x-4-ax-3-bx-2-cx-d-0-

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

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0

Commented by Frix last updated on 24/Oct/23

Usually we first try factors of d . The next

step would be substituting x = t - \frac{a}{4} to

reach t^{4} + {p t}^{2} + q t + r and now try to find

2 square factors (t^{2} - u t - v) (t^{2} + u t - w) by

matching the constants . This leads to

w^{3} + α w^{2} + β w + γ = 0; if this has at least 1

useable solution an exact solution of the

given equation makes sense; otherwise {it}^{'} s

better to approximate .

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