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Question Number 198772 by ajfour last updated on 24/Oct/23

$$x^4+{ax}^3+{bx}^2+cx+d=0$$

Commented by Frix last updated on 24/Oct/23

$$\mathrm{Usually}\mathrm{we}\mathrm{first}\mathrm{try}\mathrm{factors}\mathrm{of}d.\mathrm{The}\mathrm{next}$$$$\mathrm{step}\mathrm{would}\mathrm{be}\mathrm{substituting}x=t-\frac{a}{4}\mathrm{to}$$$$\mathrm{reach}{t}^4+{pt}^2+qt+r\mathrm{and}\mathrm{now}\mathrm{try}\mathrm{to}\mathrm{find}$$$$2\mathrm{square}\mathrm{factors}(t^2-ut-v)(t^2+ut-w)\mathrm{by}$$$$\mathrm{matching}\mathrm{the}\mathrm{constants}.\mathrm{This}\mathrm{leads}\mathrm{to}$$$$w^3+\alpha w^2+\beta w+\gamma =0;\mathrm{if}\mathrm{this}\mathrm{has}\mathrm{at}\mathrm{least}1$$$$\mathrm{useable}\mathrm{solution}\mathrm{an}\mathrm{exact}\mathrm{solution}\mathrm{of}\mathrm{the}$$$$\mathrm{given}\mathrm{equation}\mathrm{makes}\mathrm{sense};\mathrm{otherwise}{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{better}\mathrm{to}\mathrm{approximate}.$$

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