Students can Download Class 10 Maths Chapter 9 Polynomials Ex 9.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.3

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^{3} + 5x – 3, g(x) = x^{2} – 2

Answer:

p(x) = x^{3} – 3x^{3} + 5x – 3, g(x) = x^{2} + 0x – 2

∴ quotient q(x) = x – 3

Remainder r(x) = – 7x – 9

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

Answer:

p(x) = x^{4} + 0x^{3} – 3x^{3} – 3x^{2} + 4x + 5 & g(x) = x^{2} – x + 1

∴ Quotient q(x) = x^{2} + x – 3

Remainder r (x) = 8

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

Answer:

P(x) = x^{4} + 0x^{3} + 0x^{2} – 5x + 6 & g(x) = – x^{2} + 0x + 2

∴ Quotient q(x) = – x^{2} – 2

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

Answer:

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

g(t) = t^{2} – 3 (or) g(t) = t^{2} + 0t – 3

Since Remainder is zero, there fore, the first polynomial is a factor of second polynomial.

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

Answer:

P(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 & g(x) = x^{2} + 3x + 1

Since, remainder is zero, there fore first polynomial is factor of second polynomial

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

Answer:

p(x) = x^{5} + 0x^{4} – 4x^{3} + x^{2} + 3x + 1 , and g(x) = x^{3} + 0x^{2} – 3x +1

Since Remainder not equal to zero, first polynomial is not a factor of second polynomial.

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 – \(\sqrt{\frac{5}{3}}\).

Answer:

Since, two zeros are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\) there fore, By factor theorem,

x^{2} – \(\frac{5}{3}\) is a factor of

3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

Apply the division algorithm

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

g(x) = x^{2} + 0x – \(\frac{5}{3}\)

Remainder is zero,

∴ P(x) = g (x) q (x)

3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

= (x^{2} – \(\frac{5}{3}\))(3x^{2} + 6x + 3)

Other zeroes are given by the second Polynomial factors 3x^{2} + 6x + 3

Put 3x^{2} + 6x + 3 = 0

3x^{2} + 3x + 3x + 3 = 0

3x (x + 1) + 3(x + 1) = 0

(x + 1)(3x + 3) = 0

x + 1 = 0 (or) 3x + 3 = 0

x = – 1 (or) 3x = – 3

x = \(\frac{- 3}{3}\) = – 1

∴ x = – 1, – 1

Therefore other zeroes of the Polynomial , 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 are – 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).

Answer:

p(x) = x^{3} – 3x^{2} + x + 2,

q(x) = x – 2 and r(x) = – 2x + 4

By division algorithm for Polynomials.

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

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

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

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

Therefore g(x) = x^{2} – x + 1

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

Answer:

(i) Degree p(x) = Degree q(x)

p(x) = 2x^{2} + 2x + 8

q(x) = 2x^{2} = 2

q(x) = x^{2} + x + 4

r(x) = 0

(ii) Degree q(x) = Degree r(x)

p(x) = 2x^{2} + 2x + 8

q(x) = x^{2} + x + 9

q(x) = 2

r(x) = -10

(iii) Degree r(x) = 0

p(x) = x^{3} + x + 5

q(x) = x^{2} + 1

q(x) = x

r(x) = 5.