SOLVING THE CUBIC
The method we will introduce here for solving cubics is known as Cardano's method. This
method was published by Gerolamo Cardano, an Italian mathematician in the 1500s. The method
involves depressing the original cubic losing the squared term, solving the easier depressed
cubic, then re-evaluating from the initial depression.
First, let's introduce the cubic form as
we can use a substitution of the form
. In order to depress the cubic,
. Let us substitute this back in our original
equation, simplify, and discover the result.
First, the substitution:
Now, let us expand this out:
We now need to combine like terms. So, let us rewrite the equation by moving terms around:
I hope at this point it is obvious that
. So, reducing the rest of the
equation, and simplifying where needed:
Phew! That was a bit of work! So, now given any cubic equation, we can rewrite its depressed
cubic using the equation above. For example, consider the equation:
This equation only has one real root. If you were to graph this equation, you could visually see
that root. However, we're interested in doing this algebraically. So, we can rewrite our equation
as a depressed cubic using our substitution method:
using the derived equation above. If we had substituted
, we would have inserted
, and simplified. The resulting depressed equation is simpler to solve than the original
cubic. Notice our equation is in the form
.
For our example of
, we have
and
.
To solve this equation involves a two-by-two non-linear system of the form
and
So in our case,
and
.
Solving for
in the first equation gives . We can substitute this in our second equation giving
. If we multiply through by
, we get:
.
Take a look at that equation for a minute. You may think that we have made the equation much
more difficult to solve, because we have a sixth-degree polynomial. However, if you substitute
in for , you'll find that you have a simple quadratic involving , and a constant. We already
know how to find the general solution to a quadratic, so let's solve that here. Remember the
quadratic formula for solving
is:
Let's plug in our coefficients, and see where we end up:
or
Fortunately, our discriminant
was positive, and as a result, we do not have to deal with
complex numbers. However, if our discriminant was negative, how would we get a handle on
this? This is something the mathematicians 4000 years ago up through the 1500s dismissed as
either nonexistent, impossible or simply illogical.
"Imaginary numbers" just didn't exist, and they had no algebraic significance to the original
problem. However, Gerolamo Cardano and his student Lodovico Ferrari were able to study the
nature of complex roots, and see how they could be used in finding a general solution.
Remember, that a cubic can have three real roots or one real root with a complex conjugate root
pair. It was the discovering a general method for solving the cubic that lead mathematicians to
get a better grasp on the nature of complex numbers.
We won't discuss how to handle complex roots here. Suffice it to say, that the "breakthrough" on
how to handle complex numbers, was to continue working with them! If we end up with
complex roots in our solution, then we would continue to work on our calculations, treating as
any standard variable. Amazingly, at least once, the complex root will disappear through
algebraic manipulation! Of course, it's possible that the complex roots are held throughout the
duration of our manipulations, which would tell us that the equation actually contains complex
roots. And that's okay. At least with the cubic, we'll have at least one real root, and that number
may reveal itself when simplifying our expressions.
So, continuing on through our initial problem. We found
find :
Now that we have solved for both
, so now we need to
and , we can solve for :
Something to notice about the equation above. The equation tells me that there are four possible
equations, but according to the Fundamental Theorem of Algebra, there can only be at most three
to any given cubic. So, what happened? Well, if you plug all four possibilities in your calculator,
you'll find that
This is a unique property of the cube-root. Cool, eh? By the way, is it possible to manipulate the
above expression algebraically to express the correct answer, in this case 2? Well, in order to do
this, we need to assume that the expression is in the form
. If we are to manipulate this
expression, this needs to be the case.
So, let's set our expression to
and see what we can do:
So, what does this tell us? This tells us that
and
. We have a
nonlinear system of two equations and two unknowns. By eyeballing the equation, we can see
that
and
.
So, what does this mean? Going back to
, this means that
Further, if we apply the same method of simplification to
, we find that
. So, putting all this together:
So, we found a root for . We need to take this root all the way back to the beginning when we
substituted
. This means that
. To verify:
Indeed, we have found a root! We have solved the equation! Pat yourself on the back.
