p 3a = -3p a = -p Since, a is the zero of the polynomial f(x),

Therefore, f(a) = 0

f(a) = a^{3} + 3pa^{2} + 3qa + r = 0

Therefore, f(a) = 0f(a) = 0

⇒ a^{3} + 3pa^{2} + 3qa + r = 0

= ⇒ (−p)^{3 }+ 3p(−p)^{2} + 3q(−p) + r = 0

= − p^{3} + 3p^{3} – pq + r = 0

= 2p^{3} – pq + r = 0

**Question: 5**

If zeroes of the polynomial f(x) = ax^{3 }+ 3bx^{2} + 3cx + d are tin A.P., prove that 2b^{3 }– 3abc + a^{2}d = 0.

**Solution:**

f(x) = x^{3 }+ 3px^{2} + 3qx + r

Let, a – d, a, a + d be the zeroes of the polynomial.

Then,

The sum of zeroes = – b/a

a + a – d + a + d = – 3b/a

Since, f(a) = 0

⇒ a(a^{2}) + 3b(a)^{2} + 3c(a) + d = 0

⇒ a(a^{2}) + 3b(a)^{2 }+ 3c(a) + d = 0

**Question: 6**

If the zeroes of the polynomial f(x) = x^{3} – 12x^{2} + 39x + k are in A.P., find the value of k.

**Solution: **

f(x) = x^{3} – 12x^{2 }+ 39x + k

Let, a-d, a, a + d be the zeroes of the polynomial f(x).

The sum of the zeroes = 12

3a = 12

a = 4

Now,

f(a) = 0

f(a) = a^{3} – 12a^{2} + 39a + k f(4) = 4^{3} – 12(4)^{2} + 39(4) + k = 0

f(4) = 4^{3} –12(4)^{2} + 39(4) + k = 0

64 – 192 + 156 + k = 0

k = – 28

**All Chapter RD Sharma Solutions For Class10 Maths**

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