a
(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)

\( \Rightarrow \)\(\alpha = 2,\,\beta = - 3,\,\gamma = 1\);

\(\therefore Q(x) = 2{x^2} - 3x + 1\).