Question

3x⁴ + 5x³ + 3x² - 7x + 9 = 0

Answer 1

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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