H(x)=7(x5+x4+9x3+9x2-112x-112).
Another zero is -4i (to make a conjugate pair). (x-4i)(x+4i)=x2+16 is a factor.
x3+x2-7x -7
x2+16 ) x5+x4+9x3+9x2-112x-112
x5 +16x3
x4 -7x3+9x2
x4 +16x2
-7x3-7x2-112x
-7x3 -112x
-7x2 -112
-7x2 -112
0
x3+x2-7x-7=x2(x+1)-7(x+1)=(x2-7)(x+1)=(x-√7)(x+√7)(x+1).
So the five zeroes are 4i, -4i, -√7, -1, √7.