Question

How do you find the zeros of `f(x)=x^3+3x^2+6x+4`?

Answer 1

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

Explanation:

Given:

`f(x) = x^3+3x^2+6x+4`

Note that `-1+3-6+4 = 0`, so `f(-1) = 0`, `x=-1` is a zero and `(x+1)` a factor:

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

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x+1)` and `b=sqrt(3)i` as follows:

`x^2+2x+4 = x^2+2x+1+3`

`color(white)(x^2+2x+4) = (x+1)^2-(sqrt(3)i)^2`

`color(white)(x^2+2x+4) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)`

`color(white)(x^2+2x+4) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)`

Hence the other two zeros are:

`x = -1+-sqrt(3)i`