Question

How do you find all the zeros of `f(x) = x^3-4x^2+9x-36`?

Answer 1

The zeros are:

`x=4" "` and `" "x=+-3i`

Explanation:

Given:

`f(x) = x^3-4x^2+9x-36`

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

`x^3-4x^2+9x-36 = (x^3-4x^2)+(9x-36)`

`color(white)(x^3-4x^2+9x-36) = x^2(x-4)+9(x-4)`

`color(white)(x^3-4x^2+9x-36) = (x^2+9)(x-4)`

`color(white)(x^3-4x^2+9x-36) = (x^2+3^2)(x-4)`

`color(white)(x^3-4x^2+9x-36) = (x^2-(3i)^2)(x-4)`

`color(white)(x^3-4x^2+9x-36) = (x-3i)(x+3i)(x-4)`

So the zeros are:

`x=4" "` and `" "x=+-3i`