Question

2. 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

Answer:

(i) p(y)=y

2

−y+1

p(0)=(0)

2

−(0)+1=0+0+1=1

p(1)=(1)

2

−(1)+1=1−1+1=1

p(2)=(2)

2

−(2)+1=4−2+1=3

(ii) p(t)=2+t+2t

2

−t

3

p(0)=2+(0)+2(0)

2

−t

3

=2+0+0−0=2

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

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)=x

3

p(0)=(0)

3

=0

p(1)=(1)

3

=1

p(2)=(2)

3

=8

(iv) 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