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

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

(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

Solution:

(i) Given,

p(x) = x^{3} – 3x^{2} + 5x – 3 and g(x) = x^{2} – 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) = x^{4} – 3x^{2} + 4x + 5

g(x) = x^{2} + 1 – x ⇒ x^{2} – 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) = x^{2} + x – 3 and Remainder r(x) = 8

(iii) Given,

p(x) = x^{4} – 5x + 6

g(x) = 2 – x^{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)

Hence, quotient q(x) = -x^{2} – 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) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

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

Solution:

(i) On dividing 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 by t^{2} – 3

Since the remainder is zero, therefore t^{2} – 3 is a factor of the polynomial 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(ii) On dividing 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 by x^{2} + 3x + 1

Since the remainder is zero. Therefore x^{2} + 3x + 1 is a factor of the polynomial 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

(iii) On dividing x^{5} – 4x^{3} + x^{2} + 3x + 1 by x^{3} – 3x + 1

Here the remainder is 2(≠ 0). Therefore x^{3} – 3x + 1 is not a factor of the polynomial x^{5} – 4x^{3} + x^{2} + 3x + 1.

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 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,

3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 = (3x^{2} – 5) (x^{2} + 2x + 1)

Now x^{2} + 2x + 1 = x^{2} + 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 x^{3} – 3x^{2} + 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 x^{3} – 3x^{2} – 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)= x^{3} – 3x^{2} + x + 2

or (x – 2) × g(x) = x^{3} – 3x^{2} + x + 2 + 2x – 4

Hence, g(x) = x^{2} – 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) = 5x^{2} – 5x + 10,

g(x) = 5

q(x) = x^{2} – x + 2

r(x) = 0

Bv division algorithm,

5x^{2} – 5x + 10 = 5(x^{2} – 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) = 7x^{3} – 42x + 53

g(x) = x^{3} – 6x + 7

q(x) = 7, r(x) = 4

∴ By division algorithm,

7x^{3} – 42x + 53 = 7(x^{3} – 6x + 7) + 4

So degree of q(x) = 0

(iii) deg r(x) = 0

Let p(x) = x^{3} + 2

and g(x) = x^{2} – x + 1

Now dividing x^{3} + 2 by x^{2} – x + 1

By division algorithm

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

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

Also deg r(x) = 0

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