Question

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

Answer 1

Answer:

4, −3, 1 over 5

Step-by-step 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

$f(\frac{1}{5}) = f(0.2) = 5*(0.2)^{3} - 6*(0.2)^{2} - 59*(0.2) + 12 = 0$

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

$f(-\frac{1}{5}) = f(-0.2) = 5*(-0.2)^{3} - 6*(-0.2)^{2} - 59*(-0.2) + 12 = 23.52$

-1 over 5 is not a zero of the function