To test if 3 - i is a root of p=x^4-4x^2+16, substitute x = sqrt 3 - i

x^4 - 4x^2 + 16

=>(sqrt 3- i)^4 -4(sqrt 3- i)^2 + 16

=> (sqrt 3 - i)^2^2 - 4*(3 + i^2 - 2*i*sqrt 3) + 16

=> (3 + i^2 - 2*sqrt 3*i)^2 - 12 + 4 + 8*sqrt 3*i + 16

=> 4 + 4*3*i^2 - 8sqrt 3*i - 8 + 8*sqrt 3*i + 16

=> 4 - 12 - 8sqrt 3*i - 8 + 8*sqrt 3*i + 16

=> 0

**As substituting x = sqrt 3 - i gives the value of the polynomial as 0, it is a root.**

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