Q:

Find a cubic function with the given zeros. Square root of two., negative Square root of two, -2. f(x) = x3 + 2x2 - 2x + 4 f(x) = x3 + 2x2 + 2x - 4 f(x) = x3 - 2x2 - 2x - 4 f(x) = x3 + 2x2 - 2x - 4

Accepted Solution

A:
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.