Question

Given the function, f(x)=(x^3)-(3x^2)+(64x)-192 find the zeros

Answer 1

There are a few ways to find the zeros. I am going to use the grouping method.

NOTE: GCF = Greatest Common Factor

The function:
`f(x) = x^3 - 3x^2 + 64x - 192`

Lets group
`x^3 - 3x^2` and
`64x - 192`

Factor
`x^3 - 3x^2`
Find GCF
GCF =
`x^2`

Divide out the GCF
`x^2`

`(x^3 - 3x^2)/(x^2) = (x - 3)`

Factor From:

`x^2(x - 3)`

Now factor 64x -192
GCF of 64 and 192 = 64
Factor out 64 from 64x - 192

`(64x - 192)/(64) = (x - 3)`

Factor Form:

`64(x - 3)`

Now that we have factored the groups, bring them together.

`x^2(x-3)`

`64(x-3)`


`x^2(x-3) + 64(x-3)`

Since (x-3) is the GCF of
`x^2(x-3) + 64(x-3)`, factor (x-3) out.

`(x^2(x-3) + 64(x-3))/((x-3)) = x^2 + 64`

Factor Form:

`(x-3)(x^2 + 64)`

Now set
`(x^2 + 64)(x-3)` to zero and solve for x.

`(x^2 + 64) = 0`

`x^2 = - 64`

Take the square root of both sides of the equal sign:

`√(x^2) = √(-64)`

`x = √(-64)`

`x = \pm 8i`

Now for (x-3)
x - 3 = 0
x - 3 + 3 = 3
x = 3

The zeros and answers:

`x = -8i`

`x = 8i`
x = 3