## Polynomials — quartics

Let us continue our exploration of polynomials. Just as in the previous blog, let me present to you an outline of some method(s) to solve Quartics. As I said earlier, “filling up the gaps” will kindle your intellect. Above all, the main aim of all teaching is teaching “to think on one’s own”.

I) The Quartic Equation. Descartes’s Method (1637).

(a) Argue that any quartic equation can be solved once one has a method to handle quartic equations of  the form: $t^{4}+pt^{2}+qt+r=0$

(b) Show that the quartic polynomial in (a) can be written as the product of two factors $(t^{2}+ut+v)(t^{2}-ut+w)$

where u, v, w satisfy the simultaneous system $v + w - u^{2}=p$ $u(w-v)=q$ $vw=r$

Eliminate v and w to obtain a cubic equation in $u^{2}$.

(c) Show how any solution u obtained in (b) can be used to find all the roots of the quartic equation.

(d) Use Descartes’s Method to solve the following: $t^{4}+t^{2}+4t-3=0$ $t^{4}-2t^{2}+8t-3=0$

II) The Quartic Equation. Ferrari’s Method:

(a) Let a quartic equation be presented in the form: $t^{4}+2pt^{3}+qt^{2}+2rt+s=0$

The strategy is to complete the square on the left side in such a way as to incorporate the cubic term. Show that the equation can be rewritten in the form $(t^{2}+pt+u)^{2}=(p^{2}-q+2u)t^{2}+2(pu-r)t+(u^{2}-s)$

where u is indeterminate.

(b) Show that the right side of the transformed equation in (a) is the square of  a linear polynomial if u satisfies a certain cubic equation. Explain how such a value of u can be used to completely solve the quartic.

(c) Use Ferrari’s Method to solve the following: $t^{4}+t^{2}+4t-3=0$ $t^{4}-2t^{3}-5t^{2}+10t-3=0$.

Happy problem-solving 🙂