10th CBSE math solution for exercise 2.3

This page 10th CBSE math solution for exercise 2.3 is going to provide you solution for every problems that you find in the exercise no 2.3

10th CBSE math solution for exercise 2.3

(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 – 3 x² + 5 x – 3     g (x) = x² - 2         

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

(iii) p(x) = x⁴ – 5 x + 6     g (x) = 2 - x²         

(2) Check whether the first polynomial is a factor of second polynomial by dividing the second polynomial by the first polynomial

(i) t² – 3 , 2 t⁴ + 3 t³ – 2 t² – 9 t – 12

(ii)  x² + 3 x + 1     g (x) = 3 x⁴ + 5 x ³ – 7 x² + 2 x + 2  

(iii)  x³ - 3 x + 1     g (x) = x⁵ -4 x³ + x² + 3 x + 1          

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

(4) On dividing x³ – 3 x² + x + 2 by the polynomial g(x), the quotient and remainder were (x – 2) and – 2x + 4. Find g(x)

(i) p(x) = x³ – 3 x² + 5 x – 3     g (x) = x² - 2    

Quotient = x - 3

Remainder = 7 x - 9


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

Quotient = x² + x - 3

Remainder = 8


(iii) p(x) = x⁴ – 5 x + 6     g (x) = 2 - x²  

Quotient = -x² - 2

Remainder = -5 x + 10


(2) Check whether the first polynomial is a factor of second polynomial by dividing the second polynomial by the first polynomial

(i) t² – 3 , 2 t⁴ + 3 t³ – 2 t² – 9 t – 12

The remainder is zero. So we can say that the first polynomial is the factor of second polynomial.


(ii)  x² + 3 x + 1     g (x) = 3 x⁴ + 5 x ³ – 7 x² + 2 x + 2     

The remainder is zero. So we can say that the first polynomial is the factor of second polynomial.


(iii)  x³ - 3 x + 1     g (x) = x⁵ -4 x³ + x² + 3 x + 1    

The first polynomial is not a factor of second polynomial.


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

Solution:

Since the highest power of the given polynomial is 4,so there will be four zeroes for the given polynomial.  Out of the four zeroes, two zeroes are given. We are yet to find two zeroes.  For that we have to find a quadratic polynomial from the two given values of x.

x = √(5/3)   x = -√(5/3)                              

(x - √(5/3)) (x + √(5/3)) = (x² – 5/3)

If we divide the given polynomial by this quadratic polynomial, we will get the other quadratic polynomial.

By factoring the other quadratic polynomial, we will get two remaining zeroes.

3 x² + 6 x + 3 = 3 x² + 3 x + 3 x + 3

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

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

3 x + 3 = 0                x + 1 = 0

3 x = -3                    x = -1

  x = -3/3

  x = -1


(4) On dividing x³ – 3 x² + x + 2 by the polynomial g(x), the quotient and remainder were (x – 2) and – 2x + 4. Find g(x)

Division algorithm

      P (x) = g (x) x q (x) + r (x)

P (x) = x³ – 3 x² + x + 2

q (x) = (x – 2)

r (x) = - 2x + 4

   x³ – 3 x²+ x + 2 = g (x) x (x – 2) + (- 2x + 4)

   x³ – 3 x² + x + 2 + 2 x - 4 = g (x) x (x – 2)

   x³ – 3 x² + 3 x - 2 = g (x) x (x – 2)

   (x³ – 3 x² + 3 x – 2)/(x - 2) = g (x)

Therefore g (x) = x² – 1 x + 1