Question

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

Answer 1

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

Explanation:

The difference of squares identity can be written:

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

We use this later with `a=x` and `b=i`

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

Note that the ratio of the first and second terms is the same as that of the third and fourth terms, so this cubic will factor by grouping:

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

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

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

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

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

Hence zeros:

`x = +-i`

`x = 3`