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Polynomials : Exercise - 2.4 Optional (Mathematics NCERT Class 10th)


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Q.1        Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
              (i) 2{x^3} + {x^2} - 5x + 2\,;{1 \over 2},1, - 2 
              (ii) {x^3} - 4{x^2} + 5x - 2\,;2,1,1

Sol.      (i) Comparing the given polynomial with a{x^3} + b{x^2} + cx + d, we get
            a = 2 , b = 1, c = – 5 and d = 2. 
                   p\left( {{1 \over 2}} \right) = 2{\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 2}} \right)^2} - 5\left( {{1 \over 2}} \right) + 2
                     = {1 \over 4} + {1 \over 4} - {5 \over 2} + 2
                     = {{1 + 1 - 10 + 8} \over 4} = {0 \over 4} = 0
                    p\left( 1 \right) = 2{\left( 1 \right)^3} + {\left( 1 \right)^2} - 5\left( 1 \right) + 2
                     = 2 + 1 - 5 + 2 = 0
                    p\left( { - 2} \right) = 2{\left( { - 2} \right)^3} + {\left( { - 2} \right)^2} - 5\left( { - 2} \right) + 2
                     = 2\left( { - 8} \right) + 4 + 10 + 2
                     = - 16 + 16 = 0
            Therefore, {1 \over 2}, 1 and – 2 are the zeroes of 2{x^3} + {x^2} - 5x + 2.
            So, \alpha = {1 \over 2},\,\beta = 1\,and\,\gamma = - 2.
            Therefore, \alpha + \beta + \gamma = {1 \over 2} + 1+\left( { - 2} \right) = {{1 + 2 - 4} \over 2}
                      = - {1 \over 2} = {{ - b} \over a}
                     \alpha \beta + \beta \gamma + \gamma \alpha = \left( {{1 \over 2}} \right)\left( 1 \right) + \left( 1 \right)\left( { - 2} \right) + \left( { - 2} \right)\left( {{1 \over 2}} \right)
                      = {1 \over 2} - 2 - 1
                      = {{1 - 4 - 2} \over 2} = {{ - 5} \over 2} = {c \over a}
           and     \alpha \beta \gamma = {1 \over 2} \times 1 \times - 2 = 1 = {{ - 2} \over 2} = {{ - d} \over \alpha }

           (ii) Comparing the given polynomial with a{x^3} + b{x^2} + cx + d, we get
                                           a = 1, b = – 4, c = 5 and d = – 2. 
                     p\left( 2 \right) = {\left( 2 \right)^3} - 4{\left( 2 \right)^2} + 5\left( 2 \right) - 2 = 8 - 16 + 10 - 2 = 0
                     p\left( 1 \right) = {\left( 1 \right)^3} - 4{\left( 1 \right)^2} + 5\left( 1 \right) - 2 = 1 - 4 + 5 - 2 = 0
              Therefore , 2 , 1 and 1 are the zeros of  {x^3} - 4{x^2} + 5x - 2
           Thus,     \alpha = 2,\,\beta = 1\,and\,\gamma = 1.
            Now \alpha + \beta + \gamma = 2 + 1 + 1 = 4 = {{ - \left( { - 4} \right)} \over 1} = {{ - b} \over a}
                           \alpha \beta + \beta \gamma + \gamma \alpha = \left( 2 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 2 \right)
                            = 2 + 1 + 2 = 5 = {5 \over 1} = {c \over a}
           and           \alpha \beta \gamma = \left( 2 \right)\left( 1 \right)\left( 1 \right) = 2 = {{ - \left( { - 2} \right)} \over 1} = {{ - d} \over a}


