What is the prime factorization of 110?

step-by-step explanation:

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81