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Polynomials : Exercise - 2.3 (Mathematics NCERT Class 10th)


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Q.1       Divide the polynomial p(x) by the polynomial g (x) and find the quotient and remainder in each of the following : 
            (i) p\left( x \right) = {x^3} - 3{x^2} + 5x - 3,\,g\left( x \right) = {x^2} - 2
            (ii) p\left( x \right) = {x^4} - 3{x^2} + 4x + 5,\,g\left( x \right) = {x^2} + 1 - x
            (iii) p\left( x \right) = {x^4} - 5x + 6,\,g\left( x \right) = 2 - {x^2}
Sol.      (i) We have, 

7

        Therefore, the quotient is x – 3 and the remainder is 7 x – 9

        (ii) Here, the dividend is already in the standard form and the divisor is also in the standard form.
         We have,
8

         Therefore, the quotient is {x^2} + x - 3 and the remainder is 8.

        (iii) To carry out the division, we first write divisor in the standard form.
         So, divisor =  - {x^2} + 2
         We have,
22

         Therefore, the quotient is  - {x^2} + 2
         and the remainder is – 5x + 10.


Q.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,\,2{t^4} + 3{t^3} - 2{t^2} - 9t - 12
              (ii) {x^2} + 3x + 1,\,3{x^4} + 5{x^3} - 7{x^2} + 2x + 2
              (iii) {x^3} - 3x + 1,\,{x^5} - 4{x^3} + {x^2} + 3x + 1
Sol.      (i) Let us divide 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\,\,by\,{t^2} - 3.
            We have, 

10

        Since the remainder is zero, therefore, {t^2} - 3 is a factor of 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12.

        (ii) Let us divide 3{x^4} + 5{x^3} - 7{x^2} + 2x + 2\,by\,{x^2} + 3x + 1.
        We get,
18
        Since the remainder is zero, therefore {x^2} + 3x + 1 is a factor of
                                                               3{x^4} + 5{x^3} - 7{x^2} + 2x + 2

        (iii) Let us divide {x^5} - 4{x^3} + {x^2} + 3x + 1 by {x^3} - 3x + 1
         We get,
19

         Here remainder is 2( \ne 0). Therefore,  {x^3} - 3x + 1 is not a factor of
                                                                            {x^5} - 4{x^3} + {x^2} + 3x + 1.


Q.3      Obtain all the zeroes of 3{x^4} + 6{x^3} - 2{x^2} - 10x - 5 if two of its zeroes are \sqrt {{5 \over 3}} \,and\, - \sqrt {{5 \over 3}} .
Sol.      Since two zeroes are \sqrt {{5 \over 3}} \,and\, - \sqrt {{5 \over 3}} , so \left( {x - \sqrt {{5 \over 3}} \,} \right)and\,\left( {x + \sqrt {{5 \over 3}} } \right) are the factors of the given polynomial.
           Now, \left( {x - \sqrt {{5 \over 3}} \,} \right)\,\left( {x + \sqrt {{5 \over 3}} } \right) = {x^2} - {5 \over 3} 
            \Rightarrow \left( {3{x^2} - 5} \right) is a factor of the given polynomial. 
           Applying the division algorithm to the given polynomial and {3{x^2} - 5}, we have

18

         Therefore, 3{x^4} + 6{x^3} - 2{x^2} - 10x - 5\left( {3{x^2} - 5} \right)\left( {{x^2} + 2x + 1} \right)
          Now, {x^2} + 2x + 1  = {x^2} + x + x + 1
                                     = x\left( {x + 1} \right) + 1\left( {x + 1} \right)
                                    = \left( {x + 1} \right)\left( {x + 1} \right)
          So, its other zeroes are – 1 and – 1.
          Thus,  all the zeroes of the given fourth degree polynomial are
                   \sqrt {{5 \over 3}} , - \sqrt {{5 \over 3}} , - 1\,and\, - 1. 


Q.4       On dividing {x^3} - 3{x^2} + x + 2 by a polynomial g(x) , the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Sol.         Since on dividing {x^3} - 3{x^2} + x + 2 by a polynomial (x – 2) × g(x), the quotient and remainder were (x – 2) and (– 2x + 4) respectively, therefore,
              Therefore, Quotient × Divisor + Remainder = Dividend
           \Rightarrow \left( {x - 2} \right) \times g\left( x \right) + \left( { - 2x + 4} \right)   = {x^3} - 3{x^2} + x - 2
          \Rightarrow (x – 2) × g(x)  = {x^3} - 3{x^2} + x - 2 + 2x - 4
          \Rightarrow g\left( x \right) = {{{x^3} - 3{x^2} + 3x - 2} \over {x - 2}}                  ... (1)
         Let us divide {x^3} - 3{x^2} + 3x - 2\,by\,x - 2. We get

14

           Therefore, equation (1) gives g (x)  = {x^2} - x + 1


Q.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
Sol.     There can be several examples for each of (i), (ii) and (iii). 
           However, one example for each case may be taken as under : 
        (i) p\left( x \right) = 2{x^2} - 2x + 14,g\left( x \right) = 2,
             q\left( x \right) = {x^2} - x + 7,r\left( x \right) = 0
       (ii)  p\left( x \right) = {x^3} + {x^2} + x + 1,g\left( x \right) = {x^2} - 1,
             q\left( x \right) = x + 1\,,\,r\left( x \right) = 2x + 2
       (iii) p\left( x \right) = {x^3} + 2{x^2} - x + 2,
             g\left( x \right) = {x^2} - 1,q\left( x \right) = x + 2,\,r\left( x \right) = 4



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