Question

How do you find all the zeros of `f(x)=x^3-2x^2-11x+52`?

Answer 1

`f(x)` has zeros `-4` and `3+-2i`

Explanation:

`f(x) = x^3-2x^2-11x+52`

By the rational root 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 `52` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-13, +-26, +-52`

In addition, note that the signs of the coefficients are in the pattern `+ - - +`, having `2` changes of sign, so by Descartes' rule of signs, `f(x)` has `2` or `0` positive Real zeros.

The pattern of signs of coefficients of `f(-x)` is `- - + +`. With just one change of sign, `f(x)` must have exactly one negative Real zero.

So check for negative rational zeros first:

`f(-1) = -1-2+11+52 = 60`

`f(-2) = -8-8+22+52 = 58`

`f(-4) = -64-32+44+52 = 0`

So `x=-4` is a zero and `(x+4)` a factor:

`x^3-2x^2-11x+52 = (x+4)(x^2-6x+13)`

The discriminant of the remaining quadratic factor is negative, so it has no Real zeros, but we can factor it with Complex coefficients:

`x^2-6x+13`

`=(x-3)^2-9+13`

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

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

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

Hence zeros: `x=3+-2i`