Polynomials : Exercise 2.2 (Mathematics NCERT Class 9th)


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Q.1      Find the value of the polynomial  5x - 4{x^2} + 3 at
               (i) x = 0          (ii) x = – 1        (iii) x = 2
Sol.

Let p\left( x \right) = 5x - 4{x^2} + 3
(i)    At x = 0 : p\left( 0 \right) = 5\left( 0 \right) - 4{\left( 0 \right)^2} + 3
          = 0 - 0 + 3 = 3
(ii)   At x = –1 : p\left( { - 1} \right) = 5\left( { - 1} \right) - 4{\left( { - 1} \right)^2} + 3
          = - 5 - 4 + 3 = - 6
(iii)  At x = 2 : p\left( 2 \right) = 5\left( 2 \right) - 4{\left( 2 \right)^2} + 3 = 10 - 16 + 3
          = 13 - 16 = - 3


Q.2       Find p (0) , p (1) and p(2) for each of the following polynomials :
             (i) p\left( y \right) = {y^2} - y + 1
             (ii) p\left( t \right) = 2 + t + 2{t^2} - {t^3}
             (iii) p\left( x \right) = {x^3}
             (iv) p(x) = (x–1) (x + 1)

Sol.

(i)    We have , p\left( y \right) = {y^2} - y + 1
         p\left( 0 \right) = {\left( 0 \right)^2} - 0 + 1 = 0 - 0 + 1 = 1
         p\left( 1 \right) = {\left( 1 \right)^2} - 1 + 1 = 1 - 1 + 1 = 1
         and p\left( 2 \right) = {\left( 2 \right)^2} - 2 + 1 = 4 - 2 + 1 = 3
(ii)   We have p\left( t \right) = 2 + t + 2{t^2} - {t^3}
         Therefore p\left( 0 \right) = 2 + 0 + 2{\left( 0 \right)^2} - {\left( 0 \right)^3}
           = 2 + 0 + 0 - 0 = 2
          p\left( 1 \right) = 2 + 1 + 2{\left( 1 \right)^2} - {\left( 1 \right)^3}
           = 2 + 1 + 2 - 1 = 5 - 1 = 4
           and p\left( 2 \right) = 2 + 2 + 2{\left( 2 \right)^2} - {\left( 2 \right)^3}
            = 2 + 2 + 8 - 8 = 4
(iii)   We have, p\left( x \right) = {x^3}
           Therefore p\left( 0 \right) = {\left( 0 \right)^3} = 0
           p\left( 1 \right) = {\left( 1 \right)^3} = 1
           and p\left( 2 \right) = {\left( 2 \right)^3} = 8
(iv)    We have p\left( x \right) = \left( {x - 1} \right)\,\left( {x + 1} \right)
           Therefore   p\left( 0 \right) = \left( {0 - 1} \right)\,\left( {0 + 1} \right) = \left( { - 1} \right)\left( 1 \right) = - 1
            p\left( 1 \right) = \left( {1 - 1} \right)\,\left( {1 + 1} \right) = \left( 0 \right)\left( 2 \right) = 0
           and  p\left( 2 \right) = \left( {2 - 1} \right)\,\left( {2 + 1} \right) = \left( 1 \right)\left( 3 \right) = 3


Q.3        Verify whether the following are zeroes of the polynomial, indicated against them.
               (i) p\left( x \right) = 3x + 1,x = - {1 \over 3}

                   (ii) p\left( x \right) = 5x - \pi ,x = {4 \over 5}
                   (iii) p\left( x \right) = {x^2} - 1,x = 1, - 1
                   (iv) p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right),x = - 1,2
                   (v) p\left( x \right) = {x^2},x = 0
                   (vi) p\left( x \right) = l x + m,x = - {m \over \ell }
                   (vii) p\left( x \right) = 3{x^2} - 1,x = - {1 \over {\sqrt 3 }},{2 \over {\sqrt 3 }}
                   (viii) p\left( x \right) = 2x + 1,x = {1 \over 2}
Sol.

