Polynomial Coeficients

• Using the divisibility which was given in the question, we have $P(x)+1={ { Q(x) }_{ 1 }(x-1) }^{ { 4 } }$ and $P(x)-1={ { Q(x) }_{ 2 }(x+1) }^{ { 4 } }$ . Note that ${Q(x)}_{1}$ and ${Q(x)}_{2}$ are the quotients for each division.

• Derivating in both sides, $(P(x)+1)'=({ (x-1) }^{ { 4 } }{ Q(x) }_{ 1 })'\quad \rightarrow \quad P'(x)={ (x-1) }^{ { 3 } }{ Q(x) }_{ 3 }$ and $(P(x)-1)'=({ (x+1) }^{ { 4 } }{ Q(x) }_{ 2 })' \rightarrow P'(x)={ (x+1) }^{ { 3 } }{ Q(x) }_{ 4 }$ .

• Then we can concatenate expressions as, $P'(x)={ (x+1) }^{ 3 }{ (x-1) }^{ 3 }{ Q(x) }_{ 5 }$ . But this quotient ${Q(x)}_{5}$ is a constant,a pure number, because $\deg (P(x))=7$ then $\deg (P'(x))=6$ , because is the first derivative. So let ${ Q(x) }_{5} = \alpha$ .

• Expanding: $P'(x)=({ (x+1)(x-1)) }^{ 3 }\alpha =\alpha ({ x }^{ 6 }-3{ x }^{ 4 }+3{ x }^{ 2 }-1)$ .

• Integrating both sides $\int { P'(x)dx } =a\int { ({ x }^{ 6 }-3{ x }^{ 4 }+3{ x }^{ 2 }-1)dx } \rightarrow P(x)= \cfrac { a{ x }^{ 7 } }{ 7 } -\cfrac { 3a{ x }^{ 5 } }{ 5 } +a{ x }^{ 3 }-a{ x } + c$ .

• On that expression, $P(x)+1={ { Q(x) }_{ 1 }(x-1) }^{ { 4 } }$ , make $x=1$ , $P(1)+1={ { Q(1) }_{ 1 }(1-1) }^{ { 4 } }=0$ , so $P(1) = -1$ , analogously $P(-1)=1$ .

• Substituting $P(1) = -1$ and $P(-1)=1$ on $P(x)= \cfrac { a{ x }^{ 7 } }{ 7 } -\cfrac { 3a{ x }^{ 5 } }{ 5 } +a{ x }^{ 3 }-a{ x } + c$ we get $P(1)=\cfrac { a }{ 7 } -\cfrac { 3a }{ 5 } +c=-1$ and $P(-1)=-\cfrac { a }{ 7 } +\cfrac { 3a }{ 5 } +c=1$ .

• Solving the system $\alpha =\cfrac { 35 }{ 16 }$ and then, $a-\cfrac { 3a }{ 5 } =\cfrac { 35 }{ 16 } -\cfrac { 3 }{ 5 } \cfrac { 35 }{ 16 } =\cfrac { 35 }{ 16 } -\cfrac { 21 }{ 16 } =\cfrac { 14 }{ 16 } =\cfrac { 7 }{ 8 }$

Obs: The numbers that should appear from the derivation are "inside" the quotients ${Q(x)}_{3}$ and ${Q(x)}_{4}$ .

Obs: The c (integration constant) was zero, if you try to solve the system it will lead to a contradiction, so the only possibility was the constant being zero.