# An Unexpected Quartic Solution

Many years ago, while in high school, I tried my hand at solving cubic and quartic formulas. Although there are entirely systematic approaches, using Galois theory, this was not something that I was familiar with at the time. I had just heard that it is possible. Here, ‘solving’ means to find an expression for the roots of the polynomial in terms of its coefficients, involving the standard arithmetical operations of addition, subtraction, multiplication and division, as well as extracting square roots, cube roots, etc.

The solution for cubics went very well. In class one day, the teacher wrote a specific example of a quartic on the blackboard, and proceeded to solve it by reducing to two easy quadratics. The reason that his example worked so easily is because the coefficients formed a palindrome. That is, they were the same when written in reverse order. As an example, consider the equation,

 $\displaystyle x^4+2x^3-x^2+2x+1=0.$

If we divide through by ${x^2}$ then, with a little rearranging, this gives,

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

As a quadratic in ${x+1/x}$, this is easily solved. One solution is ${x+1/x=-3}$. Multiplying by x and rearranging gives a new quadratic,

 $\displaystyle x^2+3x+1=0.$

By the standard formula for quadratics, we obtain

 $\displaystyle x=(-3\pm\sqrt{5})/2.$

It can be checked that this does give two real solutions to the original quartic.

Now, the approach that I attempted for the general quartic was to apply a substitution in order to simplify it, so that a similar method can be applied. Unfortunately, this resulted in a very messy equation, which seemed to be giving a sextic. That is, I went from the original fourth order polynomial, to what was looking like a sixth order one. This was complicating the problem, and getting further away from the goal than where I had started. I am not sure why I did not give up at that point, but I continued. Then, something amazing happened. Computing the coefficients of the sixth, fifth and fourth powers in this sextic, they all vanished! In fact, I had succeeded in reducing the quartic to a cubic, which can be solved. This still seems surprising, that such a messy looking expression should cancel out like this, in just the way that was needed. See equation (2) below for what I am talking about. As this was such a surprise at the time, and is still so now, I have decided to write it up in this post. It just demonstrates that, even if something seems hopeless, if you continue regardless then everything might just fall into place.

To be complete, I include the solutions for all polynomials up to order 4, although it really gets interesting for the general quartics. As linear equations of the form ${ax+b=0}$ are immediately solved by ${x=-b/a}$, we start with the familiar quadratics.

The solution of a quadratic of the form

 $\displaystyle ax^2+bx+c=0$

for coefficients a, b and c with ${a\not=0}$, is standard high school maths, so hardly needs describing. I will still briefly go over one method of solving this, as it can be done in a very similar way as for higher degree polynomials, where we simplify it by a simple substitution. If the coefficient b was zero, then it would be a linear equation in ${x^2}$, so would have the solutions ${x=\pm\sqrt{-c/a}}$. More generally, try substituting ${x=y+k}$ for some number k to be chosen in a moment. This gives a new quadratic for y,

 $\displaystyle ay^2+(2ak+b)y+(ak^2+bk+c)=0.$

The coefficient of y gives a linear equation in k, and can be set to zero by taking ${k=-b/2a}$. So, solving gives ${y=\pm\sqrt{-(ak^2+bk+c)/a}}$ and substituting back ${x=y-b/2a}$ gives the usual quadratic formula

 $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

#### Depressed cubics

Moving up a degree, cubic equations are considerably more complicated than quadratics. However, depressed cubics, despite the name, are actually pleasantly straightforward. These have no ${x^2}$ term so, after dividing through by the leading coefficient, can be written as

 $\displaystyle x^3+3px+2q=0.$

I extract out the factors 2 and 3 from the coefficients now, as this will simplify the formulas. Consider solutions of the form ${x=u+v}$, with u and v to be chosen. We can make use of the identity ${x^3=u^3+v^3+3uvx}$ to express the cubic as,

 $\displaystyle 3(uv+p)x+(u^3+v^3+2q)=0.$

There is some freedom in how this can be solved, since adding any amount to u and subtracting the same amount from v changes nothing. I will try and make the coefficient of x and the constant term both zero, which is equivalent to solving the simultaneous equations,

 \displaystyle \begin{aligned} & u^3+v^3+2q=0,\\ & uv+p=0. \end{aligned}

It follows that ${u^3}$ and ${v^3}$ are the two solutions of the quadratic,

 $\displaystyle t^2+2qt-p^3=(t-u^3)(t-v^3)=0.$

Applying the standard formula for quadratics solves the depressed cubic,

 \displaystyle \begin{aligned} & x=u+v,\\ & u = \sqrt[3]{-q+\sqrt{q^2+p^3}},\\ & v = \sqrt[3]{-q-\sqrt{q^2+p^3}}. \end{aligned}

This is Cardano’s formula. In the case where ${q^2+p^3}$ is positive, then we can take the real cube roots and obtain a real solution to the cubic. More generally, any complex number has three cube roots. If we were to allow any of these cube roots, then there are nine possible values for x, which is too many. In fact, given any value for u, the value of v is fixed by the identity ${uv=-p}$, so we obtain the three roots of the cubic.

#### General cubics

Given an arbitrary cubic of the form

 $\displaystyle ax^3+bx^2+cx+d=0$

then, consider making a substitution of the form ${x=y+k}$, for a number k to be chosen. We obtain a cubic equation for y,

 $\displaystyle ay^3+(3ak+b)y^2+(3ak^2+2bk+c)y+(ak^3+bk^2+ck+d)=0.$

This will be a depressed cubic so long as the coefficient of ${y^2}$ is zero. As this is a linear equation for k, it is easily solved as ${k=-b/3a}$. Hence, we can solve for y using the depressed cubic formula above, then obtain ${x=y-b/3a}$.

