Posted by A while back I I began a discussion about deriving formulas for solving polynomial equations. We saw that linear and quadratic polynomials did not pose much of a challenge. But cubic polynomials are considerably more complex. The set-up was that we had a polynomial equation of this form:
\[
x^3+ax^2+bx+c=0
\]
We can assume that the leading coefficient is one, so long as we’re OK with the idea that the other coefficients might be fractions. If we have a cubic polynomial whose leading coefficient is not one, then we can simply divide through by whatever it is.
We got as far as saying that if you carry out the change of variables
\[
x=y-\frac{a}{3}
\]
the result is an equation of the form:
\[
y^3+py=q
\]
We referred to this sort of cubic polynomial, lacking a square term, as a reduced cubic.
If you work through all of the algebra you will find that
\[
p=b-\frac{a^2}{3} \phantom{xx} \textrm{and} \phantom{xx}
q= -\left( \frac{2a^3}{27}-\frac{ab}{3}+c \right)
\]
The part that’s important to us is the fact that p and q are polynomial functions in the coefficients of our original equation.
The question now becomes: where do we go from here? Why was it helpful to eliminate the quadratic term? The answer lies in the algebraic identity:
\[
\left( s-t \right)^3+3st(s-t)=s^3-t^3
\]
where s and t are arbitrary real numbers.
This identity is readily verified. If you remember the binomial theorem (and who doesn’t!) then you can check it pretty quickly in your head. But now take a good hard look at it. It actually has the same form as our reduced cubic. Imagine, for the moment, that we could find values for s and t with the properties that
\[
s^3-t^3=q \phantom{xxx} \textrm{and} \phantom{xxx} 3st=p
\]
Then our algebraic identity becomes
\[
(s-t)^3+p(s-t)=q
\]
This implies that (s-t) is a solution to our reduced cubic. It follows that finding a formula for the reduced cubic is equivalent to solving our pair of equations. This is not so hard to do using standard algebraic techniques. We begin by squaring the first equation to get
\[
s^6-2s^3t^3+t^6=q^2
\]
Then we cube the second equation to get
\[
s^3t^3=\frac{p^3}{27}
\]
When this second equation is multiplied by four and added to the first equation, the result is
\[
s^6+2s^3t^3+s^6=q^2+\frac{4p^3}{27}
\]
Not impressed? Really? Well, what if I called your attention to the fact that the left-hand side is actually a perfect square? We actually have
\[
\left( s^3+t^3)^2=s^6+2s^3t^3+t^6
\]
It follows that if we take square roots on both sides of our equation then we get
\[
s^3+t^3=\sqrt{q^2+\frac{4p^3}{27}}
\]
Time to take stock. We have one equation about the sum of the cubes of s and t. We have another that addresses the differences of the cubes of s and t (scroll up a bit to refresh your memory). Adding them together gives us
\[
2s^3=q+\sqrt{q^2+\frac{4p^3}{27}}
\]
Subtracting one from the other gives us
\[
2t^3= -q + \sqrt{q^2+\frac{4p^3}{27}}
\]
Now we divide through by 2 and take cube roots to get
\[
s=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
\]
\[
t=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
\]
Problem solved! We have found a solution to our system of equations. And since we have that
\[
y=s-t,
\]
we have successfully expressed x in terms of our coefficients p and q. In other words, we have the formula we seek.
Of course, it’s a pretty complicated formula. And it’s even worse than it looks, since p and q were just the coefficients of the reduced cubic. They were themselves complicated functions of our original coefficients a, b and c. If we tried to replace p and q with a, b and c in our formula above, we would be left with something very unwieldy indeed. If you had an actual cubic polynomial whose roots you actually wanted to find, you would do far better with a computer algebra system than you would with the cubic formula.
But it’s still pretty cool that the formula exists at all. And this particular derivation, first discovered in the sixteenth century by the Italian mathematician Tartaglia and first published by his friend/rival Cardano (it’s a long story), is really very clever.
I’m sure that, having come this far, you have already moved on to wondering if there is a formula for solving the general fourth degree polynomial. Indeed there is, but that’s a different post…
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