RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.1.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomials |

E×ercise |
Exercise 3.1 |

Number of Questions Solved |
3 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.1

Question 1.

Find the zeros of the following (RBSESolutions.com) quadratic polynomial and test the relation between zeros and coefficients.

(i) 4x^{2} + 8x

(ii) 4x^{2} – 4x + 1

(iii) 6x^{2} – x – 2

(iv) x^{2} – 15

(v) x^{2} – (√3 + 1) x + √3

(vi) 3x^{2} – x – 4

Solution

(i) Given polynomial

f(x) = 4x^{2} + 8x ⇒ f(x) = 4x(x + 2)

To find zeros of (RBSESolutions.com) polynomial f(x), f(x) will be zero.

f(x) = 0

⇒ 4x(x + 2) = 0

⇒ x = 0 or x + 2 = 0

⇒ x = 0 or x = -2

Thus zeros of f(x) are 0 and -2

Relations between zeros of polynomial and coefficient.

Sum of zeros = 0 + (-2) = -2

and product of zeros = 0 × -2 = 0

Thus, relation between zeros and polynomial is correct.

(ii) Given polynomial

f(x) = 4x^{2} – 4x + 1 = (2x)^{2} – 2 (2x). 1 + (1)^{2} = (2x – 1)^{2}

To find zeros polynomial f(x), f(x) will be zero.

⇒ (2x – 1)^{2} = 0

It (2x – 1)^{2} = 0

⇒ 2x – 1 = 0

⇒ x = [latex]\frac { 1 }{ 2 }[/latex]

Here zeros of (RBSESolutions.com) polynomial are same.

Therefore, zeros of polynomial are [latex]\frac { 1 }{ 2 }[/latex] and [latex]\frac { 1 }{ 2 }[/latex]

Relation between coefficient and zeros of polynomial

Thus, relation between zeros and polynomial is correct.

(iii) Given polynomial

f(x) = 6x^{2} – x – 2 = 6x^{2} – 4x + 3x – 2 = 2x(3x – 2) + 1(3x – 2) = (3x – 2)(2x + 1)

To find zeros of 6x^{2} – x – 2, f(x) = 0

If 3x – 2 = 0 then 3x = 2 ⇒ x = [latex]\frac { 2 }{ 3 }[/latex]

or 2x + 1 = 0 then 2x = -1 ⇒ x = [latex]\frac { -1 }{ 2 }[/latex]

Therefore zeros of (RBSESolutions.com) polynomial [latex]\frac { 2 }{ 3 }[/latex] , [latex]\frac { -1 }{ 2 }[/latex]

Relation between zeros and coefficient of polynomial

Thus, relation between zeros and polynomial is correct.

(iv) Given (RBSESolutions.com) polynomial f(x) = x^{2} – 15

To find zeros of f(x), f(x) will be zero

⇒ f(x) = 0

x^{2} = 15

or x = ±√15

Therefore zeros of polynomial x^{2} – 15 are = +√15, -√15

Relation between zeros and coefficient of polynomial

Sum of zeros = √15 – √15 = 0

Product of zeros comparing given polynomial = √15 × (-√15) = -15

Comparing given polynomial x^{2} – 15 = 0 with ax^{2} + bx + c

a = 1, b = 0 and c = -15

Thus, relation between zeros and polynomial is correct.

(v) Given (RBSESolutions.com) polynomial

f(x) = x^{2} – (√3 + 1) x + √3

= x^{2} – √3x – x + √3

= x (x – √3) – 1(x – √3)

= (x – √3)(x – 1)

Therefore, f(x) = (x – √3)(x – 1)

To find polynomial f(x), f(x) will be zero.

⇒ f(x) = o

If (x – √3) = 0 then x = √3

or (x – 1) = 0 then x = 1

Therefore, zeros of polynomial are √3, +1

Relation between zeros and coefficient of polynomial

Sum of zeros = (√3 + 1)

product of zeros = √3 × (+1) = √3

comparing given polynomial by ax^{2} + bx + c

a = 1, b = -(√3 + 1), c = √3

Thus, relation between zeros and polynomial is correct.