#### Palindromic quartics

Back in school, the teacher wrote an example of a fourth order polynomial on the blackboard. He proceeded to solve it by some simple rearrangements, which reduced to solving two quadratics. The reason that this worked, is because the coefficients formed a palindrome. That is, they were the same in reverse order as in the original order. Consider the general case,

 $\displaystyle ax^4+bx^3+cz^2+bx+a=0.$

Dividing through by ${x^2}$ gives an equation which is unchanged under replacing x by 1/x,

 $\displaystyle ax^2+bx+c+b/x+a/x^2=0.$

This reduces to a quadratic in ${y=x+1/x}$,

 $\displaystyle ay^2+by+(c-2a)=0.$

As it is quadratic, this can be solved by the standard formula so, supposing that ${y=\alpha}$ is one solution, we have ${x+1/x=\alpha}$ or, after multiplying by x,

 $\displaystyle x^2-\alpha x+1=0.$

This gives a quadratic for ${x}$, which is solved by the standard formula.

#### Quasi-palindromic quartics

Palindromic quartics reduce to a couple of quadratics, by the simple trick described above. We ask, for which general quartics of the form

 $\displaystyle ax^4+bx^3+cx^2+dx+e=0.$

does the trick still work? It turns out that it is sufficient for the coefficients to satisfy

 $\displaystyle ad^2=b^2e.$

The edge case with ${b=0}$ also has ${d=0}$, and the the quartic is actually just a quadratic in ${x^2}$. So, suppose that ${b\not=0}$. As before, divide through by ${x^2}$,

 $\displaystyle ax^2+bx+c+d/x+e/x^2=0.$

This gives a quadratic in ${y=x+d/bx}$,

 $\displaystyle ay^2+by+(c-2ad/b)=0,$

which can be solved by the standard formula. If ${y=\alpha}$ is a solution then ${x+d/bx=\alpha}$. Multiplying through by bx gives another quadratic

 $\displaystyle bx^2-\alpha bx+d=0$

which, again, can be solved for x with the standard formula.

#### General quartics

Starting with the general quartic equation

 $\displaystyle ax^4+bx^3+cx^2+dx+e=0,$

we try substituting in ${x=y+k}$ for some choice of k. This gives a new quartic for y,

 $\displaystyle \tilde ay^4+\tilde by^3+\tilde c y^2+\tilde dy+\tilde e=0$ (1)

whose coefficients are polynomials in k,

 \displaystyle \begin{aligned} & \tilde a= a,\\ & \tilde b=4ak+b,\\ & \tilde c=6ak^2+3bk+c,\\ & \tilde d=4ak^3+3bk^2+2ck+d,\\ & \tilde e=ak^4+bk^3+ck^2+dk+e. \end{aligned}

From the treatment of quasi-palindromic quartics above, we know that the equation for y can be solved, so long as we choose k such that ${\tilde a\tilde d^2=\tilde b^2\tilde e}$. Let’s try this. We just need to solve for k such that

 \displaystyle \begin{aligned} \tilde a\tilde d^2-\tilde b^2\tilde e={} & a(4ak^3+3bk^2+2ck+d)^2\\ &-(4ak+b)^2(ak^4+bk^3+ck^2+dk+e) \end{aligned}

is zero. This is starting to look a bit messy. Even worse, notice that expanding out will result in powers of k up to degree six. The attempt to simplify our quartic, seems to be giving a rather messy sextic. Hardly a simplification, and we seem to be getting further away from the goal than where we started!

Let’s plough on regardless. Expanding out the squares above, we obtain the following even messier looking expression,

 \displaystyle \begin{aligned} & a\big(16a^2k^6+24abk^5+(9b^2+16ac)k^4+4(3bc+2ad)k^3\\ &\qquad +2(2c^2+3bd)k^2+4cdk+d^2\big)\\ & -(16a^2k^2+8abk+b^2) (ak^4+bk^3+ck^2+dk+e)=0. \end{aligned} (2)

Now, let’s extract out the coefficients of powers of k. The coefficient of ${k^6}$ is,

 $\displaystyle a16a^2-16a^2a =0.$

Great! Now, the coefficient of ${k^5}$ is,

 $\displaystyle a24ab-16a^2b-8aba=0.$

Fantastic! The coefficient of ${k^4}$ is,

 $\displaystyle a(9b^2+16ac)-16a^2c-8abb-b^2a=0.$

Wow! This reduces the equation for k to a cubic, at most. Despite how unlikely it was looking a few moments ago, this works, and we obtain a solution to the general quartic. Computing the remaining coefficients gives

 \displaystyle \begin{aligned} &(4abc-8a^2d-b^3)k^3+(4ac^2-16a^2e-2abd-b^2c)k^2\\ &\ +(4acd-b^2d-8abe)k+(ad^2-b^2e)=0 \end{aligned}

which, admittedly, is still a bit messy. I will not try directly substituting this back into the formulas given above for cubics. Instead, just note that, for any coefficients of a quartic in x, we obtain at most a cubic in k. This can be solved using the methods above. Then, solve the quasi-palindromic quartic (1) by reducing it to a couple of quadratics, and we obtain the solution of the original quartic as ${y=x+k}$.