Question:
Find p(0), p( 1) and p(-2) for the following polynomials
(i) p(x) = 10x – 4x2 – 3
(ii) p(y) = (y + 2)(y – 2)
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.