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Solving cubic equations
[This thread was started in response to a query elsewhere as to whether there is a formula for solving cubic equations. - The Editor]

By Chris Stott on December 21, 1998:

From Chris Stott, age 18, Loughborough Grammar School

There is no formula to solve all cubics, like there is to solve all quadratics, so your teacher was right in a way. [Oh yes there is: read on! - The Editor]. However what there are instead are various techniques for solving them.

A cubic equation can have either:

1 real root + 2 complex roots - where it only cuts the x-axis once

2 real roots - where it cuts the x-axis once, and 'bounces' off it

or 3 real roots - where it cuts it three times.

a cubic equation can be solved, usually reasonably easily if it has integer coefficients.
there is an equation for certain types of cubic. i think there is
one for ax3 + bx + c = 0, though i cannot remember what it is,
it is not a friendly looking thing...

i will carry on with a little bit about the general techniques to
solve any polynomial p(x), such that p(x) = 0

there is a theorem in existance, by whom i can't remember, that says every polynomial can be factorized into a set of quadratics, and linear terms.

eg. consider...

f(x) = 14x6 + 93x5 + 209x4 + 189x3 + 23x2 - 21x -3

quite a daunting polynomial to try to solve...
let's see what can be done with it...

the first method of approach must be the factor theorem...which can be used to find the linear terms....

the factor theorem says that if you take a polynomial p(x) which has a factor (ax+b), then p(-b/a) = 0

so... how do we use this?
i like to look at the coefficient of the highest order term, and the lowest order term.
nb. if the lowest order term is not a constant, you can pull an x out straight away.

the highest order term is 14x6
and the lowest is -3

consider 14: 14 = 14 × 1
14 = 7 × 2
14 = -14 × -1
14 = -7 × -2
14 = -7 × 1 × -2

there are many possibilities, but one of these factors must be the a value in the factor theorem.

consider -3: -3 = 3 × -1
-3 = -3 × 1
-3 = -3 × -1 × -1

one of these factors must represent the b value in the factor theorem. so there are many possibilities. and really it is a matter of number crunching.

let's try a few combinations.

try -7 and 3, such that we are looking to see if (3-7x) exists as a factor

try f(3/7):
we find f(3/7) doesn't equal 0, so (3-7x) is not a factor (assuming i worked this out right)

try 7 and 1, looking if (7x+1) is a factor
so try f(-1/7)
with any luck: f(-1/7) = 0

and therefore (7x + 1) is a factor.

we started with f(x) = 14x6 + 93x5 + 209x4 + 189x3 + 23x2 - 21x -3
no we have f(x) = (7x + 1)g(x) (where g(x) is a new polynomial)

you will notice f(x) / (7x + 1) = g(x), so we can find g(x) using algebraic long division, which i won't go into here, but if anyone desperately wants i might be persuaded to go into at a later date.

we find g(x) = 2x5 + 13x4 + 28x3 + 23x2 - 3

therefore f(x) = (7x + 1)(2x5 + 13x4 + 28x3 + 23x2 - 3)

let's try to continue factorizing this...
first a few observations... this is a polynomial order 5, so can
be made up of:

  • 5 linear equations (of type ax+b),
  • 1 quadratic equation + 3 linear equations,
  • 2 quadratic equations + 1 linear equation


solving these is to a large extent a mixture of intuition, and number crunching. looking at the largest, and smallest powers' coefficients
2 and -3

if there are 5 factor equations we can look at five factors of these numbers, if 4 then 4, etc

2 = 2 x 1 x 1 x 1 x 1
2 = -2 x -1 x 1 x 1
2 = -2 x 1 x -1
By David Parry-Jones on Monday, February 15, 1999 - 09:57 pm:

hi Sarah

I'm Davy Boy ( David Parry-Jones at Helenia Romanes school)

I'm sorry I can't give you the actual cubic formula (I myself have looked - but haven't found anything ...... nobody would tell me anything!!!! But if you do find a formula please do e-mail me. I'd love to know it / quote it!!!)

The info I do have is mainly contained in fearful symmetry (is god a geometer) by Ian Stewart and Martin Golubitsky I hope I've spelled that correctly. On p41 it talks about group theory - it came from equations specifically solving them.....effectively cubics and quartics can be proven but quintics (degree 5!!! yes 5!!!) can't be solved via a formula - the 'symmetry' of the solutions isn't correct.

