NCERT solution for class 10th maths: chapter-2 POLYNOMIALS (Ex.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 

So, here quotient= x-3 and Remainder = 7x-9
*****************************************************************

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

So, here quotient= x²+ x -3 and Remainder = 8 
*********************************************************************************
 (iii) p(x)= x⁴ -5x +6, 
      g(x) = 2 - x²  

So, here quotient= -x²-2  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) 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 

 

                          

 Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial.
 ********************************************************************************

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

Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial. 
 *********************************************************************************

 (ii) x³-3x+1,       x⁵-4x³+x²+3x+1

since remainder ≠ 0, therefore, second polynomial is not  divisible by first polynomial.
Hence, first polynomial is not a factor of second polynomial.
 *********************************************************************************

Question-3
 Solution: 

According to division algorithm of polynomial
3x⁴+6x³-2x²-10x-5 = (3x²-5)(x²+2x+1)

x²+2x+1= x²+x+x+1
              = x(x+1)+1(x+1)
              = (x+1)(x+1)
now find x
    x+1= 0  and  x+1= 0
        x = -1           x = -1

   



Hence, the other zeroes of polynomials are -1, -1.

Question -4 On dividing x³ -3x² +x +2 by a polynomial g(x), the qotient and remainder were x-2 and -2x+4, respectively. Find g(x).
Solution : 

          Dividend f(x)= x³ -3x² +x +2
          Quotient q(x) = x-2
          Remainder r(x) = -2x+4
          Divisor g(x) = ?

According to division algorithm of polynomials
            f(x) = g(x) x q(x) + r(x) 

      f(x) - r(x) = g(x) x q(x)

         
  

     g(x) = x² - x + 1

 ********************************************************************************

Question -5 Give examples of polynomial 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:  degree of polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = q(x)
degree of quotient q(x) will be equal to degree of dividend p(x) when divisor is constant (means when any polynomial is divided by a constant) 

Let us assume a polynomial 6x² +2x +2 and divide it by 2. So,
here p(x) = 6x² +2x +2 
        g(x) = 2
By dividing we get 3x² +x +1 which means
 q(x) =  3x² +x +1
 r(x) = 0

here, degree of p(x) and q(x) is the same which is 2.

Now second part of question:-
division algorithm
p(x) = g(x) x q(x) + r(x)

 6x² +2x +2 = 2 x (3x² +x +1) + 0
                    = 6x² +2x +2 
   here both sides are equal which means this polynomial satisfy the division algorithm.

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

Let us assume  a polynomial x³+x  and divide it by x². So,
here p(x) = x³+x
        g(x) = x²
By dividing we get quotient x and remainder x. which means
 q(x) = x
 r(x) = x
 here, degree of q(x) and r(x) is the same which is 1.

Now second part of question:-
division algorithm
p(x) = g(x) x q(x) + r(x)
x³+x = x² x x + x
        = x³+x               
here both sides are equal which means this polynomial satisfy the division algorithm.

(iii) deg r(x) = 0 

degree of remainder will be 0 when remainder comes to a constant.

Let us assume a polynomial x³ +1 and divide it by x²

here p(x) = x³+1
        g(x) = x²
By dividing we get quotient x and remainder 1. means
 q(x) = x
 r(x) = 1
there is no varable in the remainder it means degree of remainder is 0.

Now second part of question:-
division algorithm
p(x) = g(x) x q(x) + r(x)
x³+1 = x² × x +1
        = x³+1
here both sides are equal which means this polynomial satisfy the division algorithm.

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