Question

Find p(0),p(1) and p(2) for each of the following polynomials:
(i) p(y)=y²−y+1
(ii) p(t)=2+t+2t²−t³
(iii) p(x)=x³
(iv) p(x)=(x−1)(x+1)

Answer 1

(i) p(y)=y²−y+1
p(0)=0²−0+1=1
p(1)=1²−1+1=1
p(2)=2²−2+1=3.
(ii) p(t)=2+t+2t²−t³
p(0)=2+0+2×0²−0³=2
p(1)=2+1+2×1²−1³=4
p(2)=2+2+2×2²−2³=4+8−8=4.
(iii) p(x)=x³
p(0)=0
p(1)=1³=1
p(2)=2³=8.
(iv) p(x)=(x−1)(x+1)
p(0)=(0−1)(0+1)=−1
p(1)=(1−1)(1+1)=0
p(2)=(2−1)(2+1)=3