RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.2.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomials |

E×ercise |
Exercise 3.2 |

Number of Questions Solved |
4 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.2

RBSE Solutions For Class 10 Maths Chapter 3 Question 1.

Using division algorithm, (RBSESolutions.com) find quotient and remainder dividing f(x) by g(x).

(i) f(x) = 3x^{3} + x^{2} + 2x + 5, g(x) = 1 + 2x + x^{2}

(ii) f(x) = x^{3} – 3x^{2} + 5x + 3, g(x) = x^{2} – 2

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

(iv) f(x) = 9x^{4} – 4x^{2} + 4, g(x) = 3x^{2} + x – 1

Solution

(i) Given : f(x) = 3x^{3} + x^{2} + 2x + 5

and g(x) = 1 + 2x + x^{2}

or g(x) = x^{2} + 2x + 1

Dividing f(x) by g(x)

Quotient q(x) = 3x – 5

Remainder r(x) = 9x + 10

Here, Quotient × Divisor + Remainder

(3x – 5) (1 + 2x + x^{2}) + 9x + 10

= 3x + 6x^{2} + 3x^{3} – 5 – 10x – 5x^{2} + 9x + 10

= 3x^{3} + x^{2} – 7x – 5 + 9x + 10

= 3x^{3} + x^{2} + 2x + 5.

= dividend

Thus division (RBSESolutions.com) algorithm is verified.

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

and g(x) = x^{2} – 2

Dividing f(x) by g(x)

Quotient q(x) = x – 3

Remainder r(x) = 7x – 9

By Euclid divison algorithm

f(x) = g(x).q(x) + r(x)

= (x^{2} – 2)(x – 3) + 7x – 9

= x^{3} – 3x^{2} – 2x + 6 + 7x – 9

= x^{3} – 3x^{2} + 5x – 3

= f(x)

Thus division (RBSESolutions.com) algorithm is verified.

(iii) Given f(x) = x^{3} – 6x^{2} + 11x – 6

and g(x) = x + 2

Dividing f(x) by g(x)

Quotient q(x) = x^{2} – 8x + 27

Remainder r(x) = -60

Here, Divisor × Quotient + Remainder

(x + 2) (x^{2} – 8x + 27) – 60

= x^{3} + 2x^{2} – 8x^{2} – 16x + 27x + 54 – 60

= x^{3} – 6x^{2} + 11x – 6

= dividend

Hence, division (RBSESolutions.com) algorithm is verified.

(iv) Given f(x) = 9x^{4} – 4x^{2} + 4

and g(x) = 3x^{2} + x – 1

Dividing f(x) by g(x)

Thus quotient q(x) = 3x^{2} – x

Remainder f(x) = -x + 4

Here : f(x).q(x) + f(x)

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

= 9x^{4} + 3x^{3} – 3x^{2} – 3x^{3} – x^{2} + x – x + 4

= 9x^{4} – 4x^{2} + 4

= f(x)

Thus, division (RBSESolutions.com) algorithm is verified.

RBSE Class 10 Maths Exercise 3.2 Solutions Question 2.

Dividing second polynomial by first polynomial and test whether first polynomial is a factor of second polynomial.

(i) f(x) = x^{2} + 3x + 1 ,f(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(ii) g(t) = t^{2} – 3, f(t) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

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

Solution

(i) Given, f(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

g(x) = x^{2} + 3x + 1

If by dividing f(x) by g(x), we get (RBSESolutions.com) remainder as 0, then g(x) will be factor of f(x).

Dividing f(x) by g(x),

Remainder is zero.

By division (RBSESolutions.com) algorithm theorem

3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 = (x^{2} + 3x + 1) (3x^{3} – 4x + 2) + 0

Thus, x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(ii) Given,

f(t) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

g(t) = t^{2} – 3

Dividing f(x) by g(x), If remainder is (RBSESolutions.com) zero then g(x) will be a factor of f(x).

Dividing f(x) by g(x)

Remainder is zero

By division (RBSESolutions.com) algorithm theorem

2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 = (t^{2} – 3) (2t^{2} + 3t + 4) + 0

Therefore t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(iii) Given

f(x) = x^{5} – 4x^{3} + x^{2} + 3x + 1

g(x) = x^{3} – 3x + 1

If dividing f(x) by g(x) we get remainder 0 then g(x) will be a factor of f(x).

Dividing f(x) by g(x),

Remainder r(x) = 2

Quotient q(x) = x^{2} – 1

Remainder is not zero.

Therefore g(x) is (RBSESolutions.com) not a factor of f(x).

Ex 3.2 Class 10 RBSE Question 3.

Following are the polynomials with their zeros, find all the other zeros.

(i) f(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2; √2 and -√2

(ii) f(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35; 2 ± √3

(iii) f(x) = x^{3} + 13x^{2} + 32x + 20; -2

Solution

(i) Given

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

Two zeros of (RBSESolutions.com) polynomial f(x) are √2 and -√2

(x – √2)(x + √2) = x^{2} – 2, will be a factor of f(x).

Dividing polynomial f(x) by x^{2} – 2

Now other (RBSESolutions.com) zeros of polynomail

If x + 5 = 0 then x = -5

or x – 7 = 0 then x = 7

Other zeros of polynomail are -5 and 7

(iii) Given

f(x) = x^{3} + 13x^{2} + 32x + 20 and one zero is -2

(x + 2) will be factor of f(x)

Now, dividing f(x) by x + 2

Thus f(x) = (x + 2)(x^{2} + 11x + 10)

= (x + 2)[x^{2} + 10x + x + 10]

= (x + 2)[x(x + 10) + 1(x + 10)]

= (x + 2)(x + 10)(x + 1)

Now, other zeros of polynomial

If x + 10 = 0 then x = -10

or x + 1 = 0 then x = -1

Thus, other zeros of polynomial are -10, -1

RBSE Class 10 Maths Chapter 3 Solutions Question 4.

Dividing polynomiaal f(x) = x^{3} – 3x^{2} + x + 2 by polynomial g(x). (RBSESolutions.com) Quotient q(x) and re-mainder r(x) are obtained as x – 2 and -2x + 4 respectively. Find polynomial g(x).

Solution

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

divisor g(x) = to find

quotient q(x) = x – 2

remainder r(x) = -2x + 4

By Euclid division theorem

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