Factoring quartics-2 special cases!

part-1. i am going to show you some special of a quartic polynomial which makes it easier to solve. in future notes i will try to generalize for every quartic polynomials. 2 cases to be presented here:

case 1: consider the polynomial f(x)=x4+ax3+bx2+acx+c2f(x)=x^4+ax^3+bx^2+acx+c^2 i think most of you know how to factor this easily, but i am going to continue regardless. af(x)=x2(x2+ax+b+acx1+c2x2)x2+2+(cx1)2+a(x+cx1)+b2=0(x+cx1)2+a(x+cx1)+b2=0x+cx1=a±a24b+82x2a±a24b+82x+c=0\begin{array}{c}a f(x)=x^2(x^2+ax+b+acx^{-1}+c^2x^{-2})\\ \rightarrow x^2+2+(cx^{-1})^2+a(x+cx^{-1})+b-2=0\\ (x+cx^{-1})^2+a(x+cx^{-1})+b-2=0\\ x+cx^{-1}=\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}\\ x^2-\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}x+c=0\end{array} i think this is the most general form, as going farther will simply make it more tedious. case 2 x42ax2x+a2a=0x^4-2ax^2-x+a^2-a=0 x42ax2+a2=a+xx^4-2ax^2+a^2=a+x x2a=±a+xx^2-a=\pm\sqrt{a+x} x=±a+a+x,±aa+xx=\pm\sqrt{a+\sqrt{a+x}},\pm\sqrt{a-\sqrt{a+x}} first one x=a+a+x=a+a+a+a+x=a+a+...=a+xx=\sqrt{a+\sqrt{a+x}}=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+x}}}}=\sqrt{a+\sqrt{a+...}}=\sqrt{a+x} x2x+a=0x^2-x+a=0 the minus sign will have the negative root, and the positive sign the positive. second one: x=aa+aa+...,y=a+aa+a...x=\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+...}}}},y=\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a-...}}}} x2=ay,y2=a+xx^2=a-y,y^2=a+x subtract both x2y2=yxxy=1x^2-y^2=-y-x\Longrightarrow x-y=-1 x2=a(x+1)x2+x+(1a)=0x^2=a-(x+1)\Longrightarrow x^2+x+(1-a)=0 we can actually simplify these two results to get our original polynomials.

Note by Aareyan Manzoor
5 years, 6 months ago

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Did'nt understand •_

SAMURAI POOP - 5 years, 5 months ago

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