The function \(\mathrm{f}\) is defined by \(\mathrm{f}(z)=az^3+bz^2+cz+d\), where \(a\), \(b\), \(c\) and \(d\) are real numbers. Given that \(2+\mathrm{i}\) and \(-3\) are roots of \(\mathrm{f}(z)=0\), find \(b\), \(c\) and \(d\) in terms of \(a\).[4]
\(2-\mathrm{i}\) is also a root (real coefficients).
\(\mathrm{f}(z)=a(z-2-\mathrm{i})(z-2+\mathrm{i})(z+3)\)
\(=a\left((z-2)^2+1\right)(z+3)=a(z^2-4z+5)(z+3)\)
\(=a(z^3-z^2-7z+15)\)
\(\therefore b=-a,\quad c=-7a,\quad d=15a\)
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