Answer:
4, −3, 1 over 5
Step-by-step explanation:
x is a zero of [tex]f(x)[/tex] if [tex]f(x) = 0[/tex]
In this problem, we have that:
[tex]f(x) = 5x^{3} - 6x^{2} - 59x + 12[/tex]
4, 3, 1 over 5
[tex]f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0[/tex]
So 4 is a zero of the function
[tex]f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84[/tex]
So 3 is not a zero of the function, and this option is incorrect
4, 3, − 1 over 5
[tex]f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84[/tex]
So 3 is not a zero of the function, and this option is incorrect
4, −3, 1 over 5
[tex]f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0[/tex]
So 4 is a zero of the function
[tex]f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0[/tex]
So -3 is a zero of the function
[tex]f(\frac{1}{5}) = f(0.2) = 5*(0.2)^{3} - 6*(0.2)^{2} - 59*(0.2) + 12 = 0[/tex]
So 1 over 5 is a zero of the function
This is the correct answer.
4, −3, −1 over 5
[tex]f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0[/tex]
So 4 is a zero of the function
[tex]f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0[/tex]
So -3 is a zero of the function
[tex]f(-\frac{1}{5}) = f(-0.2) = 5*(-0.2)^{3} - 6*(-0.2)^{2} - 59*(-0.2) + 12 = 23.52[/tex]
-1 over 5 is not a zero of the function
