Question

How do you find all the zeros of `f(x)=x^4+5x^3-2x^2-18x-12`?

Answer 1

`f(x)` has zeros `-1, 2, -3-sqrt(3)` and `-3+sqrt(3)`

Explanation:

`f(x) = x^4+5x^3-2x^2-18x-12`

Note that:

`f(-1) = 1-5-2+18-12 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

`x^4+5x^3-2x^2-18x-12`

`=(x+1)(x^3+4x^2-6x-12)`

Next try substituting `x=2` in the remaining cubic to get:

`8+4(4)-6(2)-12 = 8+16-12-12 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3+4x^2-6x-12`

`=(x-2)(x^2+6x+6)`

`=(x-2)((x+3)^2-9+6)`

`=(x-2)((x+3)^2-(sqrt(3))^2)`

`=(x-2)((x+3)-sqrt(3))((x+3)+sqrt(3))`

`=(x-2)(x+3-sqrt(3))(x+3+sqrt(3))`

Hence zeros `x = -3+-sqrt(3)`