# How do you find the number of complex, real and rational roots of #4x^3-12x+9=0#?

##### 1 Answer

This cubic has one negative irrational Real root and two Complex ones.

#### Explanation:

#f(x) = 4x^3-12x+9#

**Discriminant**

The discriminant

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

In our example,

#Delta = 0+27648+0-34992+0 = -7344#

Since

**Descartes' Rule of Signs**

The signs of the coefficients of

**Rational root theorem**

Since all of the coefficients of

That means that the only possible *rational* zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-3, +-9/2, +-9#

Note that we already know that the only Real zero is negative, so if there is a rational zero then it is one of:

#-1/4, -1/2, -3/4, -1, -3/2, -3, -9/2, -9#

We find:

#f(-3) = -63 < 0#

#f(-3/2) = 27/2 > 0#

So the Real zero is somewhere between the possible rational zeros

**Conclusion**