Given that 2i is a zero of k, write the polynomial in factored form as a product of linear factors

Question

k(x)=6x^5+11x^4+12x^3+22x^2−48x−88

Answer 1

Assuming that all the coefficients of the polynomial are real numbers, having 2i as a zero implies that -2i is also a zero. Therefore, the factors of the polynomial include (x - 2i) and (x + 2i). If we multiply those factors, we get (x^2 + 4) and we know that it, too, divides the polynomial without remainder, that is to say, it is a factor of the polynomial.

So, just long divide poly by (x^2 + 4) and factor whatever quotient you get by whatever means are most convenient

That should get you on the right path to the solution!