Question

How do you find all the real and complex roots of `x^3 - x^2 - 7x + 15=0`?

Answer 1

Use the rational root theorem to help find one root, then factor that out to find the remaining two Complex roots:

`x = -3`, `x = 2+-i`

Explanation:

`f(x) = x^3-x^2-7x+15`

By the rational root theorem, any rational roots of `f(x) = 0` must be expressible in the form `p/q` for some integers `p` and `q` where `p` is a factor of the constant term `15` and `q` is a factor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1`, `+-3`, `+-5`, `+-15`

Let's try the first few...

`f(1) = 1-1-7+15 = 8`

`f(-1) = -1-1+7+15 = 20`

`f(3) = 27-9-21+15 = 12`

`f(-3) = -27-9+21+15 = 0`

So `x=-3` is a root of `f(x) = 0` and `(x+3)` is a factor of `f(x)`

`x^3-x^2-7x+15 = (x+3)(x^2-4x+5)`

The remaining quadratic factor has negative discriminant, so no Real roots, but we can find its Complex roots using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`= (4+-sqrt((-4)^2-(4*1*5)))/(2*1)`

`=(4+-sqrt(-4))/2 = 2+-i`