# Use Cardano’s algebra method to find the roots of `x^3+6x^2+9x+4=0`

*print*Print*list*Cite

### 1 Answer

Cardano's method applies to what is called the depressed cubic, which is of the form:

`t^3+pt+q=0`

To get the equation into the depressed cubic, use the transformation `x=t-6/3=t-2` to get:

`(t-2)^3+6(t-2)^2+9(t-2)+4` expand

`=t^3-6t^2+12t-8` collect like terms

`+6t^2-24t+24`

`+9t-18+4`

`=t^3-3t+2`

This means that the cubic equation to solve using Cardano's method is:

`t^3-3t+2=0`

By inspection, we can see that this equation can be factored as

`(t+2)(t-1)^2=0`

**This equation has solutions `t=-2` and `t=1` which means that `x=-4` and `x=-1` .**

Cardano's method requires us to use the transformation `t=u+v` , which ultimately leads to a very intimidating expression for one of the roots:

`t_1=(-q/2+sqrt{q^2/4+p^3/27})^{1/3}+(-q/2-sqrt{q^2/4+p^3/27})^{1/3}`

however, in this case, we have the very pleasant condition `4q^3+27p^2=0` which means that the roots are:

`t_1=t_2=-{3q}/{2p}=-{3(2)}/{2(-3)}=1`

and

`t_3={3q}/p={3(2)}/-3=-2` which is the same as the method of inspection earlier.

**The solution to the cubic equation is `x=-4` and `x=-1` .**

**Sources:**