(i)     We have , p(x) = 3x + 1
          At x = - {1 \over 3}, p\left( { - {1 \over 3}} \right) = 3\left( { - {1 \over 3}} \right) + 1 = - 1 + 1 = 0
        
Therefore  - {1 \over 3} is a zero of polynomial = 3x + 1.
(ii)    We have p\left( x \right) = 5x - \pi

          At   x = {4\over 5}
           P\left( {{4 \over 5}}\right) = 5\left( {{4\over 5}}\right)-\pi=4-\pi
          Therefore {4\over 5} is not a zero of polynomial 5x - \pi
(iii)   We have, p\left( x \right) = {x^2} - 1

           At x = 1, p\left( 1 \right) = {\left( 1 \right)^2} - 1 = 1 - 1 = 0
           Therefore 1 is a zero of p (x).
           Also , at x = -1 p\left( { - 1} \right) = {\left( { - 1} \right)^2} - 1
            = 1 - 1 = 0
           Therefore -1 is a zero of polynomial x^2- 1
(iv)     We have , p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)

            At x = –1, p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)
             = \left( 0 \right)\left( { - 3} \right) = 0
            Therefore -1 is  a zero of p (x) .
            Also , at x = 2 , p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right) = \left( 3 \right)\left( 0 \right) = 0
           
Therefore 2 is zero of polynomial (x + 1)(x + 2).
(v)       We have, p\left( x \right) = {x^2}

            At x = 0 , p\left( 0 \right) = {\left( 0 \right)^2} = 0
            Therefore 0 is a zero of polynomial x^2 .
(vi)      We have, p\left( x \right) = \ell x + m

             At x = - {m \over \ell } p\left( { - {m \over \ell }} \right) = \ell \left( {{{ - m} \over \ell }} \right) + m
            
 = - m + m = 0
             Therefore {{ - m} \over \ell } is a zero of polynomial lx + m.
(vii)     We have p\left( x \right) = 3{x^2} - 1

             At x = - {1 \over {\sqrt 3 }} p\left( { - {1 \over {\sqrt 3 }}} \right) = 3{\left( {{1 \over {\sqrt 3 }}} \right)^2} - 1
              = 3 \times -{1 \over 3} - 1 = 1 - 1 = 0
             Therefore  {1\over {\sqrt 3 }} is a zero of polynomial 3x^2- 1
             At x = {2 \over {\sqrt 3 }} , p\left( {{2 \over {\sqrt 3 }}} \right) = 3{\left( {{2 \over {\sqrt 3 }}} \right)^2} - 1 = 3 \times {4 \over 3} - 1
              = 4 - 1 = 3
             Therefore  {2\over {\sqrt 3 }} is not a zero of polynomial 3x^2- 1
(viii)    We have p\left( x \right) = 2x + 1

              At x = {1 \over 2} p\left( {{1 \over 2}} \right) = 2\left( {{1 \over 2}} \right) + 1 = 1 + 1 = 2
              Therefore {1 \over 2} is not a zero of polynomial 2x + 1.


Q.4        Find the zero of the polynomial in each of the following cases :
              
(i) p\left( x \right) = x + 5
              
(ii) p\left( x \right) = x - 5
              
(iii) p\left( x \right) = 2x + 5
              
(iv) p\left( x \right) = 3x - 2
              
(v) p\left( x \right) = 3x
              
(vi) p\left( x \right) = ax,a \ne 0
              
(vii) p\left( x \right) = cx + d,c \ne 0,c,d are real numbers.
Sol.

(i)     We have to solve p(x) = 0
           \Rightarrow x + 5 = 0  \Rightarrow x = - 5
          Therefore - 5 is a zero of the polynomial x + 5 .
(ii)     We have to solve p(x) = 0
            \Rightarrow x – 5 = 0  \Rightarrow x = 5
           Therefore 5 is a zero of the polynomial x – 5.
(iii)    We have to solve p(x) = 0
             \Rightarrow 2x + 5 = 0  \Rightarrow x = - {5 \over 2}
            Therefore  - {5 \over 2} is a zero of the polynomial 2x + 5.
(iv)     We have to solve p(x) = 0
             \Rightarrow 3x – 2 = 0  \Rightarrow x = {2 \over 3}
            Therefore {2 \over 3} is a zero of the polynomial 3x – 2.
(v)       We have to solve p(x) = 0
             \Rightarrow 3x = 0  \Rightarrow x = 0
           
Therefore 0 is a zero of the polynomial 3x.
(vi)      We have to solve p(x) = ax, a \ne 0
              \Rightarrow ax = 0  \Rightarrow x = 0
             Therefore 0 is a zero of the polynomial ax.
(vii)     We have to solve p(x) = 0 , c \ne 0
              \Rightarrow cx + d = 0  \Rightarrow x = - {d \over c}
             Therefore  - {d \over c} is a zero of the polynomial cx + d.



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