This polynomial has no real zeroes (roots).
The complex roots may be represented by a+ib, a-ib, c+id, c-id (conjugate pairs).
Two factors are: (x-a)2+b2 are (x-c)2+d2. These are quadratics with real coefficients:
x2-2ax+a2+b2 and x2-2cx+c2+d2.
x4-2cx3+(c2+d2)x2-2ax3+4acx2-2a(c2+d2)x+(a2+b2)x2-2c(a2+b2)x+(a2+b2)(c2+d2).
(x2-2ax+a2+b2)(x2-2cx+c2+d2)=x4+5x3/3+x2-7x/3+3=0,
x4-2(a+c)x3+(a2+b2+c2+d2+4ac)x2-2(ac2+ad2+a2c+b2c)x+(a2+b2)(c2+d2)=x4+5x3/3+x2-7x/3+3=0.
Matching coefficients of the constant, x, x2 and x3:
constant: (a2+b2)(c2+d2)=3;
x term: a2c+ac2+ad2+b2c=7/6=ac(a+c)+ad2+b2c;
x2 term: a2+b2+c2+d2+4ac=1=(a+c)2+2ac+b2+d2;
x3 term: a+c=-⅚, c=-(a+⅚);
Substituting c=-(a+⅚) and (a+c)2=25/36:
2ac+b2+d2=-2a(a+⅚)+b2+d2=11/36 (1);
ac(a+c)+ad2+b2c=ad2-(a+⅚)(b2-⅚a)=7/6 (2);
(a2+b2)((a+⅚)2+d2)=3 (3).
We now have a system of equations (labelled (1), (2), (3) and 3 unknowns: a, b, d).
d2=11/36+2a(a+⅚)-b2 (from (1)). From (3): (a2+b2)((a+⅚)2+11/36+2a(a+⅚)-b2)=3 (4).
From (2): a(11/36+2a(a+⅚)-b2)-(a+⅚)(b2-⅚a)=7/6 (5).
a+2a3+5a2/2-b2(2a+⅚)=7/6, b2=(a+2a3+5a2/2-7/6)/(2a+⅚) (6).
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