Complex zeroes always come in pairs, so the complement to 2+3i is 2-3i.
f(x)=a(x-2-3i)(x-2+3i)(x+1)(x-4) where a is a constant.
f(x)=a(x²-4x+13)(x²-3x-4)=
a(x⁴-3x³ -4x²
-4x³+12x²+16x
+13x²-39x-52)=
a(x⁴-7x³+21x²-23x-52).
f(1)=a(1-7+21-23-52)=-180.
f(1)=-60a=-180, so a=3 and:
f(x)=3x⁴-21x³+63x²-69x-156.