Question 852802
Your arrangement is correct.  You would not need the factor "a" unless you need to fit some further fact about the needed function.  For your purposes, a=1.


Just know that your factors with the imaginary numbers need to be arranged so that your multiplication process can take advantage of the difference between squares.


{{{(x-(4-5i))(x-(4+5i))}}}
{{{(x-4+5i)(x-4-5i)}}}
{{{((x-4)+5i)((x-4)-5i)}}}
{{{(x-4)^2-(5i)^2}}}
{{{(x-4)^2-25*i^2}}}
{{{(x-4)^2-(-1)25}}}
{{{(x-4)^2+25}}}
. not finished, but not expected difficult
.
.