Question

How do you find all the zeros of `f(x)=x^4+4x^3-6x^2-36x-27`?

Answer 1

`f(x)` has zeros:

`-1` (with multiplicity `1`)

`3` (with multiplicity `1`)

`-3` (with multiplicity `2`)

Explanation:

`f(x) = x^4+4x^3-6x^2-36x-27`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-27` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-3, +-9, +-27`

We find:

`f(-1) = 1-4-6+36-27 = 0`

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

`x^4+4x^3-6x^2-36x-27 = (x+1)(x^3+3x^2-9x-27)`

In the remaining cubic, the ratio of 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+3x^2-9x-27`

`=(x^3+3x^2)-(9x+27)`

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

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

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

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

So the remaining zeros are `x=3` (with multiplicity `1`) and `x=-3` with multiplicity `2`.