Find P(0), P(1), P(–2) for the following polynomials:
i.
ii.
(i) P(0) = –3; P(1) = 3; P(–2) = –39
Solution: Given polynomial is P(x) = 10x – 4x2 – 3
Put x = 0,
P(0) = 10(0) – 4(0) – 3
= 0 – 0 – 3
= – 3
Put x = 1,
P(1) = 10(1) – 4(1)2 – 3
= 10 – 4 – 3
= 10 – 7
= 3
Put x = –2,
P(–2) = 10(–2) –4(–2)2 – 3
= – 20 – 16 – 3
= – 39
Hence, P(0) = –3; P(1) = 3; P(–2) = –39
(ii)
P(0) = – 4; P(1) = – 3; P(–2) = 0
Solution.
Given polynomial is P(y) = (y + 2) (y – 2)
Put y = 0,
P(0)= (0 + 2) (0 – 2)
= (2) (–2)
= – 4
Put y = 1,
P(1) = (1 + 2) (1 – 2)
= (3) (–1)
= – 3
Put y = – 2,
P(–2) = (–2 + 2) (–2 – 2)
= (0) (–4)
= 0
Hence, P(0) = – 4; P(1) = – 3; P(–2) = 0.