Question

Find p(0),p(1),p(-1) for the polynomial: p(a)=-2a³+3a²-8a+18

Answer 1

Explanation:

(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

Answer 2

Answer:

p(0)=18

p(1)=15

p(-1)=27

Explanation:

p(0)=2(0)³+3(0)²-8(0)+18

=2(0)+3(0)-8(0)+18

=0+0-0+18

=+18

p(1) =2(1)³+3(1)²-8(1)+18

=2(1)+3(1)-8(1)+18

=2+3-8+18

=+15

p(-1) =2(-1)³+3(-1)²-8(-1)+18

=2(-1) +3(1)-8(-1)+18

=-2+3+8+18

=+27

HOPE THIS HELPS