Necessary condition for a special polynomial of degree $5$ to have a special quadratic factor I am trying to solve the problem no. 41(i) on page $33$, Higher Algebra by Barnard and Child.
The problem statement is:

If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$, prove that
  $$(a^2-c^2)(a^2-c^2+bc)=a^2b^2.$$

By synthetic division, I have found the remainder of $ax^5+bx^2+c$ upon division by $x^2+px+1$ is
$$(ap^4-3ap^2-bp+a) x +(ap^3-2ap-b+c).$$
As $x^2+px+1$ is a factor of $ax^5+bx^2+c$, the memainder must be indentically zero. That is:\begin{align*}
ap^4-3ap^2-bp+a&=0 \tag{1}\\
ap^3-2ap-b+c&=0 \tag{2}
\end{align*}
Then subtracting $(1)$ from $(2)\times p$ gives
$$ap^2+cp-a=0.$$
I do not know how to eliminate $p$ from the equations. I do not even know if this approach will solve it. And I have not been able yet to think of other approach.
If it is easy, then please give me some hint. Otherwise a solution, please!

Denote $f_1(t) = at^4 - 3at^2 - bt + a$, $f_2(t) = at^3 - 2at - (b - c)$, then $f_1(p) = f_2(p) = 0$. Therefore, defining\begin{align*}
f_3(t) &= t f_2(t) - f_1(t) = at^2 + ct - a,\\
f_4(t) &= a(tf_3(t) - f_2(t)) - c f_3(t) = (a^2 - c^2)t + ab,\\
f_5(t) &= (a^2 - c^2)^2 f_3(t) - a (f_4(t))^2 + 2a^2b f_4(t) - c(a^2 - c^2) f_4(t)\\
&= a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)), 
\end{align*}
then$$
f_1(p) = f_2(p) = f_3(p) = f_4(p) = f_5(p) = 0 \Longrightarrow a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)) = 0,
$$
and $a ≠ 0$ implies$$
(a^2 - c^2)(a^2 - c^2 + bc) = a^2 b^2.
$$
