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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.

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    $\begingroup$ If it gives the answer as an approximation this makes me wonder if this was one of the questions where they ask you essentially to just graph it and look at the plot to find an approximate solution. I recall having a few like that in high school... $\endgroup$ – Dason Dec 2 '13 at 3:10
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    $\begingroup$ Especially in the newer books, look out for "Technology" icons placed next to a question. It means "stop doing math" and "punch it into your graphing calculator". $\endgroup$ – SimonT Dec 2 '13 at 4:57
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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 $$-3-\frac{1}{9} <x <-3$$

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    $\begingroup$ Nice. I like the way you think. $\endgroup$ – yroc Dec 2 '13 at 0:43
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    $\begingroup$ Very nice. It's interesting to note that these approximations can be thought of as the first terms of the iterative scheme $x_{n+1} = -3 - x_n^{-2}$ with $x_0 = -3$. This yields alternating high-low approximations which converge to the root in question. $\endgroup$ – Antonio Vargas Dec 2 '13 at 3:53
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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

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

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    $\begingroup$ Another miracle would be if I understood what you just did ;) $\endgroup$ – yroc Dec 2 '13 at 1:29
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    $\begingroup$ Hmm, maybe this is beyond high school. It's just a few substitutions that create a quadratic equation in $z^3$ (which can be easily solved via the quadratic formula). Also, there are several steps of working (expanding and simplifying) omitted. I encourage you to do this working for yourself – the miracles that occur are quite astounding! $\endgroup$ – Ben Dec 2 '13 at 1:36
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    $\begingroup$ @self. A general cubic equation $ax^3 + bx^2 + cx + d = 0$ certainly can. Take a look at wikipedia. All I've done is depressed the cubic and applied what wikipedia calls "Vieta's substitution". $\endgroup$ – Ben Dec 2 '13 at 1:43
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    $\begingroup$ @Ben Is this SE Math or SE Religion! LOL. Anyway, I follow the quadratic stuff you did, but I'm at a loss for how you decided on the substitutions you did to make the thing work, i.e., $x=y-1$ and $y=z+\frac{1}{z}$, but I guess you indicated how you did it in your response to self. $\endgroup$ – yroc Dec 2 '13 at 2:07
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    $\begingroup$ @yroc Haha yes, that's why I called them miracle substitutions - no motivation but they just work! $\endgroup$ – Ben Dec 2 '13 at 2:18
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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/

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  • $\begingroup$ Interesting reference; I'll have a look. Thanks $\endgroup$ – yroc Dec 2 '13 at 0:45
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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 $x^3 + 3x^2 + 1 = 0$.

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  • $\begingroup$ Newton's method isn't in our curriculum, but this is good to know--thanks very much. $\endgroup$ – yroc Dec 2 '13 at 0:29
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    $\begingroup$ not in most U.S. high schools. You are making a false statement. Single-variable calculus using newton's method is taught in rich U.S high schools. $\endgroup$ – ILoveMath Dec 2 '13 at 0:33
  • $\begingroup$ Do you know if Newton's method converges for this equation? $\endgroup$ – user99680 Dec 2 '13 at 0:39
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    $\begingroup$ @DonAnselmo I'd agree to your statement up to the word "rich"--financial status does not have an absolute binding to academic standards. Not all "rich" schools teach calculus with Newton's method, and not all "not-rich" schools omit calculus from their curriculum. $\endgroup$ – apnorton Dec 2 '13 at 0:41
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    $\begingroup$ @DonAnselmo: I never attended a high school in the US, but the correlation between wealth and teaching of Newton method doesn't seem to be too obvious. $\endgroup$ – Alex Dec 2 '13 at 1:22
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Since you asked to solve that equation, I would assume you need a method that can be used to get the answer to any arbitrary precision. Since Newton's method is usually not taught in high school, let us try an alternative that does not require anything like gradients or even square roots.

Observe that

$x^3+3 x^2+1=0\Leftrightarrow x^2(x+3)=-1\Leftrightarrow x=-1/x^2-3$

Now the problem becomes finding $x$ so that it equals $-1/x^2-3$.

One might be very quick to find out that $x<3$ and then plug that inequality back into the right hand side and get $x>-3-1/9$. Then you can plug that back into RHS again and get $x<-2433/784\approx-3.10332$, then do it again you get $x>-(18373123/5919489)\approx-3.10383599$. You can do it over and over again to get tighter and tighter upper and lower bounds. You can get a rational bound to arbitrary precision here.

But what if one is from one of those high school that doesn't even teach inequalities?

Well the worst one can do is to guess an arbitrary number (other than 0, of course) for $x$, you can evaluate this and see if actually $x$ do equal to $-1/x^2-3$.

Chances are, you won't be that lucky and they just happen to be the same; maybe they are even orders of magnitudes away. Well, two things that are supposed to be equal are not so, then they probably should be between that. In other words, obtain a better guess by taking the average of these two numbers: $\tilde{x}=(x-1/x^2-3)/2$

Now we have a better new guess, and keep the procedure you can refine that result to arbitrary precision.

In fact, even better than Newton's method, this kind of iterative procedure converges from any starting guess (other than 0), and the concept is extremely easy. Of course, you have to carry out the work of iterations.

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I remember the Cardano's method (simplified without extension over complex numbers) described in my high school math book. We probably skipped that chapter, however that equation could be solved applying the procedure mechanically, just like a calculus exercise.

However, given the fact that the solution is given as a simple number and not as a radical, I imagine that an approximation like the one listed above was required.

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