Q.2       Find a cubic polynomial with the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively. 
Sol.     Let zeroes be \alpha ,\beta \,\,and\,\,\gamma
We have given {\alpha + \beta + \gamma = 2}
{\alpha \beta + \beta \gamma + \gamma \alpha = - 7}
and \alpha \beta \gamma = - 14
Cubic polynomial is given by \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \,\gamma } \right) = 0
 \Rightarrow {x^3} - {x^2}\left( {\alpha + \beta + \gamma } \right) + x\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) - \alpha \beta \gamma = 0
 \Rightarrow {x^3} - 2{x^2} - 7x + 14 = 0
\alpha \beta \gamma = - 14 = {{ - 14} \over 1} = {{ - d} \over a}
             If a = 1, then b = – 2, c = – 7 and d = 14
             So, one cubic polynomial which fits the given conditions will be {x^3} - 2{x^2} - 7x + 14.


Q.3       If the zeroes of the polynomial {x^3} - 3{x^2} + x + 1 are a – b, a, a + b , find a and b
Sol.       Since (a – b) , a (a + b) are the zeroes of the polynomial {x^3} - 3{x^2} + x + 1,
             Therefore, \left( {a - b} \right) + a + \left( {a + b} \right) = {{ - \left( { - 3} \right)} \over 1} = 3 
               \Rightarrow 3a = 3  \Rightarrow a = 1 
               (a – b) a + a(a + b) + (a + b) (a – b)  = {1 \over 1} = 1
               \Rightarrow {a^2} - ab + {a^2} + ab + {a^2} - {b^2} = 1
               \Rightarrow 3{a^2} - {b^2} = 1
               \Rightarrow 3{\left( 1 \right)^2} - {b^2} = 1 [Since a = 1]
               \Rightarrow 3 - {b^2} = 1
               \Rightarrow {b^2} = 2
               \Rightarrow b = \pm \sqrt 2
            Hence , a = 1 and b = \pm \sqrt 2 .


Q.4       If two zeroes of the polynomial {x^4} - 6{x^3} - 26{x^2} + 138x - 35{\mkern 1mu} \,\,are\,\,{\mkern 1mu} 2 \pm \sqrt 3 , find other zeroes. 
Sol.       Since \,{\mkern 1mu} 2 \pm \sqrt 3 , are two zeroes of the polynomial p\left( x \right) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35
             Let x = 2 \pm \sqrt 3  \Rightarrow x - 2 = \pm \sqrt 3
             Squaring we get {x^2} - 4x + 4 = 3
               \Rightarrow {x^2} - 4x + 1 = 0
             Let us divide p(x) by {x^2} - 4x + 1 to obtain other zeroes.

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          Therefore, p\left( x \right) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35
                                  = \left( {{x^2} - 4x + 1} \right)\left( {{x^2} - 2x - 35} \right)
                                  = \left( {{x^2} - 4x + 1} \right)\left( {{x^2} - 7x + 5x - 35} \right)
                                  = \left( {{x^2} - 4x + 1} \right)\left[ {x\left( {x - 7} \right) + 5\left( {x - 7} \right)} \right]
                                  = \left( {{x^2} - 4x + 1} \right)\left( {x + 5} \right)\left( {x - 7} \right)
           \Rightarrow (x + 5) and (x – 7) are other factors of p(x).
          Therefore,  – 5 and 7 are other zeroes of the given polynomial.


Q.5       If the polynomial {x^4} - 6{x^3} + 16{x^2} - 25x + 10 is divided by another polynomial {x^2} - 2x + k, the remainder comes out to be x + a find k and a. 
Sol.       Let us divide {x^4} - 6{x^3} + 16{x^2} - 25x + 10 by {x^2} - 2x + k

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            So, remainder = (2 k – 9) x – (8 – k) k + 10 
            But the remainder is given as x + a. 
            On comparing their coefficients , we have  2 k – 9 = 1 
             \Rightarrow 2 k = 10 
             \Rightarrow k = 5 
            and                             – (8 – k)k + 10 = a 
             \Rightarrow a = – (8 – 5)5 + 10  = – 3 × 5 + 10 = – 15 + 10 = – 5
            Hence, k = 5 and a = – 5.



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