# Is there a systematic way of solving cubic equations?

In my text book, to solve cubic equations, I need to

1. find by trial & error what $f(a)$ will make the equation 0. The factor will be $(x-a)$
2. then the other factor will be $Ax^2+Bx+C$ then I can solve using compare coefficient

But for te first part, is there any other way apart from trail & error? Or is there some technique to guess the factor of an equation?

The question I am currently doing happens to be $2x^3+3x+4=9$

UPDATE: $2x^3+3x+4=9$ should be $2x^3+3x+4=9x^2$

UPDATE 2

So I am going to try substituting

$x=\pm{1}, \pm\frac{1}{2}, \pm{2}, \pm{4}$

When $x=-\frac{1}{2}$, one of the provided answers,

$2(−0.5)^3−9(−0.5)^2+3(−0.5)+4=-18.75\neq 0$

did I make a mistake?

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Have you learned the "rational root theorem"? –  The Chaz 2.0 Sep 4 '11 at 3:09
$2(-0.5)^3 - 9(0.5)^2 + 3(-0.5) + 4 =$ $-2(1/8) -9(1/4) - 3(1/2) + 4 =$ $-(1/4) -(9/4) -(3/2) + 4 = -(16/4)+4 = -4+4 = 0$. So you certainly made a mistake, but since you did not show your arithmetic, I don't know where you made your evaluation mistake. –  Arturo Magidin Sep 4 '11 at 3:51
I had 2(−0.5)^3−9(−0.5)^2+3(−0.5)+4 in my calculator ... hmm ... ya... working manually worked ... –  Jiew Meng Sep 4 '11 at 3:59

I think "trial and error" is probably referring to the Rational Roots Test. To use it, try all fractions of the form $\pm \frac{p}{q}$ where $p$ is a factor of the constant term and $q$ is a factor of the highest degree term.

In your example, we would evaluate each of the following in the function and see if any of them are roots: $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{2}{1}, \pm \frac{2}{2}, \pm\frac{4}{1}, \pm \frac{4}{2}$. (Several of these are redundant, but I include them just so you can see how the list is formed.)

Note that the vast majority of numbers in this list are not roots. All the Rational Roots Test claims is that if there is a rational root of your function, then it will appear somewhere on the list. Most of the numbers on the list are not roots, however.

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Well... there has been an edit –  The Chaz 2.0 Sep 4 '11 at 3:22
@jiewmeng I've updated my solution to correspond to your edit. –  Austin Mohr Sep 4 '11 at 3:37
@Austin Mohr, see update 2, did I make a careless mistake? I also tried the rest of the possible factors but none seem to equate to 0, ... perhaps I made some careless mistakes? –  Jiew Meng Sep 4 '11 at 3:48
@jiewmeng You've not made a mistake. The Rational Roots Test can be stated like this: If there is a rational root, it will appear somewhere on this list. Most of the numbers on the list, however, are meaningless. They certainly can't all be roots (a cubic only has 3 roots, after all). The rational roots test just gives you a relatively short list of possibilities to check. –  Austin Mohr Sep 4 '11 at 4:00
@Austin Mohr, maybe you misunderstood me, I tried all possible roots and none seem to equate to 0, at least in Ubuntu's calculator ... manually working worked ... still looking why 2(−0.5)^3−9(−0.5)^2+3(−0.5)+4 in Ubuntu gives -18.75 while manually working as shown by @Arturo Magidin gives 0 (I also verified it manually) –  Jiew Meng Sep 4 '11 at 4:05

Of course there is :-) A technique for solving cubics has been known since the 16th century.

Let's start with $x^3+ax+b=0$, which is your case. We'll see later that it's always possible to eliminate the $x^2$ term, so this is in fact the only case we really need to worry about.

The trick is to make the problem initially look harder by setting $x=p+q$. Instead of a single variable we now have two; we hope to use the extra freedom to gain something, and we shall.

Substituting, we get

$(p+q)^3+a(p+q)+b=0$

or

$p^3+3p^2q+3pq^2+q^3+a(p+q)+b=0$

and now we observe that $3p^2q+3pq^2$ can be rearranged:

$p^3+q^3+3pq(p+q)+a(p+q)+b=0$

and finally, we observe the two terms with a common $p+q$:

$p^3+q^3+(3pq+a)(p+q)=0$.

