Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:

$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$

(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor

Proving part (ii) independently is less tedious than part (i). Here it is:-
Let $f(x) = ax^5 + bx^2 + c$ 
Also $f(x) = (x^2 + px + 1)Q_3 (x)$; where $Q_3 (x)$ is a quotient polynomial of degree 3…….(1)
Let $g(x) = x^2 + px + 1$ such that g(x) = 0 has m and n as roots.
From product of roots, $n = 1/m$.
$f(n) = f(1/m) = … = (1/m)^5 . [a + b(m)^3 + c(m)^5]…....(2)$
From (1), $f(n) = f(1/m) = … = (1/m)^2 . [1 + p(m) + (m)^2]Q_3 (1/m)……(3)$
Equating (2) and (3), we have $a + bm^3 + cm^5 = (1 + pm + m^2).[m^3.Q_3 (1/m)]$
This means $a + bm^3 + cm^5$ has $(1 + pm + m^2)$ as factor.
i.e. $cx^5 + bx^3 + a$ has $(x^2 + px +1)$ as factor.
Thus, $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor.
For part(i), I still need some time to workout.
