Question

Find the zero of the given polynomial :

`p(x) = {x}^(2) + 1`​

Answer 1

+i , -i

Step-by-step explanation:

Zero of the polynomial:

p(x) =0

x² + 1 = 0

x² = -1

Take square root both sides,

`\sf x = √(-1)\\\\x =√(i^2)\\`

x = ± i

Zeros of the polynomial:

x = +i , -i

Answer 2

x = i, -i

Step-by-step explanation:

p(x) = x² + 1

To set a polynomial to zero, p(x) = 0.

⇒ x² + 1 = 0

relocate variable

⇒ x² = -1

square root both sides

⇒ x = ±√-1

breakdown

⇒ x = √-1, -√-1

simplify if x = √-a then √a i

⇒ x = i, -i