# How do you find the number of complex, real and rational roots of #x^3-6x^2-7x-12=0#?

##### 1 Answer

#### Answer:

This cubic has one positive irrational real root and two non-real complex roots.

#### Explanation:

Given:

#f(x) = x^3-6x^2-7x-12=0#

**Rational root theorem**

By the rational root theorem, any rational root of this polynomial is expressible in the form

That means that the only possible rational roots are:

#+-1, +-2, +-3, +-4, +-6, +-12#

**Descartes' Rule of Signs**

The pattern of signs of the coefficients of the polynomial is

We find that:

#f(6) = -54 < 0#

#f(12) = 768 > 0#

So the real positive root cannot be rational and must be an irrational number in

Reversing the signs on the terms of odd degree, we get the pattern

**Discriminant**

The discriminant

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example,

#Delta = 1764+1372-10368-3888-9072 = -20192#

Since