SOLVING THE QUARTIC
For solving the quartic, we find that the method is similar in nature to solving the cubic. Our
objective is to find two roots to the equation
The other two roots might be real or complex, but they can be found using polynomial long
division and the quadratic equation. As with solving the cubic, the first thing to do is to depress
the quartic by removing the cubed term. We can do this using a substitution method similar to
the cubic:
Inserting this into our original equation, we obtain:
Multiplying everything out and simplifying, we come to the compressed equation:
Consider the equation:
Our substitution will be:
Substituting back into our original equation, multiplying and simplifying, we get the following
depressed quartic:
Now, similar to solving the cubic, we are left with a depressing equation in the form:
If we move the y-term over to the right hand side of the equation, then we can manipulate the left
hand side to form a perfect square:
Now, in order to complete the square on the left hand side, we would like to end up with
. In order to make this possible, we will add the term
to both sides of
the equation:
So, now the left hand side can be written as a perfect square:
Now we need to manipulate the right hand side to form a perfect square. However, if we leave
the left hand side as written, and try manipulating the right hand side only, we'll find that we'll
undo everything on the left that we just accomplished. So, here's the trick: we need to introduce a
new unknown variable to the equation to keep our squares balanced. So, let's introduce into
our equation. We want the left hand side to be
. Multiplying this through,
means that we need to add
obtain:
to both sides of the equation. By doing so, we
At this point, we need to determine before we can complete the square on the right hand side
of the equation. So, going back to our example, let's see what would happen:
Completing the square the left hand side of the equation gives:
Now, we need to introduce our new unknown variable into the equation as before, so we can
attempt to complete the square on the right hand side. Knowing that
, we can add
to both sides
of the equation to give:
Notice, the left hand side is a perfect square and the right hand side is a quadratic. We would like
to have a perfect square on the right hand side as well. By doing so, our equation will have both a
perfect square on the right hand side and the left hand side. This tells us that our equation has a
repeated root. A quadratic polynomial has a repeated root if its discriminant is zero. So,
computing the discriminant gives:
This equation
is called the "resolvent cubic polynomial" for
the quartic equation. When solving for , we find that
. Knowing this, we can substitute
back into our equation where the left hand side was a perfect square, and we were
working on the right hand side. We wanted to make the right hand side a perfect square, just like
the left hand side. So, let's see if this is the case:
Bingo! So, where does that put us? That means both sides of our equation are perfect squares:
Now, taking the square root of both sides of the equation gives:
Let us discard the positive on the right hand side for the time being, and rearrange the equation,
producing a quadratic:
We know the quadratic formula, so its roots are:
Simplified a bit:
Going all the way back to our original quartic, we used the substitution
to our equation
, so two roots
are
As mentioned at the beginning of this section, the other two roots to our original equation can be
found by polynomial long division and the quadratic formula.
A couple words of note. As with the quadratic and the cubic, solving the quartic may mean
introducing complex numbers to our algebraic work. This is expected in many cases. The key is
not to panic! By holding on to the complex variable, and treating it like it's just part of the
number it's attached to, you may find that through algebraic manipulation, that the complex
number cancels itself out. However, they may not. Your original equation might just have
complex roots. Because we're dealing with the quartic, complex number will likely play a much
larger role in the general solution than with cubics or with the equation we demonstrated above.
Conveniently enough, I assumed was positive. Imagine how our equation would have changed
if it were negative!
This procedure will will work for the general case of any quartic. Further, when manipulating the
quartic, there will always be a resolvent cubic for your equation, when you introduce the
variable. The trick is getting a handle on solving for , either by using the Rational Zero Test or
solving the cubic directly. Because of this resolvent cubic polynomial, we are able to complete
the square on the right hand side of the equation, as outlined above.
SOLVING THE QUNTIC?
