Question

How do you find all the zeros of `x^3-4x^2-44x+96`?

Answer 1

This cubic has zeros: `2`, `8` and `-6`

Explanation:

`f(x) = x^3-4x^2-44x+96`

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 `96` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-32, +-48, +-96`

Trying each in turn, we find:

`f(2) = 8-16-88+96 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-4x^2-44x+96 = (x-2)(x^2-2x-48)`

To factor the remaining quadratic find a pair of factors of `48` which differ by `2`. The pair `8, 6` works. Hence:

`x^2-2x-48 = (x-8)(x+6)`

Hence zeros: `8` and `-6`