The remainders of the polynomial f(x) when divided by x + 1, x + 2, x – 2 are 6, 15, 3 the remainder of f(x) when divided by (x + 1) (x + 2) (x – 2) is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\mathrm{q}\left(\mathrm{x}\right)+\mathrm{r}\left(\mathrm{x}\right) \\ \mathrm{f}\left(\mathrm{x}\right) \mathrm{divided} \mathrm{by} \mathrm{x}+1, \mathrm{x}+2, \mathrm{x}-2 \mathrm{then} \mathrm{the} \mathrm{remainder} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \\ \mathrm{r}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \Rightarrow \mathrm{r}(-1)=6\Rightarrow a-b+c=6 \\ \mathrm{r}(-2)=15\Rightarrow 4a-2b+c=15 \\ \mathrm{r}\left(2\right)=3\Rightarrow 4a+2b+c=3 \\ solving a=2,b=-3,c=1 \\ r\left(x\right)=2x^2-3x+1 \end{gathered}$$