The last section of this paper involves discussing whether the general quintic has a solution
involving radicals. It turns out that in 1824, Niels Abel was able to complete a proof by Paolo
Ruffini that for general fifth-degree polynomials, you cannot write the solution involving
radicals. Évariste Galois later proved that not only can the solution to generic polynomials of
degree five not be represented using radicals, but his proof extended to polynomials of degree
five and higher.
With the previous formulas for solving the cubic and quartic equations, all that was expected of
the student was to be familiar with college-level algebra. With degree five and higher, the
student is expected to have a concrete understanding of modern algebra in undergraduate or
graduate mathematics. So, without that foundation, this proof may make little sense. As the
author has not yet taken these courses, his understanding of the proof is limited. As a result, the
following proof comes directly off of Wikipedia under the "Abel-Ruffini theorem" article.
The following proof is based on Galois theory. It should be noticed that, historically, Ruffini and
Abel's proofs precede Galois theory.
One of the fundamental theorems of Galois theory states that an equation is solvable in radicals if
and only if it has a solvable Galois group, so the proof of the Abel–Ruffini theorem comes down
to computing the Galois group of the general polynomial of the fifth degree.
Let y1 be a real number transcendental over the field of rational numbers Q, and let y2 be a real
number transcendental over Q(y1), and so on to y5 which is transcendental over Q(y1,y2,y3,y4).
These numbers are called independent transcendental elements over Q.
Let E = Q(y1,y2,y3,y4,y5) and let
Multiplying f(x) out yields the elementary symmetric functions of the yn:
s1 = y 1 + y 2 + y 3 + y 4 + y 5
and so on up to
s5 = y1y2y3y4y5.
The coefficient of xn in f(x) is thus ( − 1)5 − ns5 − n. Because our independent
transcendentals yn act as indeterminates over Q, every permutation σ in the symmetric group on
5 letters S5 induces an automorphism σ' on E that leaves Q fixed and permutes the elements yn.
Since an arbitrary rearrangement of the roots of the product form still produces the same
polynomial, e.g.:
(y − y3)(y − y1)(y − y2)(y − y5)(y − y4)
is still the same polynomial as
(y − y1)(y − y2)(y − y3)(y − y4)(y − y5)
the automorphisms σ' also leave E fixed, so they are elements of the Galois group G(E / F). Now,
since | S5 | = 5! it must be that
, as there could possibly be automorphisms
there that are not in S5. However, since the splitting field of a quintic polynomial has at most 5!
elements, | G(E / F) | = 5!, and so G(E / F) must be isomorphic to S5. Generalizing this argument
shows that the Galois group of every general polynomial of degree n is isomorphic to Sn.
And what of S5? The only composition series of S5 is
(where A5 is
thTplae alternating group on five letters, also known as the icosahedral group). However,
the quotient group A5 / {e} (isomorphic to A5 itself) is not an abelian group, and soS5 is not
solvable, so it must be that the general polynomial of the fifth degree has no solution in radicals.
Since the first nontrivial normal subgroup of the symmetric group on n letters is always the
alternating group on n letters, and since the alternating groups on n letters for
are
always simple and non-abelian, and hence not solvable, it also says that the general polynomials
of all degrees higher than the fifth also have no solution in radicals.
Note that the above construction of the Galois group for a fifth degree polynomial only applies to
the general polynomial, specific polynomials of the fifth degree may have different Galois
groups with quite different properties, e.g. x5 − 1 has a splitting field generated by a primitive
5th root of unity, and hence its Galois group is abelian and the equation itself solvable by
radicals. However, since the result is on the general polynomial, it does say that a general
"quintic formula" for the roots of a quintic using only a finite combination of the arithmetic
operations and radicals in terms of the coefficients is impossible. Q.E.D.
REFERENCES:
http://en.wikipedia.org/wiki/Cubic_formula#Cardano.27s_method
http://www.sosmath.com/algebra/factor/fac11/fac11.html
http://en.wikipedia.org/wiki/Quartic_formula#The_general_case.2C_along_Ferrari.27s_lines
http://www.sosmath.com/algebra/factor/fac12/fac12.html
http://en.wikipedia.org/wiki/Abel–Ruffini_theorem