Question

What are the real zeros of this function `f(x)=3x^3+6x^2-15x-30`?

Answer 1

`f(x)` has zeros `+-sqrt(5)` and `-2`

Explanation:

Given:

`f(x) = 3x^3+6x^2-15x-30`

Note that all of the terms are divisible by `3`. In addition, the ratio of the first and second terms is the same as that of the third and fourth terms.

So we can separate out the scalar factor `3` and factor by grouping...

`3x^3+6x^2-15x-30 = 3(x^3+2x^2-5x-10)`

`color(white)(3x^3+6x^2-15x-30) = 3((x^3+2x^2)-(5x+10))`

`color(white)(3x^3+6x^2-15x-30) = 3(x^2(x+2)-5(x+2))`

`color(white)(3x^3+6x^2-15x-30) = 3(x^2-5)(x+2)`

`color(white)(3x^3+6x^2-15x-30) = 3(x^2-(sqrt(5))^2)(x+2)`

`color(white)(3x^3+6x^2-15x-30) = 3(x-sqrt(5))(x+sqrt(5))(x+2)`

So `f(x)` has zeros `+-sqrt(5)` and `-2`