Quartic formula

We shall only sketch the rough idea, since the formula itself is rather complicated.

Let

$$p(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0.$$

Complete the square of the first few terms to get

$$\left ({x}^{2}+1/2\,{\it a_3}\,x\right )^{2} =\left (1/4\,{{\it a_3}}^{2} -{\it a_2}\right ){x}^{2}-{\it a_1}\,x-{\it a_0}.$$

Adding $2(x^2+(a_3/2)x)y+y^2$ to both sides and collecting terms, this is

$$(x^2+(a_3/2)x+y)^2= (a_3^2/4-a_2+2y)x^2+(a_3y-a_1)x+y^2-a_0.$$

Write the right-hand side in the form $ax^2+bx+c$. We want to choose $y$ so that $ax^2+bx+c=a(x-r)^2$, for some $r$. To do this we want (by the quadratic formula) $b^2-4ac=0$. This condition is

$$-8\,{y}^{3}+4\,{\it a_2}\,{y}^{2}+ \left (-2\,{\it a_3}\,{\it a_1... ...\right )y-4\,{\it a_2}\,{\it a_0}+ {{\it a_1}}^{2}+{{\it a_3}}^{2}{\it a_0}=0,$$

a cubic equation which can be solved using the cubic formula. In any case, we can now determine the root $r$ above and write

$$x^2+(a_3/2)x+y=\pm \sqrt{a}(x-r),$$

where $r$ and $y$ were solved for previously. There are $4$ roots to these two equations (one for each sign of the $\pm$ sign).

The cubic and quartic formulas were discovered in the middle ages.