Class 10 Maths

# Polynomials

## NCERT 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

Solution: Here; quotient = x – 3 and remainder = 7x – 9

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

Solution: Here, quotient = x^2 + x – 3 and remainder = 8

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

Solution: Here; quotient = - x^2 – 2 and remainder = - 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

Solution: Here, the first polynomial is a factor of the second polynomial.

(ii) x^2 + 3x + 1, 3x^4 + 5x^3 – 7x^2 + 2x + 12

Solution: Here, the first polynomial is a factor of the second polynomial.

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

Solution: Here, the first polynomial is not a factor of the second polynomial.

Question 3: Obtain all other zeroes of 3x^4 + 6x^3 – 2x^2 – 10x – 15, if two of its zeroes are (sqrt5)/(3) and (-sqrt5)/(3)

Solution: A quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence, the equation can be written as follows: Given polynomial is divided by this equation as follows: Hence;

3x^4 + 6x^3 – 2x^2 – 10x – 5

= (3x^2 – 5)(x^2 + 2x + 1)

Roots for the equation x^2 + 2x + 1 can be calculated as follows:

x^2 + 2x + 1 = 0
Or, x^2 + x + x + 1 = 0
Or, x(x + 1) + (x + 1) = 0
Or, (x + 1)(x + 1) = 0

Hence, roots 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).

Solution: Subtracting the remainder from the given polynomial we get;

x^3 – 3x^2 + x + 2 – ( - 2x + 4)

= x^3 – 3x^2 + x + 2 + 2x – 4

= x^3 – 3x^2 + 3x – 2 …… (1)

Dividing equation (1) by quotient will give the value of g(x) Hence, g(x) = x^2 – x + 1