Cardano Method Cubed Roots

Question

so I'm having a super tough time understanding this method and how to acquire the roots of this function... (Excuse my formatting; I tried using markdown with no avail)

The initial equation was as followed: Z^3 + 3Z^2 - 24Z + 28, which I depressed into x^3 - 27x +54 using traditional methodology (which was marked correct by my teacher)

You're supposed to use the equation x -> v + u where v^3 = (-c/2) + ((c^2)/4)+(b^3)/27)^(1/2) and u^3 = (-c/2) -((c^2)/4)+(b^3)/27)^(1/2)

The x roots are 3, 3, and -6 but I got decimal numbers. Therefore, I have two questions:

1) How do you find more than one root from this equation? 2) What am I doing wrong here?

Answer 1

From

$$x^3-27x+54=0$$ we have $$(u+v)^3-27(u+v)+54=u^3+v^3+(3uv-27)(u+v)+54=0.$$

Then with

$$uv=9,$$ $$u^3v^3=729$$ this reduces to

$$u^3+v^3+54=0,$$

$$u^6+u^3v^3+54u^3=u^6+54u^3+729=(u^3+27)^2=0.$$

Interstingly, there is a double root in $u^3$ and

$$u^3=-27,$$ $$u=v=-3,x=-6.$$

To find the other roots, you have two options:

• using complex numbers to find the three cubic roots of $-27$,

• factoring the initial polynomial by long division.

Taking the second option,

$$\frac{x^3-27x+54}{x+6}=x^2-6x+9=(x-3)^2.$$

Answer 2

reaching $x^3 - 27x + 54,$ we can reduce errors in this case by taking $x = 3t$ for a new variable $t.$ Now we are setting $27t^3 - 81 t + 54$ to zero, which is the same as setting $$ t^3 - 3 t + 2 $$ to zero. We can see that $t=1$ is an evident root, then factor as $$ (t-1)(t^2 + t - 2) $$ and factor again.

I suggest you also try Cardano for $t^3 - 3 t + 2.$ It is not guaranteed to be nice, might be, might not be. (Actually as nice as possible: the thing where we take the square root is zero)

Answer 3

Cheating:

If you know that there is a double root, this root is also a root of the derivative,

$$3x^3-27=0$$ so that $$x=\pm3.$$

Obviously, the double root is $3$, as it cancels the initial polynomial. Then it is an easy matter to obtain the factorization

$$x^3-27x+54=(x-3)^2(x+6).$$