x^4 +c_3 x^3 +c_2 x^2 +c_1 x+c_0 =0 for c_n ∈R this can have 4 unique zeros ∈R 2 unique zeros + 1 double zero ∈R 2 double zeros ∈R 1 triple + 1 unique zeros ∈R 1 fourfold zero ∈R 2 unique zeros ∈R + 1 pair of complex zeros 1 double zero ∈R + 1 pair of complex zeros 2 pairs of complex zeros 2 double imaginary zeros for given c_n ; can we decide which case we have without solving?


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Question Number 152625 by Dandelion last updated on 30/Aug/21

$x^{4} + c_{3} x^{3} + c_{2} x^{2} + c_{1} x + c_{0} = 0$ $for c_{n} \in R this can have$ $4 unique zeros \in R$ $2 unique zeros + 1 double zero \in R$ $2 double zeros \in R$ $1 triple + 1 unique zeros \in R$ $1 fourfold zero \in R$ $2 unique zeros \in R + 1 pair of complex zeros$ $1 double zero \in R + 1 pair of complex zeros$ $2 pairs of complex zeros$ $2 double imaginary zeros$ $for given c_{n}; can we decide which case we$ $have without solving ?$
Commented by Dandelion last updated on 30/Aug/21

$. . . is there a D (or Δ) similar to polynomes$ $of 2 nd and 3 rd degree ?$
Commented by MJS_new last updated on 31/Aug/21

${it}^{'} s a bit complicated . I^{'} ll post the answer$ $later . . .$
	Answered by MJS_new last updated on 02/Sep/21

	$x^{4} + c_{3} x^{3} + c_{2} x^{2} + c_{1} x + c_{0} = 0$ $1 . let x = t - \frac{c_{3}}{4} and rename the new constants$ $\Rightarrow$ $t^{4} + {p t}^{2} + q t + r = 0$ $D_{1} = 16 p^{4} r - 4 p^{3} q^{2} - 128 p^{2} r^{2} + 144 {p q}^{2} r - 27 q^{4} + 256 r^{3}$ $D_{2} = p^{2} + 12 r$ $D_{3} = - p^{2} + 4 r$ $(1) D_{1} < 0$ $\Rightarrow 2 unique real + 1 pair of conjugated complex solutions$ $(2) D_{1} > 0 \land p < 0 \land D_{3} < 0$ $\Rightarrow 4 unique real solutions$ $(3) D_{1} > 0 \land (p > 0 \lor D_{3} > 0)$ $\Rightarrow 2 pairs of conjugated complex solutions$ $(4) D_{1} = 0 \land p < 0 \land D_{3} < 0 \land D_{2} \neq 0$ $\Rightarrow 1 real double + 2 unique real solutions$ $(5) D_{1} = 0 \land (D_{3} > 0 \lor p > 0 \land (D_{3} \neq 0 \lor q \neq 0))$ $\Rightarrow 1 real double + 1 pair of conjugated complex solutions$ $(6) D_{1} = 0 \land D_{2} = 0 \land D_{3} \neq 0$ $\Rightarrow 1 real triple + 1 unique real solutions$ $(7) D_{1} = 0 \land D_{3} = 0 \land p < 0$ $\Rightarrow 2 real double solutions$ $(8) D_{1} = 0 \land D_{3} = 0 \land p > 0 \land q = 0$ $\Rightarrow 2 imaginary double solutions$ $(9) D_{1} = 0 \land D_{2} = 0$ $\Rightarrow 1 real fourfold solution$
	Commented by Tawa11 last updated on 02/Sep/21

	$Great sir$
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