Finding remaining zeros of linear phase FIR filter Question given is: if $h(n)$ is a linear phase causal FIR of order $10$ with real coefficients, find the remaining zeros of this filter if the zeros given are

$$q_{1,2} = -2 \pm 2j$$

$$q_{3} = -\frac12 + \frac{j}{2} $$

$$q_{4} = -1$$

I know for sure that $-\frac {1} 2 - \frac j 2 $ has to be another zero, but there are 5 more zeros unaccounted for. How am I supposed to find them?
For linear phase FIR filters, each zero at z = z0 will have a matching reciprocal zero at z = 1/zo. And for real-valued coefficients each zero at z = zo will have a matching conjugate zero at z = *zo. Thus for linear phase real-valued coefficients, when you place one zero on the z-plane you determine the location of the other three zeros.