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

 (i) p(y)=y2−y+1

Solution:

p(y) = y2–y+1

∴p(0) = (0)2−(0)+1=1

p(1) = (1)2–(1)+1=1

p(2) = (2)2–(2)+1=3

(ii) p(t)=2+t+2t2−t3

Solution:

p(t) = 2+t+2t2−t3

∴p(0) = 2+0+2(0)2–(0)3=2

p(1) = 2+1+2(1)2–(1)3=2+1+2–1=4

p(2) = 2+2+2(2)2–(2)3=2+2+8–8=4

(iii) p(x)=x3

Solution:

p(x) = x3

∴p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) P(x) = (x−1)(x+1)

Solution:

p(x) = (x–1)(x+1)

∴p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3