Class 10 NCERT 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) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

R = 7x-9

Q = x-3

ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

R = 8

Q = x²+x-3

iii) p(x) = x⁴– 5x + 6, g(x) = 2 – x²

Q = -x²-2

R = -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 + 3x + 1

Solution:

i) t² – 3, 2t⁴ + 3t³ – 2t²– 9t – 12

Q = 2t³+3t+4

R = 0

Yes 1^st polynomial is factor of 2^nd polynomial. 

ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

R = 0

Q = 3x²-4x+2

iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

R = x²-1

Q = 2

Question 3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are and √(5/3) and -√(5/3).

Solution:

R = 0

Q = 3x²+6x+3

∴ we are factorizing

3x²+6x+3

x²+2x+1

(x+1)²

(x+1)  (x+1) = 0

∴ x = -1 and x = -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:

Dividend = Divisor * Quotient + Remainder

x³-3x²+3x-2/x-2

R = 0

Q = x² -x +1

Answer: g(x)=x²-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

Solution:

i) deg p(x) = deg q(x)

p(x)=2x²-2x+14, g(x)=2

p(x)/g(x)=2x²-2x+14/2=(x²-x+7)

=x²-x+7=q(x)

=q(x)=x²-x+7

r(x)=0

ii) deg q(x)=deg r(x)

p(x)=4x²+4x+4, g(x)=x²+x+1

q(x) = 4

r(x) = 0

∴Here deg q(x)=deg r(x)

iii) deg r(x)=0

p(x)=x³+2x²-x+2 ,g(x)=x²-1

q(x) = x+2

r(x) = 4

deg of r(x) = 0