What type and how many zeros does the function f(x)=x3−8 have? How do you know? What is/are the real zero(s) of f(x)=x3−8? How can you determine the real zero(s) from the graph? How can you determine the real zero(s) by using the factored form of the equation?

Question

What type and how many zeros does the function f(x)=x3−8 have? How do you know? What is/are the real zero(s) of f(x)=x3−8? How can you determine the real zero(s) from the graph? How can you determine the real zero(s) by using the factored form of the equation?

Extrinix · Answer

not enough specification to solve this sir....

Extrinix · Answer

or maam*

person321 · Answer

well thats the question *shrugs*

Extrinix · Answer

what is the subject?

person321 · Answer

algebra II

Extrinix · Answer

my grade level hasn't gotten to Alg. II yet sorry mane

person321 · Answer

ok

jimthompson5910 · Answer

x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)

I'm using the difference of cubes factoring formula
a^3 - b^3 = (a-b)(a^2 + ab + b^2)

jimthompson5910 · Answer

Solving x^3 - 8 = 0 is the same as solving (x-2)(x^2 + 2x + 4) = 0

person321 · Answer

I know how to factor it but why are they talking about zeros?

person321 · Answer

what do they mean by "real zeros"

jimthompson5910 · Answer

"zeros" is the same as saying "roots"

The root is the x value that makes the entire thing equal to 0

In this case, x = 2 is a real root since plugging it in leads to
(x-2)(x^2 + 2x + 4) = (2-2)(2^2 + 2*2 + 4) = (0)*(12) = 0
Or you could say
x^3 - 8 = 2^3 - 8 = 8-8 = 0
Either way shows that x = 2 is a root

The other two roots are found by solving x^2 + 2x + 4 = 0
Use the quadratic formula. You should find that these two roots aren't real numbers (instead they are complex numbers).

person321 · Answer

thank you so much <3

jimthompson5910 · Answer

No problem