# RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.2

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) = 3x3 + x2 + 2x + 5, g(x) = 1 + 2x + x2
(ii) f(x) = x3 – 3x2 + 5x + 3, g(x) = x2 – 2
(iii) f(x) = x3 – 6x2 + 11x – 6, g(x) = x + 2
(iv) f(x) = 9x4 – 4x2 + 4, g(x) = 3x2 + x – 1
Solution
(i) Given : f(x) = 3x3 + x2 + 2x + 5
and g(x) = 1 + 2x + x2
or g(x) = x2 + 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 + x2) + 9x + 10
= 3x + 6x2 + 3x3 – 5 – 10x – 5x2 + 9x + 10
= 3x3 + x2 – 7x – 5 + 9x + 10
= 3x3 + x2 + 2x + 5.
= dividend
Thus division (RBSESolutions.com) algorithm is verified.

(ii) given f(x) = x3 – 3x2 + 5x – 3
and g(x) = x2 – 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)
= (x2 – 2)(x – 3) + 7x – 9
= x3 – 3x2 – 2x + 6 + 7x – 9
= x3 – 3x2 + 5x – 3
= f(x)
Thus division (RBSESolutions.com) algorithm is verified.
(iii) Given f(x) = x3 – 6x2 + 11x – 6
and g(x) = x + 2
Dividing f(x) by g(x)

Quotient q(x) = x2 – 8x + 27
Remainder r(x) = -60
Here, Divisor × Quotient + Remainder
(x + 2) (x2 – 8x + 27) – 60
= x3 + 2x2 – 8x2 – 16x + 27x + 54 – 60
= x3 – 6x2 + 11x – 6
= dividend
Hence, division (RBSESolutions.com) algorithm is verified.
(iv) Given f(x) = 9x4 – 4x2 + 4
and g(x) = 3x2 + x – 1
Dividing f(x) by g(x)

Thus quotient q(x) = 3x2 – x
Remainder f(x) = -x + 4
Here : f(x).q(x) + f(x)
= (3x2 + x – 1) (3x2 – x) + (-x) + 4
= 9x4 + 3x3 – 3x2 – 3x3 – x2 + x – x + 4
= 9x4 – 4x2 + 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) = x2 + 3x + 1 ,f(x) = 3x4 + 5x3 – 7x2 + 2x + 2
(ii) g(t) = t2 – 3, f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
(iii) g(x) = x3 – 3x + 1, f(x) = x5 – 4x3 + x2 + 3x + 1
Solution
(i) Given, f(x) = 3x4 + 5x3 – 7x2 + 2x + 2
g(x) = x2 + 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
3x4 + 5x3 – 7x2 + 2x + 2 = (x2 + 3x + 1) (3x3 – 4x + 2) + 0
Thus, x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2
(ii) Given,
f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
g(t) = t2 – 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
2t4 + 3t3 – 2t2 – 9t – 12 = (t2 – 3) (2t2 + 3t + 4) + 0
Therefore t2 – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12
(iii) Given
f(x) = x5 – 4x3 + x2 + 3x + 1
g(x) = x3 – 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) = x2 – 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) = 2x4 – 3x3 – 3x2 + 6x – 2; √2 and -√2
(ii) f(x) = x4 – 6x3 – 26x2 + 138x – 35; 2 ± √3
(iii) f(x) = x3 + 13x2 + 32x + 20; -2
Solution
(i) Given
f(x) = 2x4 – 3x3 – 3x2 + 6x – 2
Two zeros of (RBSESolutions.com) polynomial f(x) are √2 and -√2
(x – √2)(x + √2) = x2 – 2, will be a factor of f(x).
Dividing polynomial f(x) by x2 – 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) = x3 + 13x2 + 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)(x2 + 11x + 10)
= (x + 2)[x2 + 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) = x3 – 3x2 + 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) = x3 – 3x2 + 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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