NCRT CLASS 9TH MATH CHAPTER 2( POLYNOMIALS EXERSICE 2.2 SOLUTION )

              EXERSICE 2.2 

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
 *    let  p(x) = 5x – 4x2 + 3
(i) so, p(x=0) = 5(0) – 4(0)2 + 3

                       = 0 – 0 + 3  =3

     Thus, p(x=o) is 3 . 

(ii) p(x=-1) = 5(-1) – 4(-1)2 + 3
                   = – 5x – 4x2 + 3 

                   = -9 + 3 = -6

     hence , the value of 5x – 4x2 + 3 at 

     x =   -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 

               = 10 – 4(4) + 3
               = 10 – 16 + 3 = -3

                hence , the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:
(i) Given that p(y) = y2 – y + 1.

    so,
∴  P(y=0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
    p(y=1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1
    p(Y=2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴ p(t=0) = 2 + 0 + 2(0)2 – (0)3

    = 2 + 0 + 0 – 0=2
   P(t=1) = 2 + 1 + 2(1)2 – (1)3
   = 2 + 1 + 2 – 1 = 4
  p(t=2) = 2 + 2 + 2(2)2 – (2)3
   = 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(x=0) = (0)3 = 0, p(1) = (1)3 = 1
    p(x=2) = (2)3 = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(x=0) = (0 – 1)(0 + 1) = (-1)(1) = -1
   p(x=1) = (1 – 1)(1 +1) = (0)(2) = 0
   P(x=2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –
(ii) p (x) = 5x – π, x = 
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – 
(vii) P (x) = 3x2 – 1, x = – ,
(viii) p (x) = 2x + 1, x = 
Solution:
(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π
∴ 
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x2 – 1

(viii) We have, p(x) = 2x + 1
∴ 
Since,  ≠ 0, so, x =  is not a zero of 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.
Solution:
(i)  p(x) = x + 5. 

     Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
    Thus, zero of x + 5 is  -5.

ii) P(x)=x - 5

    Since, p(x)=0.

     x-5=0 ⇒x=5.

     hence, zero of x - 5 is 5.

 iii)p(x) = 2x+5

      Since, p(x)=0.

      2x+5=0⇒2x=-5

       x=-5/2.

      hence zero of 2x+5 is -5/2.

iv)  p(x)= 3x-2

      Since, p(x)=0

       3x-2=0⇒x=2/3

       hence zero of 3x-2  is 2/3.

v)   p(x)= 3x

       Since, p(x)=0

       3x=0 ⇒x=0

       hence zero of 3x is 0.

vi)   p(x) =ax

        since ,p(x) =0

        ax=o ⇒x=0

         hence zero of ax is 0.

vii)  p(x)=cx+d

        since ,p(x) =0

        cx+d =0 ⇒ cx=-d

       x=-d/c

        hence zero of cx+d is -d/c