ok, so if a+bi is a root then a-bi isi also a root
since 1+3i is a root then 1-3i is also a root
so
if a function has roots r1,r2 then it factors to
f(x)=(x-r1)(x-r2)
so
roots are 2 and -4 and 1+3i and 1-3i
so
f(x)=(x-2)(x-(-4))(x-(1+3i))(x-(1-3i))
f(x)=(x-2)(x+4)(x-1-3i)(x-1+3i)
expand
f(x)=x⁴-2x²+36x-80