The cubic equation was found by Scipio del Ferro and Niccolo Fontana;
published by Girolamo Cardano in Ars Magna of 1545 - the book also contained a formula for a quartic (a horrendous formula found by Cardano's student Ludovico Ferrari.)

The story continues on to dicuss who found that the quadratic cubic quartic was solved by a certain trick that didn't work for quintics..... eventually somebody built a prove that showed it wasn't possible
(this was a complicated prove and didn't really deal with the heart of the problem this came to the person who built group theory.)


I hope that's of some help.... I'd love to know any of the formulas -

Good Luck

By David Seery (djs61) on January 24, 1999:

Hello Sarah

Your maths teacher is almost certainly right -- there is no formula for the solution of the general cubic.

Explicit formulae do exist for the solution of special cases (usually one of the coefficients zero), although they are probably not of as much use as you might think for computation.

Let's have a look at the equation x3 = px + q. We
can cast any cubic of the form ax3 = p'x + q' into
this form, so this is the general cubic without a
quadratic term.

If you don't understand anything that I'm doing, please
get back to me and I will try to explain it better.

The solution relies on two clever tricks. The first is to make the substituion x = u + v.
(u + v)3 = p(u + v) + q,
so u3 + 3vu2 + 3uv2 + v3 = p(u + v) + q
and u3 + v3 + (3uv - p)(u + v) = q.

At this point we observe that
3uv - p = 0
u3 + v3 = q
is an assignment that makes the equation true. This is the second clever trick. Substitution of v = p/3u into the second equation gives

u3 + p3/27u3 = q

and then

u6 - qu3 = p3/27

This is a quadratic in u3 which is easily solved.

u = (q/2 + (q2/4 - p3/27)1/2)1/3

Although the square root in that formula can be positive or negative, we lose no generality by taking the positive root, since v always works out to have the other one.

u = (q/2 - (q2/4 - p3/27)1/2)1/3

Adding these together gives the general solution. It might look as if we only get one solution this way, but in fact the cube root has three solutions of the complex numbers.

I'm not sure what made the originator of the u+v trick
try it in the first place, but equations like this were
solved geometrically until algebraic approaches were
available, and it's possible it has a geometric inspiration. An explicit formula is also available for cubics of the form x3 = rx2 + q, although I have never seen a demonstration of it. It is possible it relies on a similar device.

By David Sanders (Dps24) on Monday, March 15, 1999 - 10:55 pm:

Hi, I'm another David, and I can tell you that there is a general formula for all cubics. The other David got the crucial bit, but left out the (comparatively) easy bit!

Let's consider a cubic
a' x3 + b' x2 + c' x + d' = 0

Since it's a cubic (and not a quadratic), we know that a is not 0, so we can divide by a, leaving something of the form

x3 + bx2 + cx + d = 0

Now, put y = x - z. We don't know what z is yet -- we'll decide on that later. All we're doing is shifting the curve horizontally (right if z > 0, left if z < 0).

Then x = y + z, so:

(y+z)3 + b(y+z)2 + c(y+z) + d = 0.

Expand this:

y3 + 3y2z + 3yz2 + z3 + b(y2 + 2yz + z2) + cy + cz + d = 0

so

y3 + y2.[3z + b] + y.[3z2 + 2bz + c] + [z3 + bz2 + cz + d] = 0.

Now, if we could somehow get rid of that term in y2, we would reduce it to something of the form x3 + px + q = 0, which is (except for a couple of minus signs) what the other David solved.

But we can get rid of that y2 term, simply by setting the bracket that it's multiplied by to zero. That is, put 3z + b = 0, i.e. z = -b/3.

Notice that b is not 0, else we are already in the case we've solved!

So we have solved any cubic by reducing it to the form x3 + px + q = 0 via a shift of the origin.

There is a site with historical info. at St. Andrews.

It also has a solution of the general cubic, AND tells you how to do the quartic (4th powers), which you can also do exactly.

Since we can do everything up to quartics, it seems logical that we should be able to carry on -- but unfortunately we can't, as one of the other correspondents pointed out. There's info. on that in St. Andrews too.

Finally, I should point out that you almost never actually use the formula to solve cubic equations! Either they factorise, you do it approximately on a computer, or you don't bother. Having said that, I did use the formula last year in some work on Van der Waals' equation (which has to do with the relationship between pressure, volume and temperature of a gas -- it's an extension / modification of Boyle's law etc.), since a cubic equation crops up there.

Hope that's been helpful,

David.