(a) ${{f(x)} \over {x + 1}} = {\phi _1}(x) + {6 \over {x + 1}},\,{{f(x)} \over {x – 2}} = {\phi _2}(x) + {3 \over {x – 2}}$

and ${{f(x)} \over {x + 2}} = {\phi _3}(x) + {{15} \over {x + 2}}$

${{f(x)} \over {(x + 1)\,(x + 2)\,(x – 2)}} = \phi (x) + {{Q(x)} \over {(x + 1)\,(x + 2)\,(x – 2)}}$

We have to find $Q(x)$, which will be a second degree polynomial. When $Q(x)$ is divided by $(x + 1)$, we should get the same remainder as being obtained by dividing $f(x)$ by $(x + 1)$ i.e., $6$. Similarly when $Q(x)$ is divided by $(x – 2)$, remainder should be $3$ and when $f(x)$ is divided by $x + 2,$ the remainder should be $15.$

$\therefore Q( – 1) = 6$

$Q(2) = 3$, $Q( – 2) = 15$

Let $Q(x) = \alpha {x^2} + \beta x + \gamma $,

$\therefore \alpha – \beta + \gamma = 6$…..(i)

$4\alpha + 2\beta + \gamma = 3$…..(ii); $4\alpha – 2\beta + \gamma = 15$ …..(iii)

$ \Rightarrow $$\alpha = 2,\,\beta = – 3,\,\gamma = 1$;

$\therefore Q(x) = 2{x^2} – 3x + 1$.