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statuscvo [17]
3 years ago
15

Find a cubic function with the given zeros.

Mathematics
1 answer:
Fed [463]3 years ago
8 0

Answer:

The correct option is D) f(x) = x^3 + 2x^2 - 2x - 4 .

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 f(x) = x^3 + 2x^2 - 2x + 4 .

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

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 + 2x^2 + 2x - 4 .

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

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 - 2x^2 - 2x - 4 .

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

Therefore, the option is incorrect.

Substitute x=-2 in f(x) = x^3 + 2x^2 - 2x - 4 .

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

Now check for other roots as well.

Substitute x=√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

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

Substitute x=-√2 in f(x) = x^3 + 2x^2 - 2x - 4 .

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

Therefore, the option is correct.

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