Question Number 164019 by ajfour last updated on 13/Jan/22
$$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$$${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$$$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} +{ax}+{p}=\mathrm{0} \\ $$$$\&\:\:\:\:\:\:{dt}^{\mathrm{2}} +{ct}+{q}=\mathrm{0} \\ $$$$\:\:\:\:{p}+{q}={b} \\ $$$${x}=−\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\:\: \\ $$$${t}=−\frac{{c}}{\mathrm{2}{d}}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}} \\ $$$${tx}=\mathrm{1}\:\:\:\:\Rightarrow \\ $$$$\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\mp\frac{{c}}{\mathrm{2}{d}}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}} \\ $$$$\:\:\:\:\:\:\:\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{p}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{q}}{{d}}}\:=\mathrm{1} \\ $$$${let}\:\:\:{p}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow \\ $$$$\:\:\:\frac{{ac}}{\mathrm{4}{d}}\mp\frac{{a}}{\mathrm{2}}\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{{b}}{{d}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}{d}}=\frac{\mathrm{4}}{{a}^{\mathrm{2}} }+\cancel{\frac{{c}^{\mathrm{2}} }{\mathrm{4}{d}^{\mathrm{2}} }}−\frac{\mathrm{2}{c}}{{ad}} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{a}^{\mathrm{2}} {b}−{a}^{\mathrm{4}} +\mathrm{16}{d}−\mathrm{8}{ac}=\mathrm{0} \\ $$$${if}\:{x}={s}+{h} \\ $$$${s}^{\mathrm{4}} +\mathrm{4}{hs}\left({s}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)+\mathrm{6}{h}^{\mathrm{2}} {s}^{\mathrm{2}} +{h}^{\mathrm{4}} \\ $$$$+{a}\left\{{s}^{\mathrm{3}} +\mathrm{3}{hs}\left({s}+{h}\right)+{h}^{\mathrm{2}} \right\} \\ $$$$\:\:+{b}\left\{{s}^{\mathrm{2}} +\mathrm{2}{hs}+{h}^{\mathrm{2}} \right\}+{c}\left({s}+{h}\right)+{d}=\mathrm{0} \\ $$$${A}=\mathrm{4}{h}+{a} \\ $$$${B}=\mathrm{6}{h}^{\mathrm{2}} +\mathrm{3}{ah}+{b} \\ $$$${C}=\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c} \\ $$$${D}={h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{4}{h}+{a}\right)^{\mathrm{2}} \left(\mathrm{8}{h}^{\mathrm{2}} +\mathrm{4}{ah}+\mathrm{4}{b}−{a}^{\mathrm{2}} \right) \\ $$$$+\mathrm{16}\left({h}^{\mathrm{4}} +{ah}^{\mathrm{3}} +{bh}^{\mathrm{2}} +{ch}+{d}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{8}\left(\mathrm{4}{h}+{a}\right)\left(\mathrm{4}{h}^{\mathrm{3}} +\mathrm{3}{ah}^{\mathrm{2}} +\mathrm{2}{bh}+{c}\right) \\ $$$$….. \\ $$$$\:\:\:\:\:\: \\ $$