You have a problem

As seen on the image above, where d is the pentagon diagonal and l is the side, $\triangle ABF\simeq \triangle CEF$ . Let l = 1; $\frac { 1 }{ d-1 } =\frac { d }{ 1 }$ . Solving this equation will lead to d = $\frac { (1+\sqrt { 5 } ) }{ 2 }$ . We have just defined the ratio Side/Diagonal.

The Length of the sides of the regular pentagon is 2, and the ratio Side/Diagonal can be expressed as $\frac { l(1+\sqrt { 5 } ) }{ 2 }$ . We can admit now that the side of the hexagon is equal to $1+\sqrt { 5 }$ .

A regular hexagon can be split into 6 equilateral triangles. That said, the area of it is equal to $\frac { 6{ (1+\sqrt { 5 } ) }^{ 2 }\sqrt { 3 } }{ 4 }$ Solving the equation above you will end up with $3(3\sqrt { 3 } +\quad \sqrt { 15 } )$

The ratio (Area of Hexagon)/(Perimeter of Pentagon) will be $\frac { 3(3\sqrt { 3 } +\quad \sqrt { 15 } ) }{ 10 }$

A = 3; B = 10; C = 15;

To find the Polynomial equation you simply need to multiply: (x-3)(x-10)(x-15)

That will give you the final result of ${ x }^{ 3 }-28{ x }^{ 2 }+225x-450$

a = -28 b= 225 c = -450

a+b+c = -253.

Sin 54= $\dfrac{\sqrt5} 4$ . In the isosceles triangle with equal sides 2 and angle between as 108 =54 * 2, the base, that is diagonal of P = 2 * side*Sin54= $\sqrt5 +1$ . This is the side of Q. $So ~ area ~ of ~ Q =6*\frac{\sqrt3}4*(\frac{\sqrt5} 4)^2=3(3\sqrt3+\sqrt{15}). ~~~~~ P_{per}=2*5=10.\\$ $\therefore ~ the ~ ratio = \frac 3{10}*\{ 3\sqrt3+\sqrt{15}\}.\\$ $Using ~Vieta's ~with~A=3, ~B=10, ~ C=15,\\ a= - (A+B+C),.. b=A*B+B*C+C*A,.. c= - A*B*C,... ~~a+b+c= -253.$