METHOD 1
substituting
\( – 5 + 12{\text{i}} + a(2 + 3{\text{i}}) + b = 0\) (A1)
equating real or imaginary parts (M1)
\(12 + 3a = 0 \Rightarrow a = – 4\) A1
\( – 5 + 2a + b = 0 \Rightarrow b = 13\) A1
METHOD 2
other root is \(2 – 3{\text{i}}\) (A1)
considering either the sum or product of roots or multiplying factors (M1)
\(4 = – a\) (sum of roots) so \(a = – 4\) A1
\(13 = b\) (product of roots) A1
[4 marks]