use given information about polynomial to find remaining zeros

Question

Degree: 6  zeros: -10+isqrt11, 13i, -12, -7a^3i ( where a is a real number)

Answer 1

A degree 6 polynomial can have 6 real zeroes; 2 real zeroes and 4 complex zeroes; 4 complex zeroes and 2 real zeroes; 6 complex zeroes.

The given information shows one real zero only, so there must be another to make a pair.

Of the remaining 4 complex zeroes there have to be 2 conjugate pairs. I guess that 13i and -7a³i are conjugates, so -7a³i=-13i, 7a³=13, a³=13/7, a=1.2291 approx.

(x-13i)(x+13i)=x²+169.

-10+i√11 and -10-i√11 is a conjugate pair, so (x+10-i√11)(x+10+i√11)=x²+20x+221.

(x²+169)(x²+20x+221)=x⁴+20x³+390x²+3380x+37349.

x+12 is another factor, so:

x⁵+20x⁴+390x³+3380x²+37349x+12x⁴+240x³+4680x²+40560x+448188=

x⁵+32x⁴+630x³+8060x²+77909x+448188.

We need another factor, x-p, polynomial is:

x⁶+32x⁵+630x⁴+8060x³+77909x²+448188x-p(x⁵+32x⁴+630x³+8060x²+77909x+448188)=

x⁶+(32-p)x⁵+(630-32p)x⁴+(8060-630p)x³+(77909-8060p)x²+(448188-77909p)x+448188p, where p is a real number. There doesn't appear to be any way to find p, the missing zero, unless there is an error in the question.