Can $x^3+3x^2+1=0$ be solved using high school methods?

I encountered the following problem in a high-school math text, which I wasn't able to solve using factorization/factor theorem:

Solve $x^3+3x^2+1=0$

Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic? The answer provided was $x\doteq-3.1$, which I was only able to confirm using CAS.

Solution 1:

It is easy to show that the equation has no rational roots, using high-school level math. Indeed, if $x =\frac{p}{q}$ is a root, you can easily show that $p \mid 1$ and $q \mid 1$ thus the only potential rational roots are $\pm 1$, and they don't work. This shows that we cannot probably calculate the solution in a simple way, we have to approximate.

To approximate, the simplest idea would be to write the equation as

$$x^2(x+3)=-1$$

As $x^2 \geq 0$ it follows that $$x+3 <0 \,.$$ Thus, $x <-3$. But then, $$x^2 >9 \,.$$ Multiplying this by $x+3$, which is negative, we get

$$-1=x^2(x+3)< 9(x+3) \,.$$

This yields

$$x+3 > \frac{-1}{9}$$

Thus

$$ \frac{-1}{9} < x+3 <0 \,,$$ or $$-3-\frac{1}{9} <x <-3$$

Solution 2:

You might be able to get away with using some "miracle" substitutions that solve the cubic (which can be generalised to solve a general cubic).

Begin with

$$x^3+3x^2+1=0 \tag{$\dagger$}$$

1. Substitute $x = y - 1$ into $(\dagger)$ to obtain

   $$y^3 - 3y + 3 = 0. \tag{$\star$}$$

2. Let $y = z + \frac{1}{z}$ so that $(\star)$ becomes $z^3+\frac{1}{z^3}+3 = 0$ or by multiplying through by $z^3$,

   $$\left(z^3\right)^2 + 3\left(z^3\right) + 1 = 0. \tag{$\ast$}$$

3. Solve $(\ast)$ for $z^3$ with the quadratic equation formula,

   $$z^3 = \frac{-3\pm\sqrt{5}}{2}.$$

   The two real solutions of $(\ast)$ are thus $z_1 = -\sqrt[3]{(3+\sqrt{5})/2}$ and $z_2 = -\sqrt[3]{(3-\sqrt{5})/2}$.

4. Unravel the substitutions: $x = y - 1 = z + \frac{1}{z} - 1$ so solutions to $(\dagger)$ are

   $$\begin{align*}x_1&= z_1 + 1/z_1 - 1 = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1\\ x_2&= z_2 + 1/z_2 - 1 =-\sqrt[3]{(3-\sqrt{5})/2} - \sqrt[3]{2/(3-\sqrt{5})} - 1\end{align*}$$

It actually turns out that $x_1 = x_2$ (another miracle!). So we may write the final solution in exact form as

$$x = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1 \approx -3.1$$

Solution 3:

Tartaglia:

When the cube and things together
Are equal to some discreet number,
Find two other numbers differing in this one.
Then you will keep this as a habit
That their product should always be equal
Exactly to the cube of a third of the things.
The remainder then as a general rule
Of their cube roots subtracted
Will be equal to your principal thing
In the second of these acts,
When the cube remains alone,
You will observe these other agreements:
You will at once divide the number into two parts
So that the one times the other produces clearly
The cube of the third of the things exactly.
Then of these two parts, as a habitual rule,
You will take the cube roots added together,
And this sum will be your thought.
The third of these calculations of ours
Is solved with the second if you take good care,
As in their nature they are almost matched.
These things I found, and not with sluggish steps,
In the year one thousand five hundred, four and thirty.
With foundations strong and sturdy
In the city girdled by the sea.

See here the math http://talkmath.wordpress.com/2010/12/31/the-del-ferro-tartaglia-cardano-solution-of-the-cubic/

Solution 4:

Single-variable calculus involving some form of Newton's method is taught in some U.S. high schools.

Newton's method can be used to find an arbitrarily accurate solution to $x^3 + 3x^2 + 1 = 0$.