The polynomial must be of the form f(x) = a(x – 2)²(x + 1) because it has a double zero at 2 and a zero at – 1.
To solve for 'a', note that f (0) = a(0 – 2)²(0 + 1) = – 2.
It follows that a = – 1/2
Hence $\mathrm{f}\left(\mathrm{x}\right)=-\frac{1}{2}\left[{\mathrm{x}}^3-3{\mathrm{x}}^2+4\right.$
$\therefore \mathrm{a}=-\frac{1}{2},\mathrm{b}=\frac{3}{2},\mathrm{c}=0;\mathrm{d}=-2\Rightarrow \mathrm{c}+\mathrm{b}+\mathrm{c}+\mathrm{d}=-1$Ans.