Question

find p(0),p(1)and p(2) for each of the following polynomials: (1)p(t)=2+t+2t²-t³​

Answer 1

Question:-

Find p(0) , p(1) and p(2) of the polynomial

$p(t) = 2 + t + 2 {t}^{2} - {t}^{3}$

Solution:-

1) Substitute t = 0

$t(0) = 2 + (0) + 2( {0)}^{2} + {(0)}^{3}$

$t(0) = 2 + 0 + 0 + 0$

$t(0) = 2$

2) Substitute t = 1

$p(1) = 2+ (1) + 2( {1)}^{2} + {(1)}^{3}$

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

$p(1) = 6$

3) Substitute t = 2

$p(2) = 2 + (2) + 2 {(2)}^{2} - {(2)}^{3}$

$p(2) = 2 + 2 + 8 - 8$

$p(2) = 4$

Answer 2

Step-by-step explanation:

1) Substitute t = 0

t(0) = 2 + (0) + 2( {0)}^{2} + {(0)}^{3}t(0)=2+(0)+2(0)

2

+(0)

3

t(0) = 2 + 0 + 0 + 0t(0)=2+0+0+0

t(0) = 2t(0)=2

2) Substitute t = 1

p(1) = 2+ (1) + 2( {1)}^{2} + {(1)}^{3}p(1)=2+(1)+2(1)

2

+(1)

3

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

p(1) = 6p(1)=6

3) Substitute t = 2

p(2) = 2 + (2) + 2 {(2)}^{2} - {(2)}^{3}p(2)=2+(2)+2(2)

2

−(2)

3

p(2) = 2 + 2 + 8 - 8p(2)=2+2+8−8

p(2) = 4p(2)=4