Question

Which of the following represents the zeros of f(x) = 5x3 − 6x2 − 59x + 12? (2 points)

Group of answer choices

4, 3, 1 over 5

4, 3, − 1 over 5

4, −3, 1 over 5

4, −3, −1 over 5

Answer 1

4, −3, 1 over 5

Explanation:

x is a zero of
`f(x)` if
`f(x) = 0`

In this problem, we have that:

`f(x) = 5x^(3) - 6x^(2) - 59x + 12`

4, 3, 1 over 5

`f(4) = 5*4^(3) - 6*4^(2) - 59*4 + 12 = 0`

So 4 is a zero of the function

`f(3) = 5*3^(3) - 6*3^(2) - 59*3 + 12 = -84`

So 3 is not a zero of the function, and this option is incorrect

4, 3, − 1 over 5

`f(3) = 5*3^(3) - 6*3^(2) - 59*3 + 12 = -84`

So 3 is not a zero of the function, and this option is incorrect

4, −3, 1 over 5

`f(4) = 5*4^(3) - 6*4^(2) - 59*4 + 12 = 0`

So 4 is a zero of the function

`f(-3) = 5*(-3)^(3) - 6*(-3)^(2) - 59*(-3) + 12 = 0`

So -3 is a zero of the function

`f((1)/(5)) = f(0.2) = 5*(0.2)^(3) - 6*(0.2)^(2) - 59*(0.2) + 12 = 0`

So 1 over 5 is a zero of the function

This is the correct answer.

4, −3, −1 over 5

`f(4) = 5*4^(3) - 6*4^(2) - 59*4 + 12 = 0`

So 4 is a zero of the function

`f(-3) = 5*(-3)^(3) - 6*(-3)^(2) - 59*(-3) + 12 = 0`

So -3 is a zero of the function

`f(-(1)/(5)) = f(-0.2) = 5*(-0.2)^(3) - 6*(-0.2)^(2) - 59*(-0.2) + 12 = 23.52`

-1 over 5 is not a zero of the function