(vi) Given (RBSESolutions.com) polynomial f(x) = 3x^{2} – x – 4

= 3x^{2} – (4 – 3) x – 4

= 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

3x^{2} – x – 4 = (3x – 4) (x + 1)

To find zeros of f(x),

⇒ f(x) = 0

⇒ (3x – 4)(x + 1) = 0

If 3x – 4 = 0 then 3x = 4 ⇒ x = [latex]\frac { 4 }{ 3 }[/latex]

or x + 1 = 0 then x = -1

Thus, zeros of polynomial are [latex]\frac { 4 }{ 3 }[/latex] and – 1.

The relation between zeros and coefficient of the polynomial

Thus, relation between zeros and polynomial is correct.

Question 2.

Find quadratic polynomial. Sum and product (RBSESolutions.com) of whose zeros are given numbers respectively.

(i) -3, 2

(ii) √2, [latex]\frac { 1 }{ 3 }[/latex]

(iii) [latex]\frac { -1 }{ 4 }[/latex] , [latex]\frac { 1 }{ 4 }[/latex]

(iv) 0, √5

(v) 4, 1

(vi) 1, 1

Solution

If zeros of quadratic polynomial f(x) are known, then find required polynomial by following formula

Let f(x) = k{x^{2} – (sum of zeros) x + product of zeros}, where k = a real number

(i) Lef f(x) be a polynomial.

Sum and product of whose zeros are -3 and 2 respectively.

f(x) = k[x^{2} – (-3)x + 2] = k[x^{2} + 3x + 2] (∴ k = real number)

Thus required polynomial f(x) = x^{2} + 3x + 2

(ii) Let f(x) is quadratic polynomial.

Sum and product of whose zeros are √2 and [latex]\frac { 1 }{ 3 }[/latex] respectively.

where [latex]\frac { k }{ 3 }[/latex] is a constant term, real number.

Hence, required polynomial is 3x^{2} – 3√2x + 1

(iii) Let f(x) is a quadratic (RBSESolutions.com) polynomial sum and product of whose zeros are [latex]\frac { -1 }{ 4 }[/latex] and [latex]\frac { 1 }{ 4 }[/latex]

respectively.

Hence, required (RBSESolutions.com) polynomial is 4x^{2} + x + 1.

(iv) Let f(x) is a quadratic polynomial.

Sum and product of whose zeros are 0 and √5 respectively.

f(x) = k[ x^{2} – 0. x + √5 ] = k[x^{2} + √5 ] (where k is a constant term)

Hence, required polynomial is x^{2} + √5

(v) Let f(x) is a quadratic polynomial.

Sum and product of whose zeros are 4 and 1 respectively

f(x) = k[x^{2} – 4x + 1], (where k is constant term)

Hence, required polynomial is x^{2} – 4x + 1

(vi) Let f(x) is a quadratic polynomial.

Sum and product of whose zeros are 1 and 1 respectively.

f(x) = k{x^{2} – 1.x + 1) = k(x^{2} – x + 1), where k is constant term

Hence, required polynomial is x^{2} – x + 1

Question 3.

If sum of square of zeros of (RBSESolutions.com) quadratic equation f(x) = x^{2} – 8x + k is 40 then find the value of k.

Solution

Given polynomial f(x) = x^{2} – 8x + k

Let α and β are zeros of polynomial f(x) then

Now, from equation (i)

(α + β) = 8

Squaring both sides,

(α + β)^{2} = 8^{2}

⇒ a^{2} + β^{2} + 2αβ = 64 …(iii)

It is given that sum of square of zeros is 40.

i.e., α^{2} + β^{2} = 40

Putting values from equation (i) and (ii) in (iii)

40 + 2k = 64

⇒ 2k = 64 – 40

⇒ 2k = 24

⇒ k = 12

Thus k = 12

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.1, drop a comment below and we will get back to you at the earliest.

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