Very complicated, isn't it?

Using calculus , i got $\large\ f(x) = \frac { -3{ x }^{ 5 } }{ 8 } + \frac { 5{ x }^{ 3 } }{ 4 } - \frac { 15x }{ 8 }$ ,

forcing

$\large\ ace = \frac { -3 }{ 8 } \times \frac { 5 }{ 4 } \times \frac { -15 }{ 8 } =\frac { 225 }{ 256 }$ .

Thus, $p = 225$ and $q = 256$ .

So, $\sqrt { pq } = \sqrt { 225\times 256 } = \boxed{240}$

You were thinking that how i got $f(x)$ , the proof is here:

$f(x) + 1 = (x - 1)^{ 3 }.r(x)$ $⇒ f'(x) = (x - 1)^{ 2 }.r_{ 1 }(x)$ $...(1)$

and

$f(x) - 1 = (x + 1)^{ 3 }s(x)$ $⇒ f'(x) = (x + 1)^{ 2 }s_{ 1 }(x)$ $...(2)$

From $(1)$ and $(2)$ , $f '(x)$ has double root $1$ and double root $-1$

which implies

$f'(x) = k{ (x + 1) }^{ 2 }{ (x - 1) }^{ 2 } = k.{ ({ x }^{ 2 } - 1) }^{ 2 }=k.({ x }^{ 4 } + 1 - 2{ x }^{ 2 })$ , $k \in R$

Then

$\large\ f(x) = \int { (k.f'(x))dx } = k.\int { ({ x }^{ 4 } + 1 - 2{ x }^{ 2 })dx } = k.\frac { { x }^{ 5 } }{ 5 } - \frac { 2{ x }^{ 3 } }{ 3 } + x + C$ .

Now, as

$\large\ f(x) + 1 = (x - 1)^{ 3 }r(x) ⇒ f(1) + 1 = 0 ⇒ k.\left( \frac { 8 }{ 15 } + C \right) + 1 = 0$ $...(3)$

and

$\large\ f(x) - 1 = (x + 1)^{ 3 }r(x) ⇒ f(-1) - 1 = 0 ⇒ k.\left( -\frac { 8 }{ 15 } + C \right) - 1 = 0$ $...(4)$

we have from $(3)$ and $(4)$

$⇒ C = 0$ , $\large\ k = \left( -\frac { 15 }{ 8 } \right)$ .

Then we have the answer

$\large\ f(x) = \left( -\frac { 15 }{ 8 } \right) .\left( \frac { { x }^{ 5 } }{ 5 } - \frac { 2{ x }^{ 3 } }{ 3 } + x \right)$

or

$\large\ f(x) = \frac { -3{ x }^{ 5 } }{ 8 } + \frac { 5{ x }^{ 3 } }{ 4 } - \frac { 15x }{ 8 }$ as desired.