Question

How do you find all roots for `x^3 - x^2 - 14x + 24 = 0`?

Answer 1

`x=2`, `x=-4` and `x=3`

Explanation:

`f(x) = x^3-x^2-14x+24`

By the rational root 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 `24` 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, +-24`

Trying each in turn, we find:

`f(1) = 1-1-14+24 = 10`

`f(-1) = -1-1+14+24 = 36`

`f(2) = 8-4-28+24 = 0`

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

`x^3-x^2-14x+24 = (x-2)(x^2+x-12)`

We can factor the remaining quadratic by finding a pair of factors of the absolute value `12` of the constant term which differ by the coefficient `1` of the middle term. The pair `4,3` works.

Hence we find:

`x^2+x-12 = (x+4)(x-3)`

Putting it together:

`x^3-x^2-14x+24 = (x-2)(x+4)(x-3)`

which has zeros `x=2`, `x=-4` and `x=3`