Question

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

Answer 1

Answer is:
x = 0, x = –1 with multiplicity 2

Answer 2

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

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) = (√(2))^3+2(√(2))^2 - 2(√(2)) - 4\\f(x) =2√(2)+4-2√(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) = (-√(2))^3+2(-√(2))^2 - 2(-√(2)) - 4\\f(x) =-2√(2)+4+2√(2)-4\\f(x) =0`

Therefore, the option is correct.