Question

if poly 2+ t + 2t² - 3 then fond p(0) , p(1) , p(2) ​

Answer 1

Question

if polynomial, 2+ t + 2t² - 3= 0, then found p(0) , p(1) , p(2) .

Solution

Given:-

• Polynomial , 2t² + t - 1 = 0

Find:-

• p(0) , p(1) & p(2)

Explanation

case(1):-

when,

• x = 0, keep in polynomial equation

==> p(0) = 2 × (0)² + 0 - 1

==>p(0) = 2 × 0 - 1

==> p(0) = -1

Case(2):-

When,

• x = 1 , keep in polynomial

==> p(1) = 2 × (1)² + 1 - 1

==>p(1) = 2 × 1

==>p(1) = 2

Case(3):-

when,

• x = 2 , keep in polynomial

==>p(2) = 2 + (2)² + 2 - 1

==> p(2) = 2 × 4 + 1

==>p(2) = 8 + 1.

==> p(2) = 9

Hence

• Value of p(0) = -1
• Value of p(1) = 2
• Value of p(2) = 9

___________________

Answer 2

Step-by-step explanation:

Given:

• Polynomial is 2 + t + 2t² – 3
• Values of p = (0) , (1) and (2)

To Find:

• Values of the polynomial with p(0) , p(1) and p(2).

Solution: Taking p(0)

$\implies{\rm }$ p(0) = 2 + 0 + 2(0)² – 3

$\implies{\rm }$ p(0) = 2 + 0 + 0 – 3

$\implies{\rm }$ p(0) = – 1

★ Taking p(1) ★

$\implies{\rm }$ p(1) = 2 + 1 + 2(1)² – 3

$\implies{\rm }$ p(1) = 3 + 2 – 3

$\implies{\rm }$ p(1) = 5 – 3

$\implies{\rm }$ p(1) = 2

★ Taking p(2) ★

$\implies{\rm }$ p(2) = 2 + 2 + 2(2)² – 3

$\implies{\rm }$ p(2) = 4 + 2(4) – 3

$\implies{\rm }$ p(2) = 4 + 8 – 3

$\implies{\rm }$ p(2) = 12 – 3

$\implies{\rm }$ p(2) = 9

So,

• p(0) = – 1
• p(1) = 2
• p(2) = 9