Question

How do you find all the zeros of `f(x) = x^3 + 13x^2 + 57x + 85`?

Answer 1

`x = 5` or `x = -4+-i`

Explanation:

`f(x) = x^3+13x^2+57x+85`

By the rational roots theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `85` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-5`, `+-17`, `+-85`

In addition, since all of the coefficients of `f(x)` are positive, it has no positive zeros, so that leaves:

`-1`, `-5`, `-17`, `-85`

Trying each of these in turn we find:

`f(-5) = -125+325-285+85 = 0`

So `x=-5` is a zero and `(x+5)` a factor:

`x^3+13x^2+57x+85 = (x+5)(x^2+8x+17)`

We can find the remaining two zeros by completing the square:

`0 = x^2+8x+17`

`=(x+4)^2-16+17`

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

`=(x+4)^2-i^2`

`=((x+4)-i)((x+4)+i)`

`=(x+4-i)(x+4+i)`

So `x = -4+-i`