RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Solution:
(i) Given,
p(x) = x³ – 3x² + 5x – 3 and g(x) = x² – 2
Let the quotient be q(x) and the remainder be r(x). Then by Euclid’s Division Algorithm,
p(x) = g(x) . q(x) + r(x)

∴ Quotient q(x) = x – 3, Remainder r(x) = 7x – 9

(ii) Given,
p(x) = x⁴ – 3x² + 4x + 5
g(x) = x² + 1 – x ⇒ x² – x + 1
Let the quotient be q(x) and the remainder be r(x). Then by Euclid’s Division Algorithm,
p(x) = g(x) . q(x) + r(x)

∴ Quotient q(x) = x² + x – 3 and Remainder r(x) = 8

(iii) Given,
p(x) = x⁴ – 5x + 6
g(x) = 2 – x²
Let the quotient be q(x) and the remainder be r(x). Then by Euclid’s Division Algorithm,
p(x) = g(x) . q(x) + r(x)

Hence, quotient q(x) = -x² – 2 and remainder r(x) = -5x + 10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3² + 1
Solution:
(i) On dividing 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3

Since the remainder is zero, therefore t² – 3 is a factor of the polynomial 2t⁴ + 3t³ – 2t² – 9t – 12

(ii) On dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1

Since the remainder is zero. Therefore x² + 3x + 1 is a factor of the polynomial 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) On dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1

Here the remainder is 2(≠ 0). Therefore x³ – 3x + 1 is not a factor of the polynomial x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\).
Solution:
Since two zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\), therefore \(\left(x-\sqrt{\frac{5}{3}}\right)\) and \(\left(x+\sqrt{\frac{5}{3}}\right)\) will be a factor of the given polynomial.

Hence,
3x⁴ + 6x³ – 2x² – 10x – 5 = (3x² – 5) (x² + 2x + 1)
Now x² + 2x + 1 = x² + x + x + 1
= x (x + 1) + 1 (x + 1)
= (x + 1) (x + 1)
Therefore its other zeroes will be – 1 and – 1.
Thus all the zeroes of the biquadratic polynomial will be \(\sqrt{\frac{5}{3}}\), \(-\sqrt{\frac{5}{3}}\), – 1 and – 1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
On dividing the polynomial x³ – 3x² – x + 2 by a polynomial g(x), thequotient (x – 2) and the remainder (- 2x + 4) are obtained.
∴ Dividend = Divisor × Quotient + Remainder
or (x – 2) × 0) + (- 2x + 4)= x³ – 3x² + x + 2
or (x – 2) × g(x) = x³ – 3x² + x + 2 + 2x – 4

Hence, g(x) = x² – x + 1 is obtained.

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) We require p(x) and q(x) such that
deg p(x) = deg q(x)
Then,
deg p(x) = deg q(x)
⇒ The degree of g(x) must be zero.
Let p(x) = 5x² – 5x + 10,
g(x) = 5
q(x) = x² – x + 2
r(x) = 0

Bv division algorithm,
5x² – 5x + 10 = 5(x² – x + 2) + 0
p(x) = g(x) . q(x) + r(x)
Hence, deg p(x) = deg q(x) = 2

(ii) deg q(x) = deg r(x)
P(x) = g(x)∙q(x) + r(x) degree of p(x) must be equal to the sum of degree of g(x) and degree of q(x).
Suppose p(x) = 7x³ – 42x + 53
g(x) = x³ – 6x + 7
q(x) = 7, r(x) = 4

∴ By division algorithm,
7x³ – 42x + 53 = 7(x³ – 6x + 7) + 4
So degree of q(x) = 0

(iii) deg r(x) = 0
Let p(x) = x³ + 2
and g(x) = x² – x + 1
Now dividing x³ + 2 by x² – x + 1

By division algorithm
x³ + 2 = (x² – x + 1) (x + 1) + 1
p(x) = g(x) . q(x) + r(x)
Also deg r(x) = 0