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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 \[f(x)=x^4+ax^3+bx^2+acx+c^2\] i think most of you know how to factor this easily, but i am going to continue regardless. \[\begin{array}a f(x)=x^2(x^2+ax+b+acx^{-1}+c^2x^{-2})\\ \rightarrow x^2+2+(cx^{-1})^2+a(x+cx^{-1})+b-2=0\\ (x+cx^{-1})^2+a(x+cx^{-1})+b-2=0\\ x+cx^{-1}=\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}\\ x^2-\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}x+c=0\end{array}\] i think this is the most general form, as going farther will simply make it more tedious. \[\] case 2 \[x^4-2ax^2-x+a^2-a=0\] \[x^4-2ax^2+a^2=a+x\] \[x^2-a=\pm\sqrt{a+x}\] \[x=\pm\sqrt{a+\sqrt{a+x}},\pm\sqrt{a-\sqrt{a+x}}\] first one \[x=\sqrt{a+\sqrt{a+x}}=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+x}}}}=\sqrt{a+\sqrt{a+...}}=\sqrt{a+x}\] \[x^2-x+a=0\] the minus sign will have the negative root, and the positive sign the positive. second one: \[x=\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+...}}}},y=\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a-...}}}}\] \[x^2=a-y,y^2=a+x\] subtract both \[x^2-y^2=-y-x\Longrightarrow x-y=-1\] \[x^2=a-(x+1)\Longrightarrow x^2+x+(1-a)=0\] we can actually simplify these two results to get our original polynomials.

Note by Aareyan Manzoor
9 months, 3 weeks ago

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Did'nt understand •_ Samurai Poop · 8 months, 4 weeks ago

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