Question

How do you find the zeroes of `f(x)=x^3+10x^2-13x-22`?

Answer 1

Use the rational roots theorem to find possible roots, evaluate `f(x)` for those candidates to find roots `x=-1` and `x=2`. Deduce the third root `x=-11`.

Explanation:

By the rational roots theorem, any rational roots of `f(x) = 0` must be of the form `p/q` where `p`, `q` are integers, `q != 0`, `p` a divisor of the constant term `22` and `q` a divisor of the coefficient `1` of the term of highest degree `x^3`.

So the only possible rational roots are:

`+-1`, `+-2`, `+-11`, `+-22`

`f(1) = 1+10-13-22 = -24`
`f(-1) = -1+10+13-22 = 0`
`f(2) = 8+40-26-22 = 0`

So far, that gives `(x+1)` and `(x-2)` as factors of `f(x)`.

So the remaining factor must be `(x+11)` to get the correct coefficient for `x^3` and the constant term `-22`.

`f(x) = (x+1)(x-2)(x+11)`

Let us check:

`f(-11) = -1331+1210-143-22 = 0`

So the roots of `f(x) = 0` are `x = -1`, `x=2` and `x=-11`