Find p(0), p( 1) and p(-2) for the following polynomials

Solution:

(i) Given, polynomial is

p(x) = 10x – 4x2 – 3

On putting x = 0,1 and – 2, respectively in Eq. (i), we get p(0) = 10(0)-4(0)2 -3 = 0-0-3= -3

p(1) = 10 (1) — 4 (1 )2 -3

= 10-4-3= 10-7= 3

and p(-2) =10 (-2)- 4 (-2)2 – 3

= -20-4×4-3 =-20-16-3=-39

Hence, the values of p(0), p(1) and p(-2) are respectively, -3,3 and – 39.

(ii) Given, polynomial isp(y) = (y+2)(y-2)

On putting y =0,1 and -2, respectively in Eq. (i), we get p(0) =(0+2)(0-2)= -4

p(1) = (1 + 2)(1-2)

= 3 x (-1) = -3

and p(-2) = (-2 + 2)(-2 -2)

=0 (-4) = 0

Hence, the values of p(0),p(1) and p(-2) are respectively,-4,-3 and 0.