This seems about as bad as where we started - however, we have this extra degree of freedom: we can force $3pq+a$ to be 0. If we do that, by setting $q = -\frac{a}{3p}$, we get

$p^3 - \frac{a^3}{27p^3} + b = 0$

and we're basically done: renaming $p^3$ as $z$, this is a quadratic equation in $z$. Solve it, determine $p$, calculate $q$, and the solution to the original equation is $x=p+q$ (of course there could be multiple solutions).

All of this can be recast as explicit equations for the solution, but I never remember those; the only way I can remember this is by recalling the $x=p+q$ trick.

Finally, if you have

$x^3+ax^2+bx+c=0$

you can make a simple linear change of variables $x=y+r$, whereby this becomes

$y^3+(3r+a)y^2+\ldots=0$

so choosing $r = -a/3$ makes the $y^2$ term vanish. Solve for $y$ as before and remember the shift $x=y-a/3$ to get the solution for the original equation.

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In light of the "update", it might be a good time to include the reduction to the form without a quadratic term  :) –  The Chaz 2.0 Sep 4 '11 at 3:26
This is what I do in the last paragraph, right? –  Alon Amit Sep 4 '11 at 3:52
There's a nice geometric interpretation of the "depression" of a cubic: by Vieta, we know that the mean of the three roots is $-a/3$, so performing the translation implied by the depression corresponds to shifting the roots such that the shape formed by the three roots in the complex plane is "centered" on the origin. –  Guess who it is. Sep 4 '11 at 5:40

Yes there is, but it won't be much use in an exam:

Given the cubic equation:

For the general cubic equation (1) with real coefficients, the general formula for the roots, in terms of the coefficients, is as follows if $(2 b^3-9 a b c+27 a^2 d)^2-4 (b^2-3 a c)^3=-27 a^2 \Delta>0$, i.e. if there are two non real roots:

However, this formula is wrong if the operand of the square root is negative or if the coefficients belong to a field which is not contained in the field of the real numbers: When this operand is real and positive, the cubic roots are real and well defined. In the other case, the square root is not real and one has to choose, once for all a determination for it, for example the one with positive imaginary part. For extracting the cubic roots we have also to choose a determination for the cubic roots, and this gives nine possible values for the first root of an equation which has only three roots.

A correct solution may be obtained by remarking that the proof of above formula shows that the product of the two cubic roots is rational. This gives the following formula in which $\sqrt{ }$ or $\sqrt[3]{ }$ stand for any determination of the square or cubic root, if $b^2-3ac \mbox{ } \neq \mbox{ } 0$.

If $Q \mbox{ } \neq \mbox{ } 0$ and $b^2 − 3ac = 0$, the sign of Q has to be chosen for having $C \mbox{ } \neq \mbox{ } 0$.

If $Q = 0$ and $b^2 − 3ac = 0$, the three roots are equal:

$$x_1=x_2=x_3=-\frac{b}{3a}$$.

If $Q = 0$ and $b^2-3ac \mbox{ } \neq \mbox{ } 0$, above expression for the roots is correct but misleading, hiding the fact that no radical is needed to represent the roots. In fact, in this case, there is a double root

$$x_1=x_2=\frac{bc-9ad}{2(3ac-b^2)}$$.

and a simple root

$$x_3=\frac{9a^2d-4abc+b^3}{a(3ac-b^2)}$$.

(continued)

source: Wikipedia

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In view of the extraordinary length of the quotation, it could well be suggested that you simply link that page, and then explain what you want to express??Per chance, I am making some unconscious mistake, but I really see no reason to post such a long quotation, instead of one simple link(as you already did), without the content; in any case, what is the link for?? –  awllower Sep 4 '11 at 9:24
@awllower: it's a standard practice in Stack Exchange to quote the most important bits of the source site due to the possibility of link rot (although with a popular article in Wikipedia, the chance of that happening is probably closer to nil). The link was there since I originally intended for a (much) shorter quote, but ended up converting the whole section. –  Lie Ryan Sep 4 '11 at 9:51
Thanks for your explanation!! I see the reason now, and also appreciate your good answer. –  awllower Sep 4 '11 at 10:12