Answer :
Answer:
The correct option is D) [tex]f(x) = x^3 + 2x^2 - 2x - 4[/tex] .
Step-by-step explanation:
Consider the provided cubic function.
We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.
A "zero" of a given function is an input value that produces an output of 0.
Substitute the value of zeros in the provided options to check.
Substitute x=-2 in [tex]f(x) = x^3 + 2x^2 - 2x + 4[/tex] .
[tex]f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8[/tex]
Therefore, the option is incorrect.
Substitute x=-2 in [tex]f(x) = x^3 + 2x^2 + 2x - 4[/tex] .
[tex]f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8[/tex]
Therefore, the option is incorrect.
Substitute x=-2 in [tex]f(x) = x^3 - 2x^2 - 2x - 4[/tex] .
[tex]f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16[/tex]
Therefore, the option is incorrect.
Substitute x=-2 in [tex]f(x) = x^3 + 2x^2 - 2x - 4[/tex] .
[tex]f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0[/tex]
Now check for other roots as well.
Substitute x=√2 in [tex]f(x) = x^3 + 2x^2 - 2x - 4[/tex] .
[tex]f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0[/tex]
Substitute x=-√2 in [tex]f(x) = x^3 + 2x^2 - 2x - 4[/tex] .
[tex]f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0[/tex]
Therefore, the option is correct.
