This article explains how to solve roots of a quartic polynomial in the complex plane. I use the depressed form of the quartic polynomial and Descartes’ method. Additionally, I demonstrate an interactive demo which visualizes the roots of a quartic polynomial.
There is a quartic polynomial which has a single variable, \(x\), and its highest exponent is 4:
\[f(x): a x^4 + b x^3 + c x^2 + d x + e = 0,\]where \(a \neq 0\), \(b\), \(c\), \(d\), \(e\) are real numbers.
Using Tschirnhaus transformation can eliminate \(x^3\) term and depress the quartic polynomial for simplicity. Let \(x\) be \((y-p)\) and substitute \(x\) in the quartic polynomial.
\[\begin{align} f(y-p) =&~ a (y-p)^4 + b (y-p)^3 + c (y-p)^2 + d (y-p) + e \\ =&~ a y^4 + (-4ap + b) y^3 + (6ap^2 -3 bp + c)y^2 +\\ &~ (-4ap^3 + 3bp^2 -2cp + d)y + (ap^4 - bp^3 + cp^2 - dp + e) \\ =&~ 0 \end{align}\]When we set \(p=\frac{b}{4a}\), the coefficient of \(y^3\) term becomes zero. Then, let’s rearrange the polynomial, making the coefficient of \(y^4\) term be 1:
\[g(y): y^4 + q y^2 + r y + s = 0,\]where
\[\begin{align} q &= \frac{-3b^2+8ac}{8a^2},\\ r &= \frac{b^3-4abc+8a^2d}{8a^3},\\ s &= \frac{-3b^4+16ab^2c-64a^2bd+256a^3e}{256a^4}. \\ \end{align}\]Let the above quartic polynomial can be represented as the multiplication of two quadratic polynomials:
\[\begin{align} &y^4 + q y^2 + r y + s \\ &= (y^2 + ky + m)(y^2 + ly + n) \\ &= y^4 + (k+l)y^3 + (kl+m+n) y^2 + (kn+lm)y + mn \end{align}\]Using that the coefficient of \(y^3\) is zero, we obtain \(k+l=0\). Thus, after eliminating \(l\), the coefficients satisfy
\[\begin{align} m+n-k^2&=q, \\ k(n-m)&=r, \\ mn&=s. \end{align}\]For \(k=0\), \(m+n=q\) and \(mn=s\) satisfy, thus \(m\) and \(n\) are the roots of the following quadratic polynomial:
\[t^2 - qt + s = 0\]Consequently,
\[m, n = \frac{q}{2} \pm \frac{\sqrt{q^2-4s}}{2}.\]Also, the above polynomial of \(y\) can be rearranged as
\[(y^2 + m)(y^2 + n) = 0.\]Therefore, we obtain four roots:
\[\begin{align} y_1 &= \sqrt{-m},\\ y_2 &= -\sqrt{-m}, \\ y_3 &= \sqrt{-n}, \\ y_4 &= -\sqrt{-n}, \end{align}\]and
\[\begin{align} x_1 &= \sqrt{-\frac{q}{2}+ \frac{\sqrt{q^2-4s}}{2}}-p,\\ x_2 &= -\sqrt{-\frac{q}{2}+ \frac{\sqrt{q^2-4s}}{2}}-p, \\ x_3 &= \sqrt{-\frac{q}{2}- \frac{\sqrt{q^2-4s}}{2}}-p, \\ x_4 &= -\sqrt{-\frac{q}{2} - \frac{\sqrt{q^2-4s}}{2}}-p, \end{align}\]On the other hands, for a non-zero \(k \neq 0\), let us solve the polynomial. Please remind the below equations.
\[\begin{align} m+n-k^2&=q, \\ k(n-m)&=r, \\ mn&=s. \end{align}\]By multiplying the first row by \(k\) and adding/subtracting the second row, we obtain
\[\begin{align} 2mk &= k^3 + qk - r, \\ 2nk &= k^3 + qk + r. \end{align}\]Then, by multiplying the third row by \(k^2\) and multiplying \(m\) and \(n\) above, a cubic polynomial of \(k^2\) is derived.
\[\begin{align} (k^3 + qk - r)(k^3 + qk + r) &= 4sk^2\\ (k^2)^3 + 2q (k^2)^2 + (q^2-4s)(k^2) -r^2 &= 0 \end{align}\]By using Cardano’s formula, we can obtain the roots of the above cubic polynomial. Let the roots be \(u^2\), \(v^2\), \(w^2\). Then, the roots of six-degree polynomial of \(k\) are \(\pm u\), \(\pm v\), \(\pm w\).
Let’s get back to the depressed quartic polynomial:
\[\begin{align} &(y^2 + ky + m)(y^2 + ly + n) \\ &= (y^2 + ky + m)(y^2 - ky + n) = 0. \end{align}\]From two quadratic polynomials, the roots are
\[\begin{align} y_1 &= -\frac{k}{2}+\frac{\sqrt{k^2-4m}}{2} \\ y_2 &= -\frac{k}{2}-\frac{\sqrt{k^2-4m}}{2} \\ y_3 &= \frac{k}{2}+\frac{\sqrt{k^2-4n}}{2} \\ y_4 &= \frac{k}{2}-\frac{\sqrt{k^2-4n}}{2} \\ \end{align}\]Since \(k\) is the root of the above six-degree polynomial, and is \(u\), \(v\), or \(w\). For now, let \(k=u\). Then,
\[\begin{align} m &= \frac{u^3 + qu - r}{2u}, \\ n &= \frac{u^3 + qu + r}{2u}. \end{align}\]and
\[\begin{align} y_1 &= -\frac{u}{2}+\frac{\sqrt{u^2-4m}}{2} \\ &= \frac{u}{2}+\frac{\sqrt{u^3-2(u^3+qu-r)}}{2u} \\ &= \frac{u}{2}+\frac{\sqrt{u^3-2u^3+(u^2+v^2+w^2)u\pm2uvw)}}{2u} \\ &= \frac{u}{2}+\frac{\sqrt{v^2+w^2\pm2vw}}{2} \\ &= \frac{u \pm v \pm w}{2}, \\ \end{align}\]where \(u^2+v^2+w^2=-2q\) and \(u^2v^2w^2=-r\) from the relationship between coefficients and roots. Similarly, we can obtain
\[\begin{align} y_2 &= \frac{u \mp v \pm w}{2}, \\ y_3 &= \frac{-u \pm v \pm w}{2}, \\ y_4 &= \frac{—u \mp v \pm w}{2}. \\ \end{align}\]Even if we assume \(k=v\) or \(w\), the roots are derived as the same with the above. Because of three \(\pm\) signs, there are 8 combinations of \(u\), \(v\), \(w\). We can obtain the roots by computing the value of the quartic polynomial with 8 candidates and finding roots that produce zero.
In summary, for a quartic polynomial,
\[a x^4 + b x^3 + c x^2 + d x + e = 0,\]The roots are among the values below.
\[x_* = \frac{\pm u \pm v \pm w}{2} - p\]where \(u^2\), \(v^2\), \(w^2\) are the roots of the cubic polynomial, \(h(t): t^3 + 2q t^2 + (q^2-4s)t -r^2 = 0\), and
\[\begin{align} q &= \frac{-3b^2+8ac}{8a^2},\\ r &= \frac{b^3-4abc+8a^2d}{8a^3},\\ s &= \frac{-3b^4+16ab^2c-64a^2bd+256a^3e}{256a^4}. \\ \end{align}\]